Comprehensive Linear Algebra: Vector Spaces, Systems of Equations, and Linear Transformations

Vector Space Fundamentals and Axioms

• Definition of a Vector Space:

  • A vector space $V$ over the field of real numbers $\mathbb{R}$ is defined as a non-empty set of elements called vectors, equipped with an equivalence relation (equality of vectors) and two closed operations:

    • Vector Addition: A rule assigning to each pair of vectors $a, b \in V$ a sum $a + b \in V$.

    • Scalar Multiplication: A rule assigning to each scalar $k \in \mathbb{R}$ and each vector $a \in V$ an element $k a \in V$.

  • To qualify as a vector space, the set $V$ and these operations must satisfy the following ten axioms for all vectors $a, b, c \in V$ and all real scalars $k, p, q \in \mathbb{R}$:

    • (i) Closure of Vector Addition: If $a, b \in V$, then $a + b \in V$ .

    • (ii) Closure of Scalar Multiplication: For every $a \in V$ and $k \in \mathbb{R}$, the product $k a \in V$.

    • (iii) Associative Law of Addition: $(a + b) + c = a + (b + c)$.

    • (iv) Existence of an Additive Identity (Zero Vector): There exists a unique vector called zero, denoted by $0$, such that $0 + a = a$.

    • (v) Existence of an Additive Inverse (Minus $a$): For every vector $a \in V$, there exists an element denoted by $-a$ such that $a + (-a) = 0$.

    • (vi) Commutative Law of Addition: $a + b = b + a$.

    • (vii) Distributive Law of Scalar Multiplication over Vector Addition: $k(a + b) = k a + k b$.

    • (viii) Distributive Law of Scalar Multiplication over Scalar Addition: $(p + q)a = p a + q a$.

    • (ix) Associativity of Scalar Multiplication: $(p q)a = p(q a)$.

    • (x) Scalar Identity Property: $1 a = a$.

• Canonical Examples of Vector Spaces:

  • The Two-Dimensional Real Space $\mathbb{R}^2$:

    • Comprises all real column vectors $x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$.

    • Vector addition operates component-wise: $\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix} x_1 + y_1 \\ x_2 + y_2 \end{pmatrix}$.

    • Scalar multiplication is defined by: $k \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} k x_1 \\ k x_2 \end{pmatrix} \in \mathbb{R}^2$.

    • The zero vector is $0 = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \in \mathbb{R}^2$.

    • The additive inverse of $x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$ is $-x = \begin{pmatrix} -x_1 \\ -x_2 \end{pmatrix}$.

    • Associativity follows directly from the real numbers: $\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} + \begin{pmatrix} z_1 \\ z_2 \end{pmatrix} = \begin{pmatrix} x_1 + y_1 + z_1 \\ x_2 + y_2 + z_2 \end{pmatrix}$.

  • The $n$-Dimensional Euclidean Vector Space $\mathbb{R}^n$:

    • Comprises all column vectors of size $n \times 1$: $x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}$.

    • Vector addition: $x + y = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} + \begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{pmatrix} = \begin{pmatrix} x_1 + y_1 \\ x_2 + y_2 \\ \vdots \\ x_n + y_n \end{pmatrix}$.

    • Scalar multiplication: $k x = \begin{pmatrix} k x_1 \\ k x_2 \\ \vdots \\ k x_n \end{pmatrix}$ for any scalar $k \in \mathbb{R}$.

    • The zero element is $0 = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}$.

  • The Matrix Space $M(2)$:

    • Consists of all $2 \times 2$ matrices with real entries.

    • Addition is performed component-wise: $\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} + \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} = \begin{pmatrix} a_{11} + b_{11} & a_{12} + b_{12} \\ a_{21} + b_{21} & a_{22} + b_{22} \end{pmatrix}$.

    • The zero vector is the zero matrix: $0 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.

  • The Polynomial Space $P_n$:

    • The space of all polynomials in an indeterminate $x$ having degree less than or equal to $n$.

    • Vector addition is defined by combining corresponding coefficients:             $(a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0) + (b_n x^n + b_{n-1} x^{n-1} + \dots + b_1 x + b_0) = (a_n + b_n)x^n + (a_{n-1} + b_{n-1})x^{n-1} + \dots + (a_0 + b_0)$.

