Comprehensive Study Guide on Unit 1: Matrices

Characteristic Equation and Characteristic Polynomial

  • Characteristic Equation Definition:

    • For a square matrix AA, the equation ∣A−λI∣=0|A - \lambda I| = 0 is called the characteristic equation of matrix AA, where II is the identity matrix of the same order as AA and λ\lambda is a scalar parameter.

  • Characteristic Roots / Eigenvalues / Latent Values:

    • Solving the characteristic equation ∣A−λI∣=0|A - \lambda I| = 0 for an n×nn \times n matrix AA yields nn roots for λ\lambda.
    • These roots are called the characteristic roots, eigenvalues, or latent values of matrix A$.\n\n* **Eigenvectors / Latent Vectors:**\n * Corresponding to each value of \lambda,thematrixequation, the matrix equationAX = \lambda X(orequivalently(or equivalently(A - \lambda I)X = 0)hasanon−zerosolutionvector) has a non-zero solution vectorX$.
    • If XrX_r is a non-zero vector satisfying AX=λrXAX = \lambda_r X corresponding to the eigenvalue λ=λr\lambda = \lambda_r, then XrX_r is called the latent vector or eigenvector of matrix AA corresponding to \lambda_r$.\n\n* **Characteristic Polynomial:**\n * When the determinant |A - \lambda I|isexpandedwithoutsettingitequaltozero,ityieldsapolynomialinis expanded without setting it equal to zero, it yields a polynomial in\lambda,whichisdefinedasthecharacteristicpolynomialofmatrix, which is defined as the characteristic polynomial of matrixA$.

  • Working Rules to Find the Characteristic Equation:

    • For a 3×33 \times 3 Matrix:
      • Method 1: Direct expansion of ∣A−λI∣=0|A - \lambda I| = 0.
      • Method 2: Using the polynomial formula:             λ3−S1λ2+S2λ−S3=0\lambda^3 - S_1 \lambda^2 + S_2 \lambda - S_3 = 0
        • S1=sum of the main diagonal elements=Trace(A)S_1 = \text{sum of the main diagonal elements} = \text{Trace}(A)
        • S2=sum of the minors of the main diagonal elementsS_2 = \text{sum of the minors of the main diagonal elements}
        • S3=Determinant of A=∣A∣S_3 = \text{Determinant of } A = |A|
    • For a 2×22 \times 2 Matrix:
      • Method 1: Direct expansion of ∣A−λI∣=0|A - \lambda I| = 0.
      • Method 2: Using the polynomial formula:             λ2−S1λ+S2=0\lambda^2 - S_1 \lambda + S_2 = 0
        • S1=sum of the main diagonal elements=Trace(A)S_1 = \text{sum of the main diagonal elements} = \text{Trace}(A)
        • S2=Determinant of A=∣A∣S_2 = \text{Determinant of } A = |A|

  • Worked Problems on Characteristic Equations and Polynomials:

    • Problem 1: Find the characteristic equation of the matrix A=(1202)A = \begin{pmatrix} 1 & 2 \\ 0 & 2 \end{pmatrix}.
      • S1=1+2=3S_1 = 1 + 2 = 3
      • S2=∣A∣=1(2)−2(0)=2S_2 = |A| = 1(2) - 2(0) = 2
      • Characteristic equation: λ2−3λ+2=0\lambda^2 - 3\lambda + 2 = 0
    • Problem 2: Find the characteristic equation of A=(8−62−67−42−43)A = \begin{pmatrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{pmatrix}.
      • S1=8+7+3=18S_1 = 8 + 7 + 3 = 18
      • S2=∣7−4−43∣+∣8223∣+∣8−6−67∣=(21−16)+(24−4)+(56−36)=5+20+20=45S_2 = \begin{vmatrix} 7 & -4 \\ -4 & 3 \end{vmatrix} + \begin{vmatrix} 8 & 2 \\ 2 & 3 \end{vmatrix} + \begin{vmatrix} 8 & -6 \\ -6 & 7 \end{vmatrix} = (21 - 16) + (24 - 4) + (56 - 36) = 5 + 20 + 20 = 45
      • S3=∣A∣=8(21−16)+6(−18+8)+2(24−14)=8(5)+6(−10)+2(10)=40−60+20=0S_3 = |A| = 8(21 - 16) + 6(-18 + 8) + 2(24 - 14) = 8(5) + 6(-10) + 2(10) = 40 - 60 + 20 = 0
      • Characteristic equation: λ3−18λ2+45λ=0\lambda^3 - 18\lambda^2 + 45\lambda = 0
    • Problem 3: Find the characteristic polynomial of A=(31−12)A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}.
      • S1=3+2=5S_1 = 3 + 2 = 5
      • S2=∣A∣=3(2)−1(−1)=6+1=7S_2 = |A| = 3(2) - 1(-1) = 6 + 1 = 7
      • Characteristic polynomial: λ2−5λ+7\lambda^2 - 5\lambda + 7

Cayley-Hamilton Theorem

  • Theorem Statement:

    • Every square matrix satisfies its own characteristic equation.
    • If the characteristic equation of an n×nn \times n square matrix AA is λn+cn−1λn−1+⋯+c1λ+c0=0\lambda^n + c_{n-1}\lambda^{n-1} + \dots + c_1 \lambda + c_0 = 0, then:         An+cn−1An−1+⋯+c1A+c0I=0A^n + c_{n-1}A^{n-1} + \dots + c_1 A + c_0 I = 0

  • Primary Uses of Cayley-Hamilton Theorem:

    1. To calculate positive integral powers of matrix AA (e.g., A2,A3,A4,AnA^2, A^3, A^4, A^n).
    2. To calculate the matrix inverse A−1A^{-1} of a non-singular square matrix AA.

  • Worked Problems on Cayley-Hamilton Theorem:

    • Problem 1: Show that the matrix A=(1−221)A = \begin{pmatrix} 1 & -2 \\ 2 & 1 \end{pmatrix} satisfies its own characteristic equation.
      • S1=1+1=2S_1 = 1 + 1 = 2
      • S2=∣A∣=1−(−4)=5S_2 = |A| = 1 - (-4) = 5
      • Characteristic equation: λ2−2λ+5=0\lambda^2 - 2\lambda + 5 = 0
      • Proof that A2−2A+5I=0A^2 - 2A + 5I = 0:             A2=(1−221)(1−221)=(−3−44−3)A^2 = \begin{pmatrix} 1 & -2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} -3 & -4 \\ 4 & -3 \end{pmatrix}A2−2A+5I=(−3−44−3)−(2−442)+(5005)=(0000)=0A^2 - 2A + 5I = \begin{pmatrix} -3 & -4 \\ 4 & -3 \end{pmatrix} - \begin{pmatrix} 2 & -4 \\ 4 & 2 \end{pmatrix} + \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = 0
    • Problem 2: If A=(1005)A = \begin{pmatrix} 1 & 0 \\ 0 & 5 \end{pmatrix}, write A2A^2 in terms of AA and II using Cayley-Hamilton theorem.
      • S1=1+5=6S_1 = 1 + 5 = 6
      • S2=∣A∣=5S_2 = |A| = 5
      • Characteristic equation: λ2−6λ+5=0\lambda^2 - 6\lambda + 5 = 0
      • By Cayley-Hamilton theorem: A2−6A+5I=0  ⟹  A2=6A−5IA^2 - 6A + 5I = 0 \implies A^2 = 6A - 5I
    • Problem 3: Verify Cayley-Hamilton theorem, find A4A^4 and A−1A^{-1} when A=(2−12−12−11−12)A = \begin{pmatrix} 2 & -1 & 2 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix}.
      • S1=2+2+2=6S_1 = 2 + 2 + 2 = 6
      • S2=(4−1)+(4−2)+(4−1)=3+2+3=8S_2 = (4 - 1) + (4 - 2) + (4 - 1) = 3 + 2 + 3 = 8
      • S3=∣A∣=2(4−1)+1(−2+1)+2(1−2)=2(3)−1−2=3S_3 = |A| = 2(4 - 1) + 1(-2 + 1) + 2(1 - 2) = 2(3) - 1 - 2 = 3
      • Characteristic equation: λ3−6λ2+8λ−3=0\lambda^3 - 6\lambda^2 + 8\lambda - 3 = 0
      • Verification (A3−6A2+8A−3I=0A^3 - 6A^2 + 8A - 3I = 0):             A2=(7−69−56−65−57),A3=(29−2838−2223−2822−2229)A^2 = \begin{pmatrix} 7 & -6 & 9 \\ -5 & 6 & -6 \\ 5 & -5 & 7 \end{pmatrix}, \quad A^3 = \begin{pmatrix} 29 & -28 & 38 \\ -22 & 23 & -28 \\ 22 & -22 & 29 \end{pmatrix}A3−6A2+8A−3I=(29−2838−2223−2822−2229)−(42−3654−3036−3630−3042)+(16−816−816−88−816)−(300030003)=(000000000)=0A^3 - 6A^2 + 8A - 3I = \begin{pmatrix} 29 & -28 & 38 \\ -22 & 23 & -28 \\ 22 & -22 & 29 \end{pmatrix} - \begin{pmatrix} 42 & -36 & 54 \\ -30 & 36 & -36 \\ 30 & -30 & 42 \end{pmatrix} + \begin{pmatrix} 16 & -8 & 16 \\ -8 & 16 & -8 \\ 8 & -8 & 16 \end{pmatrix} - \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0
      • To find A4A^4:             A3=6A2−8A+3IA^3 = 6A^2 - 8A + 3IA4=6A3−8A2+3A=6(6A2−8A+3I)−8A2+3A=28A2−45A+18IA^4 = 6A^3 - 8A^2 + 3A = 6(6A^2 - 8A + 3I) - 8A^2 + 3A = 28A^2 - 45A + 18IA4=28(7−69−56−65−57)−45(2−12−12−11−12)+18(100010001)=(124−123162−9596−12395−95124)A^4 = 28 \begin{pmatrix} 7 & -6 & 9 \\ -5 & 6 & -6 \\ 5 & -5 & 7 \end{pmatrix} - 45 \begin{pmatrix} 2 & -1 & 2 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} + 18 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 124 & -123 & 162 \\ -95 & 96 & -123 \\ 95 & -95 & 124 \end{pmatrix}
      • To find A−1A^{-1}:             A2−6A+8I−3A−1=0  ⟹  3A−1=A2−6A+8IA^2 - 6A + 8I - 3A^{-1} = 0 \implies 3A^{-1} = A^2 - 6A + 8I3A−1=(7−69−56−65−57)−(12−612−612−66−612)+(800080008)=(30−3120−113)3A^{-1} = \begin{pmatrix} 7 & -6 & 9 \\ -5 & 6 & -6 \\ 5 & -5 & 7 \end{pmatrix} - \begin{pmatrix} 12 & -6 & 12 \\ -6 & 12 & -6 \\ 6 & -6 & 12 \end{pmatrix} + \begin{pmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{pmatrix} = \begin{pmatrix} 3 & 0 & -3 \\ 1 & 2 & 0 \\ -1 & 1 & 3 \end{pmatrix}A−1=13(30−3120−113)A^{-1} = \frac{1}{3} \begin{pmatrix} 3 & 0 & -3 \\ 1 & 2 & 0 \\ -1 & 1 & 3 \end{pmatrix}
    • Problem 4: Verify that A=(122−1)A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} satisfies its own characteristic equation and find A4A^4
      • S1=1−1=0S_1 = 1 - 1 = 0
      • S2=∣A∣=−1−4=−5S_2 = |A| = -1 - 4 = -5
      • Characteristic equation: λ2−5=0  ⟹  A2−5I=0\lambda^2 - 5 = 0 \implies A^2 - 5I = 0
      • A2=(122−1)(122−1)=(5005)A^2 = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}
      • A2−5I=(5005)−(5005)=0A^2 - 5I = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} - \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = 0
      • A4=A2(5I)=5A2=5(5005)=(250025)A^4 = A^2(5I) = 5A^2 = 5 \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 25 & 0 \\ 0 & 25 \end{pmatrix}
    • Problem 5: Find A−1A^{-1} if A=(1−1432−121−1)A = \begin{pmatrix} 1 & -1 & 4 \\ 3 & 2 & -1 \\ 2 & 1 & -1 \end{pmatrix} using Cayley-Hamilton theorem.
      • S1=1+2−1=2S_1 = 1 + 2 - 1 = 2
      • S2=(−2+1)+(−1−8)+(2+3)=−1−9+5=−5S_2 = (-2 + 1) + (-1 - 8) + (2 + 3) = -1 - 9 + 5 = -5
      • S3=∣A∣=1(−2+1)+1(−3+2)+4(3−4)=−1−1−4=−6S_3 = |A| = 1(-2 + 1) + 1(-3 + 2) + 4(3 - 4) = -1 - 1 - 4 = -6
      • Characteristic equation: λ3−2λ2−5λ+6=0\lambda^3 - 2\lambda^2 - 5\lambda + 6 = 0
      • By Cayley-Hamilton theorem: A3−2A2−5A+6I=0A^3 - 2A^2 - 5A + 6I = 0
      • 6A−1=−A2+2A+5I  ⟹  A−1=16(−A2+2A+5I)6A^{-1} = -A^2 + 2A + 5I \implies A^{-1} = \frac{1}{6}(-A^2 + 2A + 5I)
      • A2=(1−1432−121−1)(1−1432−121−1)=(61170113−18)A^2 = \begin{pmatrix} 1 & -1 & 4 \\ 3 & 2 & -1 \\ 2 & 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & -1 & 4 \\ 3 & 2 & -1 \\ 2 & 1 & -1 \end{pmatrix} = \begin{pmatrix} 6 & 1 & 1 \\ 7 & 0 & 11 \\ 3 & -1 & 8 \end{pmatrix}
      • −A2+2A+5I=(−6−1−1−70−11−31−8)+(2−2864−242−2)+(500050005)=(1−37−19−1313−5)-A^2 + 2A + 5I = \begin{pmatrix} -6 & -1 & -1 \\ -7 & 0 & -11 \\ -3 & 1 & -8 \end{pmatrix} + \begin{pmatrix} 2 & -2 & 8 \\ 6 & 4 & -2 \\ 4 & 2 & -2 \end{pmatrix} + \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} = \begin{pmatrix} 1 & -3 & 7 \\ -1 & 9 & -13 \\ 1 & 3 & -5 \end{pmatrix}
      • A−1=16(1−37−19−1313−5)A^{-1} = \frac{1}{6} \begin{pmatrix} 1 & -3 & 7 \\ -1 & 9 & -13 \\ 1 & 3 & -5 \end{pmatrix}
    • Problem 6: If A=(1202)A = \begin{pmatrix} 1 & 2 \\ 0 & 2 \end{pmatrix}, find AnA^n in terms of AA
      • Characteristic equation: λ2−3λ+2=0  ⟹  λ=1,2\lambda^2 - 3\lambda + 2 = 0 \implies \lambda = 1, 2
      • Let λn=(λ2−3λ+2)Q(λ)+aλ+b\lambda^n = (\lambda^2 - 3\lambda + 2) Q(\lambda) + a\lambda + b
      • When λ=1  ⟹  1n=a+b\lambda = 1 \implies 1^n = a + b
      • When λ=2  ⟹  2n=2a+b\lambda = 2 \implies 2^n = 2a + b
      • Solving gives a=2n−1na = 2^n - 1^n, b=2(1)n−2nb = 2(1)^n - 2^n
      • By Cayley-Hamilton theorem, A2−3A+2I=0A^2 - 3A + 2I = 0, so An=aA+bIA^n = aA + bI
      • An=(2n−1n)(1202)+[2(1)n−2n](1001)A^n = (2^n - 1^n) \begin{pmatrix} 1 & 2 \\ 0 & 2 \end{pmatrix} + [2(1)^n - 2^n] \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
    • Problem 7: Express as a linear polynomial in AA for A=(1423)A = \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix}:
      • (i) A5−4A4−7A3+11A2−A−10IA^5 - 4A^4 - 7A^3 + 11A^2 - A - 10I
      • (ii) A4−4A3−5A2+A+2IA^4 - 4A^3 - 5A^2 + A + 2I
      • Solution: S1=4,S2=−5  ⟹  λ2−4λ−5=0  ⟹  A2−4A−5I=0S_1 = 4, S_2 = -5 \implies \lambda^2 - 4\lambda - 5 = 0 \implies A^2 - 4A - 5I = 0
      • Dividing λ5−4λ4−7λ3+11λ2−λ−10\lambda^5 - 4\lambda^4 - 7\lambda^3 + 11\lambda^2 - \lambda - 10 by λ2−4λ−5\lambda^2 - 4\lambda - 5 gives quotient λ3−2λ+3\lambda^3 - 2\lambda + 3 and remainder λ+5\lambda + 5.
      • Therefore, A5−4A4−7A3+11A2−A−10I=(A2−4A−5I)(A3−2A+3I)+A+5I=A+5IA^5 - 4A^4 - 7A^3 + 11A^2 - A - 10I = (A^2 - 4A - 5I)(A^3 - 2A + 3I) + A + 5I = A + 5I
      • For (ii): A4−4A3−5A2+A+2I=A2(A2−4A−5I)+A+2I=A+2IA^4 - 4A^3 - 5A^2 + A + 2I = A^2(A^2 - 4A - 5I) + A + 2I = A + 2I
    • Problem 8: Find A−1A^{-1} when A=(10321−11−11)A = \begin{pmatrix} 1 & 0 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{pmatrix} using Cayley-Hamilton theorem.
      • S1=3,S2=−1,S3=−9S_1 = 3, S_2 = -1, S_3 = -9
      • Characteristic equation: λ3−3λ2−λ+9=0  ⟹  A3−3A2−A+9I=0\lambda^3 - 3\lambda^2 - \lambda + 9 = 0 \implies A^3 - 3A^2 - A + 9I = 0
      • Pre-multiplying by A−1A^{-1}: 9A−1=−A2+3A+I9A^{-1} = -A^2 + 3A + I
      • A2=(4−363240−25)A^2 = \begin{pmatrix} 4 & -3 & 6 \\ 3 & 2 & 4 \\ 0 & -2 & 5 \end{pmatrix}
      • −A2+3A+I=(−43−6−3−2−402−5)+(30963−33−33)+(100010001)=(03332−73−1−1)-A^2 + 3A + I = \begin{pmatrix} -4 & 3 & -6 \\ -3 & -2 & -4 \\ 0 & 2 & -5 \end{pmatrix} + \begin{pmatrix} 3 & 0 & 9 \\ 6 & 3 & -3 \\ 3 & -3 & 3 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 3 & 3 \\ 3 & 2 & -7 \\ 3 & -1 & -1 \end{pmatrix}
      • A−1=19(03332−73−1−1)A^{-1} = \frac{1}{9} \begin{pmatrix} 0 & 3 & 3 \\ 3 & 2 & -7 \\ 3 & -1 & -1 \end{pmatrix}
    • Problem 9: Verify Cayley-Hamilton theorem for A=(137423121)A = \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix}.
      • S1=4,S2=−20,S3=35S_1 = 4, S_2 = -20, S_3 = 35
      • Characteristic equation: λ3−4λ2−20λ−35=0\lambda^3 - 4\lambda^2 - 20\lambda - 35 = 0
      • A2=(20232315223710914),A3=(1351522321401632086076111)A^2 = \begin{pmatrix} 20 & 23 & 23 \\ 15 & 22 & 37 \\ 10 & 9 & 14 \end{pmatrix}, \quad A^3 = \begin{pmatrix} 135 & 152 & 232 \\ 140 & 163 & 208 \\ 60 & 76 & 111 \end{pmatrix}
      • A3−4A2−20A−35I=0A^3 - 4A^2 - 20A - 35I = 0 verified.
    • Problem 10: Verify Cayley-Hamilton theorem for:
      • (i) A=(3−1−15)A = \begin{pmatrix} 3 & -1 \\ -1 & 5 \end{pmatrix}: Characteristic equation λ2−8λ+14=0\lambda^2 - 8\lambda + 14 = 0. A2=(10−8−826)A^2 = \begin{pmatrix} 10 & -8 \\ -8 & 26 \end{pmatrix}. Verification A2−8A+14I=0A^2 - 8A + 14I = 0 holds.
      • (ii) A=(1423)A = \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix}: Characteristic equation λ2−4λ−5=0\lambda^2 - 4\lambda - 5 = 0. A2=(916817)A^2 = \begin{pmatrix} 9 & 16 \\ 8 & 17 \end{pmatrix}. Verification A2−4A−5I=0A^2 - 4A - 5I = 0 holds.

Eigenvalues and Eigenvectors of Real Matrices

  • Working Rule to Find Eigenvalues and Eigenvectors:

    1. Find the characteristic equation ∣A−λI∣=0|A - \lambda I| = 0
    2. Solve the characteristic equation to get the characteristic roots (eigenvalues λ1,λ2,…\lambda_1, \lambda_2, \dots).
    3. To find the eigenvector XX corresponding to an eigenvalue λ\lambda, solve the linear homogeneous system [A−λI]X=0[A - \lambda I]X = 0

  • Important Notes & Definitions:

    • Corresponding to nn distinct eigenvalues, there exist nn linearly independent eigenvectors.
    • If two or more eigenvalues are equal, it may or may not be possible to obtain linearly independent eigenvectors corresponding to the repeated eigenvalues.
    • If XiX_i is a solution vector for an eigenvalue λi\lambda_i, then cXic X_i (where cc is an arbitrary non-zero constant) is also a solution. Thus, an eigenvector is not unique.
    • Algebraic Multiplicity (AM): The order of an eigenvalue as a root of the characteristic polynomial (e.g., if λ\lambda is a double root, its algebraic multiplicity is 2).
    • Geometric Multiplicity (GM): The maximum number of linearly independent eigenvectors corresponding to an eigenvalue λ\lambda.

  • Non-Symmetric Matrix Characteristics (A≠ATA \neq A^T):

    • If eigenvalues are non-repeated, a set of linearly independent eigenvectors is always obtained.
    • If eigenvalues are repeated, linearly independent eigenvectors may or may not exist.
    • If a full linearly independent set of eigenvectors exists, diagonalization is possible through similarity transformation.

  • Symmetric Matrix Characteristics (A=ATA = A^T):

    • If eigenvalues are non-repeated, a linearly independent and pairwise orthogonal set of eigenvectors is obtained.
    • If eigenvalues are repeated, a linearly independent and pairwise orthogonal set of eigenvectors can still be constructed.
    • Diagonalization of a symmetric matrix is always possible through orthogonal transformation.

  • Worked Problems on Finding Eigenvalues and Eigenvectors:

    • Problem 1: Find eigenvalues and eigenvectors of A=(113−1)A = \begin{pmatrix} 1 & 1 \\ 3 & -1 \end{pmatrix} (Non-symmetric).
      • Characteristic equation: λ2−4=0  ⟹  λ=±2\lambda^2 - 4 = 0 \implies \lambda = \pm 2.
      • Eigenvalues: 2,−22, -2
      • For λ=−2\lambda = -2: Solve [A+2I]X=0  ⟹  (3131)(x1x2)=(00)  ⟹  3x1+x2=0  ⟹  x11=x2−3[A + 2I]X = 0 \implies \begin{pmatrix} 3 & 1 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 3x_1 + x_2 = 0 \implies \frac{x_1}{1} = \frac{x_2}{-3}. Eigenvector X1=(1−3)X_1 = \begin{pmatrix} 1 \\ -3 \end{pmatrix}.
      • For λ=2\lambda = 2: Solve [A−2I]X=0  ⟹  (−113−3)(x1x2)=(00)  ⟹  x1−x2=0  ⟹  x11=x21[A - 2I]X = 0 \implies \begin{pmatrix} -1 & 1 \\ 3 & -3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies x_1 - x_2 = 0 \implies \frac{x_1}{1} = \frac{x_2}{1}. Eigenvector X2=(11)X_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}.
    • Problem 2: Find eigenvalues and eigenvectors of A=(221131122)A = \begin{pmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix} (Non-symmetric with repeated roots).
      • S1=7,S2=11,S3=5S_1 = 7, S_2 = 11, S_3 = 5
      • Characteristic equation: λ3−7λ2+11λ−5=0  ⟹  (λ−1)(λ2−6λ+5)=0  ⟹  λ=1,1,5\lambda^3 - 7\lambda^2 + 11\lambda - 5 = 0 \implies (\lambda - 1)(\lambda^2 - 6\lambda + 5) = 0 \implies \lambda = 1, 1, 5
      • Eigenvalues: 1,1,51, 1, 5
      • For λ=5\lambda = 5: Solve [A−5I]X=0[A - 5I]X = 0−3x1+2x2+x3=0-3x_1 + 2x_2 + x_3 = 0x1−2x2+x3=0x_1 - 2x_2 + x_3 = 0             Using cross-multiplication on first two equations: x12+2=x21+3=x36−2  ⟹  x14=x24=x34  ⟹  X1=(111)\frac{x_1}{2 + 2} = \frac{x_2}{1 + 3} = \frac{x_3}{6 - 2} \implies \frac{x_1}{4} = \frac{x_2}{4} = \frac{x_3}{4} \implies X_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.
      • For λ=1\lambda = 1: Solve [A−I]X=0  ⟹  x1+2x2+x3=0[A - I]X = 0 \implies x_1 + 2x_2 + x_3 = 0             Put x1=0  ⟹  2x2+x3=0  ⟹  take x3=2,x2=−1  ⟹  X2=(0−12)x_1 = 0 \implies 2x_2 + x_3 = 0 \implies \text{take } x_3 = 2, x_2 = -1 \implies X_2 = \begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix}.             Put x2=0  ⟹  x1+x3=0  ⟹  take x1=1,x3=−1  ⟹  X3=(10−1)x_2 = 0 \implies x_1 + x_3 = 0 \implies \text{take } x_1 = 1, x_3 = -1 \implies X_3 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}.
    • Problem 3: Find eigenvalues and eigenvectors of A=(2−2211113−1)A = \begin{pmatrix} 2 & -2 & 2 \\ 1 & 1 & 1 \\ 1 & 3 & -1 \end{pmatrix}.
      • S1=2,S2=−4,S3=−8S_1 = 2, S_2 = -4, S_3 = -8
      • Characteristic equation: λ3−2λ2−4λ+8=0  ⟹  (λ−2)(λ2−4)=0  ⟹  λ=2,2,−2\lambda^3 - 2\lambda^2 - 4\lambda + 8 = 0 \implies (\lambda - 2)(\lambda^2 - 4) = 0 \implies \lambda = 2, 2, -2
      • Eigenvalues: 2,2,−22, 2, -2
      • For λ=−2\lambda = -2: Solve [A+2I]X=0[A + 2I]X = 04x1−2x2+2x3=04x_1 - 2x_2 + 2x_3 = 0x1+3x2+x3=0x_1 + 3x_2 + x_3 = 0             By cross-multiplication: x1−2−6=x22−4=x312+2  ⟹  x1−8=x2−2=x314  ⟹  x14=x21=x3−7  ⟹  X1=(41−7)\frac{x_1}{-2 - 6} = \frac{x_2}{2 - 4} = \frac{x_3}{12 + 2} \implies \frac{x_1}{-8} = \frac{x_2}{-2} = \frac{x_3}{14} \implies \frac{x_1}{4} = \frac{x_2}{1} = \frac{x_3}{-7} \implies X_1 = \begin{pmatrix} 4 \\ 1 \\ -7 \end{pmatrix}.
      • For λ=2\lambda = 2: Solve [A−2I]X=0[A - 2I]X = 00x1−2x2+2x3=00x_1 - 2x_2 + 2x_3 = 0x1−x2+x3=0x_1 - x_2 + x_3 = 0             By cross-multiplication: x1−2+2=x22−0=x30+2  ⟹  x10=x22=x32  ⟹  x10=x21=x31  ⟹  X2=(011)\frac{x_1}{-2 + 2} = \frac{x_2}{2 - 0} = \frac{x_3}{0 + 2} \implies \frac{x_1}{0} = \frac{x_2}{2} = \frac{x_3}{2} \implies \frac{x_1}{0} = \frac{x_2}{1} = \frac{x_3}{1} \implies X_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}.             Only one linearly independent eigenvector is obtained for the repeated root \lambda = 2$.\n * *Problem 4:* Find eigenvalues and eigenvectors of symmetric matrix A = \begin{pmatrix} 1 & 1 & 3 \ 1 & 5 & 1 \ 3 & 1 & 1 \end{pmatrix}.\n * S_1 = 7, S_2 = 0, S_3 = -36\n * Characteristic equation: \lambda^3 - 7\lambda^2 - 36 = 0 \implies (\lambda + 2)(\lambda^2 - 9\lambda + 18) = 0 \implies \lambda = -2, 3, 6\n * For \lambda = -2::X_1 = \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}\n * For \lambda = 3::X_2 = \begin{pmatrix} 1 \ -1 \ 1 \end{pmatrix}\n * For \lambda = 6::X_3 = \begin{pmatrix} 1 \ 2 \ 1 \end{pmatrix}\n * *Problem 5:* Find eigenvalues, eigenvectors, algebraic multiplicity, and geometric multiplicity of A = \begin{pmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \end{pmatrix}.\n * S_1 = 0, S_2 = -3, S_3 = 2\n * Characteristic equation: \lambda^3 - 3\lambda - 2 = 0 \implies (\lambda + 1)^2(\lambda - 2) = 0\n * Eigenvalues: 2, -1, -1\n * Algebraic multiplicity of \lambda = -1 is 2.\n * For \lambda = 2::X_1 = \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix}.\n * For \lambda = -1::x_1 + x_2 + x_3 = 0.\n * Put x_1 = 0 \implies x_2 + x_3 = 0 \implies X_2 = \begin{pmatrix} 0 \ 1 \ -1 \end{pmatrix}.\n * To find X_3 = \begin{pmatrix} l \ m \ n \end{pmatrix}orthogonaltoorthogonal toX_1andandX_2:\n 1. l + m + n = 0\n 2. 0l + m - n = 0\n                By cross-multiplication: \frac{l}{-1 - 1} = \frac{m}{0 + 1} = \frac{n}{1 - 0} \implies \frac{l}{-2} = \frac{m}{1} = \frac{n}{1} \implies X_3 = \begin{pmatrix} -2 \ 1 \ 1 \end{pmatrix}.\n * Two linearly independent eigenvectors X_2andandX_3correspondtocorrespond to\lambda = -1.Thus,geometricmultiplicityof. Thus, geometric multiplicity of\lambda = -1 is 2.\n\n# Properties of Eigenvalues and Eigenvectors\n\n* **Fundamental Properties:**\n 1. **Sum of Eigenvalues:** The sum of the eigenvalues of a matrix Aequalsthesumofthemaindiagonalelements(Traceofequals the sum of the main diagonal elements (Trace ofA).\n        \sum_{i=1}^n \lambda_i = \text{Trace}(A)\n 2. **Product of Eigenvalues:** The product of the eigenvalues of a matrix Aequalsthedeterminantofequals the determinant ofA\n        \prod_{i=1}^n \lambda_i = |A|\n 3. **Transpose Property:** A square matrix Aanditstransposeand its transposeA^T have identical eigenvalues.\n 4. **Triangular/Diagonal Matrix Property:** The eigenvalues of a triangular (upper or lower) matrix or a diagonal matrix are simply its main diagonal elements.\n 5. **Inverse Property:** If \lambda_1, \lambda_2, \dots, \lambda_nareeigenvaluesofare eigenvalues ofA,thentheeigenvaluesof, then the eigenvalues ofA^{-1} are:\n        \frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \dots, \frac{1}{\lambda_n}\n 6. **Power Property:** The eigenvalues of A^k(where(wherekisapositiveinteger)areis a positive integer) are\lambda_1^k, \lambda_2^k, \dots, \lambda_n^k\n 7. **Scalar Multiple Property:** The eigenvalues of k Aarearek\lambda_1, k\lambda_2, \dots, k\lambda_n\n 8. **Polynomial Matrix Property:** The eigenvalues of a matrix polynomial f(A)arearef(\lambda_1), f(\lambda_2), \dots, f(\lambda_n)\n 9. **Adjugate Property:** The eigenvalues of \text{adj } A are:\n        \frac{|A|}{\lambda_1}, \frac{|A|}{\lambda_2}, \dots, \frac{|A|}{\lambda_n}\n 10. **Orthogonality Property:** The eigenvectors of a real symmetric matrix corresponding to distinct eigenvalues are mutually orthogonal in pairs.\n\n* **Short Problems on Eigenvalue Properties:**\n * *Problem 1:* Find sum and product of eigenvalues of A = \begin{pmatrix} -1 & 1 & 1 \ 1 & -1 & 1 \ 1 & 1 & -1 \end{pmatrix}.\n * Sum = -1 + (-1) + (-1) = -3\n * Product = |A| = -1(1 - 1) - 1(-1 - 1) + 1(1 - (-1)) = 0 + 2 + 2 = 4\n * *Problem 2:* Product of two eigenvalues of A = \begin{pmatrix} 6 & -2 & 2 \ -2 & 3 & -1 \ 2 & -1 & 3 \end{pmatrix} is 16. Find the third eigenvalue.\n * \lambda_1 \lambda_2 \lambda_3 = |A| = 6(9 - 1) + 2(-6 + 2) + 2(2 - 6) = 48 - 8 - 8 = 32\n * Given \lambda_1 \lambda_2 = 16 \implies 16 \lambda_3 = 32 \implies \lambda_3 = 2\n * *Problem 3:* Find sum and product of eigenvalues of A = \begin{pmatrix} a & b \ c & d \end{pmatrix} without solving.\n * Sum = \text{Trace}(A) = a + d\n * Product = |A| = ad - bc\n * *Problem 4:* If 3 and 15 are two eigenvalues of A = \begin{pmatrix} 8 & -6 & 2 \ -6 & 7 & -4 \ 2 & -4 & 3 \end{pmatrix},find, find|A| without expanding.\n * Sum of eigenvalues = 8 + 7 + 3 = 18 \implies 3 + 15 + \lambda_3 = 18 \implies \lambda_3 = 0\n * |A| = \text{product of eigenvalues} = 3 \times 15 \times 0 = 0\n * *Problem 5:* If 2, 2, 3areeigenvaluesofare eigenvalues ofA,findeigenvaluesof, find eigenvalues ofA^T\n * Eigenvalues of A^Tareare2, 2, 3\n * *Problem 6:* Find eigenvalues of lower triangular matrix A = \begin{pmatrix} 2 & 0 & 0 \ 1 & 3 & 0 \ 0 & 4 & 4 \end{pmatrix}.\n * Eigenvalues are diagonal elements: 2, 3, 4\n * *Problem 7:* Two eigenvalues of A = \begin{pmatrix} 3 & 1 & 1 \ 1 & 5 & -1 \ 1 & -1 & 3 \end{pmatrix}are3and6.Findeigenvaluesofare 3 and 6. Find eigenvalues ofA^{-1}.\n * Sum = 3 + 5 + 3 = 11 \implies 3 + 6 + k = 11 \implies k = 2\n * Eigenvalues of Aareare3, 6, 2\n * Eigenvalues of A^{-1}areare\frac{1}{3}, \frac{1}{6}, \frac{1}{2}\n * *Problem 8:* Find eigenvalues of A = \begin{pmatrix} 1 & 5 \ 4 & 4 \end{pmatrix}.Formmatrixwhoseeigenvaluesare. Form matrix whose eigenvalues are\frac{1}{6}andand-1\n * Characteristic equation: \lambda^2 - 5\lambda - 6 = 0 \implies \lambda = 6, -1\n * Eigenvalues of Aareare6, -1\n * Matrix with eigenvalues \frac{1}{6}andand-1isisA^{-1} = \frac{1}{|A|} \text{adj } A = -\frac{1}{6} \begin{pmatrix} 4 & -5 \ -4 & 1 \end{pmatrix}\n * *Problem 9:* Find eigenvalues of inverse of upper triangular matrix A = \begin{pmatrix} 2 & 1 & 0 \ 0 & 3 & 4 \ 0 & 0 & 4 \end{pmatrix}.\n * Eigenvalues of Aareare2, 3, 4\n * Eigenvalues of A^{-1}areare\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\n * *Problem 10:* Find eigenvalues of A^3givengivenA = \begin{pmatrix} 1 & 2 & 3 \ 0 & 2 & -7 \ 0 & 0 & 3 \end{pmatrix}.\n * Eigenvalues of Aareare1, 2, 3\n * Eigenvalues of A^3areare1^3, 2^3, 3^3 \implies 1, 8, 27\n * *Problem 11:* If 1 and 2 are eigenvalues of 2 \times 2matrixmatrixA,findeigenvaluesof, find eigenvalues ofA^2andandA^{-1}.\n * Eigenvalues of A^2::1^2, 2^2 \implies 1, 4\n * Eigenvalues of A^{-1}::1, \frac{1}{2}\n * *Problem 12:* If 1, 1, 5areeigenvaluesofare eigenvalues ofA,findeigenvaluesof, find eigenvalues of5A\n * Eigenvalues of 5A::5(1), 5(1), 5(5) \implies 5, 5, 25\n * *Problem 13:* Given upper triangular matrix A = \begin{pmatrix} 2 & 3 \ 0 & 5 \end{pmatrix},findeigenvaluesof, find eigenvalues ofA, A^2, A^3, A^4, 3A, A^{-1}, A - I,and, and3A^3 + 5A^2 - 6A + 2I\n * Eigenvalues of A::2, 5\n * Eigenvalues of A^2::4, 25\n * Eigenvalues of A^3::8, 125\n * Eigenvalues of A^4::16, 625\n * Eigenvalues of 3A::6, 15\n * Eigenvalues of A^{-1}::\frac{1}{2}, \frac{1}{5}\n * A - I = \begin{pmatrix} 1 & 3 \ 0 & 4 \end{pmatrix} \impliesEigenvaluesareEigenvalues are1, 4\n * Eigenvalues of 3A^3 + 5A^2 - 6A + 2I:\n * For \lambda_1 = 2::3(2)^3 + 5(2)^2 - 6(2) + 2 = 24 + 20 - 12 + 2 = 34\n * For \lambda_2 = 5::3(5)^3 + 5(5)^2 - 6(5) + 2 = 375 + 125 - 30 + 2 = 472\n * *Problem 14:* Find eigenvalues of \text{adj } AififA = \begin{pmatrix} 3 & 2 & 1 \ 0 & 4 & 2 \ 0 & 0 & 1 \end{pmatrix}.\n * Eigenvalues of A::3, 4, 1\n * |A| = 3 \times 4 \times 1 = 12\n * Eigenvalues of \text{adj } A = |A| A^{-1}areare\frac{12}{3}, \frac{12}{4}, \frac{12}{1} \implies 4, 3, 12\n * *Problem 15:* Two eigenvalues of A = \begin{pmatrix} 2 & 2 & 1 \ 1 & 3 & 1 \ 1 & 2 & 2 \end{pmatrix}areequalandare equal and\frac{1}{5} times the third. Find them.\n * Sum = 2 + 3 + 2 = 7\n * \lambda_1 + \lambda_2 + \lambda_3 = 7 \implies \frac{1}{5}\lambda_3 + \frac{1}{5}\lambda_3 + \lambda_3 = 7 \implies \frac{7}{5}\lambda_3 = 7 \implies \lambda_3 = 5\n * \lambda_1 = \lambda_2 = 1\n * Eigenvalues: 1, 1, 5\n * *Problem 16:* If 2, 3areeigenvaluesofare eigenvalues ofA = \begin{pmatrix} 2 & 0 & 1 \ 0 & 2 & 0 \ a & 0 & 2 \end{pmatrix},find, finda\n * Let eigenvalues be 2, 3, k\n * Sum = 2 + 2 + 2 = 6 \implies 2 + 3 + k = 6 \implies k = 1\n * Product = 2 \times 3 \times 1 = 6 = |A|\n * |A| = 2(4) - 0 + 1(-2a) = 8 - 2a\n * 6 = 8 - 2a \implies 2a = 2 \implies a = 1\n * *Problem 17:* Prove eigenvectors of real symmetric matrix A = \begin{pmatrix} 1 & 1 & 3 \ 1 & 5 & 1 \ 3 & 1 & 1 \end{pmatrix} are orthogonal in pairs.\n * Characteristic equation: \lambda^3 - 7\lambda^2 + 36 = 0 \implies \lambda = -2, 3, 6\n * Eigenvectors: X_1 = \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}, X_2 = \begin{pmatrix} 1 \ -1 \ 1 \end{pmatrix}, X_3 = \begin{pmatrix} 1 \ 2 \ 1 \end{pmatrix}\n * X_1^T X_2 = -1 + 0 + 1 = 0\n * X_2^T X_3 = 1 - 2 + 1 = 0\n * X_3^T X_1 = -1 + 0 + 1 = 0\n * All pairs are orthogonal.\n * *Problem 18:* Find sum and product of all eigenvalues of A = \begin{pmatrix} 1 & 2 & 3 \ 2 & 2 & 4 \ 1 & 2 & 7 \end{pmatrix}.Is. IsA singular?\n * Sum = 1 + 2 + 7 = 10\n * Product = |A| = 1(14 - 8) - 2(14 - 4) + 3(4 - 2) = 6 - 20 + 6 = -8\n * Since |A| eq 0,matrix, matrixA is non-singular.\n * *Problem 19:* Find product of eigenvalues of A = \begin{pmatrix} 1 & 2 & -2 \ 1 & 0 & 3 \ -2 & -1 & -3 \end{pmatrix}.\n * Product = |A| = 1(3) - 2(-3 + 6) - 2(-1) = 3 - 6 + 2 = -1\n\n# Orthogonal Transformation and Diagonalization of Symmetric Matrices\n\n* **Orthogonal Matrices:**\n * A real square matrix A is called an orthogonal matrix if:\n        A A^T = A^T A = I \quad \text{or equivalently} \quad A^T = A^{-1}\n * *Verification Problems:*\n 1. Check if B = \begin{pmatrix} \cos\theta & \sin\theta & 0 \ -\sin\theta & \cos\theta & 0 \ 0 & 0 & 1 \end{pmatrix} is orthogonal.\n            B B^T = \begin{pmatrix} \cos\theta & \sin\theta & 0 \ -\sin\theta & \cos\theta & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta & 0 \ \sin\theta & \cos\theta & 0 \ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \cos^2\theta + \sin^2\theta & 0 & 0 \ 0 & \sin^2\theta + \cos^2\theta & 0 \ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} = I\n            Hence, B is orthogonal.\n 2. Show that P = \begin{pmatrix} \cos\theta & \sin\theta \ -\sin\theta & \cos\theta \end{pmatrix} is orthogonal.\n            P P^T = \begin{pmatrix} \cos^2\theta + \sin^2\theta & -\sin\theta\cos\theta + \sin\theta\cos\theta \ -\sin\theta\cos\theta + \sin\theta\cos\theta & \sin^2\theta + \cos^2\theta \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = I\n\n* **Working Rule for Diagonalization via Orthogonal Transformation:**\n * **Step 1:** Find the characteristic equation |A - \lambda I| = 0.\n * **Step 2:** Solve the characteristic equation to get eigenvalues \lambda_1, \lambda_2, \lambda_3\n * **Step 3:** Obtain mutually orthogonal eigenvectors X_1, X_2, X_3\n * **Step 4:** Normalize each eigenvector by dividing by its length (norm):\n        e_i = \frac{X_i}{|X_i|}\n        Form the normalized modal matrix N = \begin{pmatrix} e_1 & e_2 & e_3 \end{pmatrix}.\n * **Step 5:** Compute N^T\n * **Step 6:** Compute matrix product A N\n * **Step 7:** Compute diagonal matrix D = N^T A N,wherethediagonalelementsof, where the diagonal elements ofDaretheeigenvaluesofare the eigenvalues ofA\n\n* **Worked Diagonalization Problems:**\n * *Problem 1:* Diagonalize A = \begin{pmatrix} 3 & -1 & 1 \ -1 & 5 & -1 \ 1 & -1 & 3 \end{pmatrix}.\n * S_1 = 11, S_2 = 36, S_3 = 36\n * Characteristic equation: \lambda^3 - 11\lambda^2 + 36\lambda - 36 = 0 \implies \lambda = 2, 3, 6\n * Eigenvectors: X_1 = \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}, X_2 = \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix}, X_3 = \begin{pmatrix} 1 \ -2 \ 1 \end{pmatrix}\n * Lengths: |X_1| = \sqrt{2}, |X_2| = \sqrt{3}, |X_3| = \sqrt{6}\n * Normalized Matrix N = \begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \ 0 & \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \end{pmatrix}\n * D = N^T A N = \begin{pmatrix} 2 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 6 \end{pmatrix}\n * *Problem 2:* Diagonalize A = \begin{pmatrix} 8 & -6 & 2 \ -6 & 7 & -4 \ 2 & -4 & 3 \end{pmatrix}.\n * S_1 = 18, S_2 = 45, S_3 = 0\n * Characteristic equation: \lambda^3 - 18\lambda^2 + 45\lambda = 0 \implies \lambda = 0, 3, 15\n * Eigenvectors: X_1 = \begin{pmatrix} 1 \ 2 \ 2 \end{pmatrix}, X_2 = \begin{pmatrix} 2 \ 1 \ -2 \end{pmatrix}, X_3 = \begin{pmatrix} 2 \ -2 \ 1 \end{pmatrix}\n * Lengths: |X_1| = 3, |X_2| = 3, |X_3| = 3\n * Normalized Matrix N = \frac{1}{3} \begin{pmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ 2 & -2 & 1 \end{pmatrix}\n * D = N^T A N = \begin{pmatrix} 0 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 15 \end{pmatrix}\n\n# Quadratic Forms and Canonical Forms\n\n* **Definition of Quadratic Form:**\n * A homogeneous polynomial of second degree in any number of variables is called a quadratic form.\n * *Example:* 2x_1^2 + 3x_2^2 - x_3^2 + 4x_1 x_2 + 5x_1 x_3 - 6x_2 x_3isaquadraticforminthreevariablesis a quadratic form in three variablesx_1, x_2, x_3\n\n* **Matrix Corresponding to a Quadratic Form:**\n * For a three-variable quadratic form, the symmetric matrix Q is:\n        Q = \begin{pmatrix} \text{coeff of } x_1^2 & \frac{1}{2}\text{coeff of } x_1 x_2 & \frac{1}{2}\text{coeff of } x_1 x_3 \ \frac{1}{2}\text{coeff of } x_2 x_1 & \text{coeff of } x_2^2 & \frac{1}{2}\text{coeff of } x_2 x_3 \ \frac{1}{2}\text{coeff of } x_3 x_1 & \frac{1}{2}\text{coeff of } x_3 x_2 & \text{coeff of } x_3^2 \end{pmatrix}\n\n* **Problems on Finding Matrix of Quadratic Form and Vice Versa:**\n * *Problem 1:* Write matrix of quadratic form 2x_1^2 - 2x_2^2 + 4x_3^2 + 2x_1 x_2 - 6x_1 x_3 + 6x_2 x_3\n * Q = \begin{pmatrix} 2 & 1 & -3 \ 1 & -2 & 3 \ -3 & 3 & 4 \end{pmatrix}\n * *Problem 2:* Write matrix of quadratic form 2x^2 + 8z^2 + 4xy + 10xz - 2yz\n * Q = \begin{pmatrix} 2 & 2 & 5 \ 2 & 0 & -1 \ 5 & -1 & 8 \end{pmatrix}\n * *Problem 3:* Write quadratic form corresponding to symmetric matrix A = \begin{pmatrix} 0 & -1 & 2 \ -1 & 1 & 4 \ 2 & 4 & 3 \end{pmatrix}\n * Q = 0x_1^2 + x_2^2 + 3x_3^2 - 2x_1 x_2 + 4x_1 x_3 + 8x_2 x_3\n\n* **Classification and Nature of Quadratic Forms:**\n * **Rank (r):** The number of non-zero terms (square terms) in the canonical form.\n * **Index (s):** The number of positive square terms in the canonical form.\n * **Signature:** The difference between the number of positive and negative square terms:\n        \text{Signature} = s - (r - s) = 2s - r\n * **Classifications based on Eigenvalues:**\n 1. **Positive Definite:** All eigenvalues are strictly positive (> 0).\n 2. **Negative Definite:** All eigenvalues are strictly negative (< 0).\n 3. **Positive Semi-Definite:** All eigenvalues are \ge 0withatleastoneeigenvalueequaltowith at least one eigenvalue equal to0\n 4. **Negative Semi-Definite:** All eigenvalues are \le 0withatleastoneeigenvalueequaltowith at least one eigenvalue equal to0\n 5. **Indefinite:** Eigenvalues contain both positive and negative values.\n\n* **Problems on Determining Nature:**\n * *Problem 1:* Determine nature of f(x_1, x_2, x_3) = x_1^2 + 2x_2^2\n * Matrix Q = \begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 0 \end{pmatrix}\n * Eigenvalues: 1, 2, 0\n * Nature: Positive Semi-definite.\n * *Problem 2:* Discuss nature of 2x^2 + 3y^2 + 2z^2 + 2xy without reducing to canonical form.\n * Matrix Q = \begin{pmatrix} 2 & 1 & 0 \ 1 & 3 & 0 \ 0 & 0 & 2 \end{pmatrix}\n * Principal Minors:\n * D_1 = |2| = 2 > 0\n * D_2 = \begin{vmatrix} 2 & 1 \ 1 & 3 \end{vmatrix} = 5 > 0\n * D_3 = |Q| = 2(6 - 0) - 1(2 - 0) + 0 = 10 > 0\n * Since D_1, D_2, D_3 > 0, the quadratic form is Positive Definite.\n\n* **Working Rule for Reduction of Quadratic Form to Canonical Form via Orthogonal Transformation:**\n * **Step 1:** Write the symmetric matrix A of the given quadratic form.\n * **Step 2:** Find the characteristic equation |A - \lambda I| = 0\n * **Step 3:** Solve for eigenvalues \lambda_1, \lambda_2, \lambda_3\n * **Step 4:** Find mutually orthogonal eigenvectors X_1, X_2, X_3\n * **Step 5:** Form normalized matrix N\n * **Step 6:** Compute N^T\n * **Step 7:** Compute A N\n * **Step 8:** Compute D = N^T A N\n * **Step 9:** Write canonical form as:\n        \begin{pmatrix} y_1 & y_2 & y_3 \end{pmatrix} D \begin{pmatrix} y_1 \ y_2 \ y_3 \end{pmatrix} = \lambda_1 y_1^2 + \lambda_2 y_2^2 + \lambda_3 y_3^2\n\n* **Worked Problems on Reduction to Canonical Form:**\n * *Problem 1:* Reduce Q = x^2 + 3y^2 + 3z^2 - 2yz to canonical form using orthogonal transformation.\n * Matrix A = \begin{pmatrix} 1 & 0 & 0 \ 0 & 3 & -1 \ 0 & -1 & 3 \end{pmatrix}\n * S_1 = 7, S_2 = 14, S_3 = 8\n * Characteristic equation: \lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0 \implies \lambda = 1, 2, 4\n * Eigenvectors: X_1 = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}, X_2 = \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix}, X_3 = \begin{pmatrix} 0 \ -1 \ 1 \end{pmatrix}\n * Normalized Matrix N = \begin{pmatrix} 1 & 0 & 0 \ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}\n * D = N^T A N = \begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 4 \end{pmatrix}\n * Canonical Form: y_1^2 + 2y_2^2 + 4y_3^2\n * *Problem 2:* Reduce 2x_1 x_2 + 2x_1 x_3 - 2x_2 x_3 to canonical form by orthogonal transformation and determine its nature.\n * Matrix A = \begin{pmatrix} 0 & 1 & 1 \ 1 & 0 & -1 \ 1 & -1 & 0 \end{pmatrix}\n * S_1 = 0, S_2 = -3, S_3 = -2\n * Characteristic equation: \lambda^3 - 3\lambda + 2 = 0 \implies \lambda = 1, 1, -2\n * For \lambda = -2::X_1 = \begin{pmatrix} -1 \ 1 \ 1 \end{pmatrix}\n * For x_1 - x_2 - x_3 = 0((\lambda = 1 repeated):\n * Choose X_2 = \begin{pmatrix} 0 \ -1 \ 1 \end{pmatrix}\n * Choose X_3 = \begin{pmatrix} 2 \ 1 \ 1 \end{pmatrix}orthogonaltoorthogonal toX_1andandX_2\n * Normalized Matrix N = \begin{pmatrix} -\frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{6}} \ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \end{pmatrix}\n * D = N^T A N = \begin{pmatrix} -2 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}\n * Canonical Form: -2y_1^2 + y_2^2 + y_3^2\n * Nature: Eigenvalues are -2, 1, 1$$ (both positive and negative values exist), hence Indefinite in nature.