    • Scalar multiplication distributes to every coefficient:             $k(a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0) = (k a_n)x^n + (k a_{n-1})x^{n-1} + \dots + (k a_0)$.

    • The zero element is the zero polynomial ($0 x^n + \dots + 0$).

  • The Continuous Function Space $C[0,1]$:

    • Consists of all continuous, real-valued functions defined on the closed interval $[0,1]$.

    • Addition: $(f + g)(x) = f(x) + g(x)$ for all $x \in [0,1]$.

    • Scalar multiplication: $(k f)(x) = k \cdot f(x)$ for all $k \in \mathbb{R}$ and $x \in [0,1]$.

    • Continuity of $f + g$ and $k f$ is established in calculus.

    • The zero element is the constant zero function: $0(x) = 0$ for all $x \in [0,1]$.

    • $C[0,1]$ is an infinite-dimensional vector space.

• Foundational Theorems and Proofs on Vector Spaces:

  • Theorem (Uniqueness of Zero): The zero vector in a vector space $V$ is unique.

    • Proof: Let $0$ and $0'$ both be zero elements in $V$.

    • Since $0$ is an additive identity: $0 + 0' = 0'$.

    • Since $0'$ is an additive identity: $0 + 0' = 0$.

    • Equating the two expressions yields $0' = 0 + 0' = 0$.

  • Theorem ($0 a = 0$): For all vectors $a \in V$, the product of the scalar zero and $a$ is the zero vector.

    • Proof: Using scalar addition, $0 + 0 = 0$.

    • Multiplying by vector $a$ gives $0 a = (0 + 0)a = 0 a + 0 a$.

    • Adding the additive inverse $-(0 a)$ to both sides:             $0 a + (-(0 a)) = (0 a + 0 a) + (-(0 a))$.             $0 = 0 a + (0 a + (-(0 a))) = 0 a + 0 = 0 a$.

    • By uniqueness of zero, $0 a = 0$.

  • Theorem ($k 0 = 0$): For all scalars $k \in \mathbb{R}$, $k 0 = 0$.

    • Proof: In $V$, $0 + 0 = 0$.

    • Scaling by $k$ yields $k 0 = k(0 + 0) = k 0 + k 0$.

    • Subtracting $k 0$ from both sides establishes $k 0 = 0$.

  • Theorem ($-a = (-1)a$): For every vector $a \in V$, the additive inverse is obtained by scaling $a$ by $-1$.

    • Proof: Evaluate the sum $a + (-1)a$:             $a + (-1)a = 1 a + (-1)a = (1 + (-1))a = 0 a = 0$.

    • By the definition of the additive inverse $-a$, it follows that $(-1)a = -a$.

  • Theorem (Uniqueness of $-a$): The additive inverse of any vector $a \in V$ is unique.

    • Proof: Suppose $b$ and $b'$ are both additive inverses of $a$, such that $a + b = 0$ and $a + b' = 0$.

    • Then $b = b + 0 = b + (a + b') = (b + a) + b' = 0 + b' = b'$.

    • Hence, $-a$ is uniquely determined.

  • Theorem (Cancellation Law): For all vectors $a, b, c \in V$, $a = b$ if and only if $a + c = b + c$.

    • Proof: If $a = b$, then $(a + c) - (b + c) = (a + c) - (a + c) = 0$, so $a + c = b + c$.

    • Conversely, if $a + c = b + c$, then:             $a = a + 0 = a + (c + (-c)) = (a + c) + (-c) = (b + c) + (-c) = b + (c + (-c)) = b + 0 = b$.

The Geometry of Euclidean Space and the Dot Product

• Definition of the Dot Product:

  • For vectors $x, y \in \mathbb{R}^n$, the dot product (scalar product), denoted $x \cdot y$, is defined algebraically as:         $x \cdot y = x_1 y_1 + x_2 y_2 + \dots + x_n y_n$

  • Worked Numerical Examples:

    • Let $x = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ and $y = \begin{pmatrix} 2 \\ 4 \\ 10 \end{pmatrix}$.             $x \cdot y = 1(2) + (-1)(4) + 2(10) = 2 - 4 + 20 = 18$.

    • Let $x = \begin{pmatrix} -1 \\ 2 \end{pmatrix}$ and $y = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$.             $x \cdot y = (-1)(2) + 2(-1) = -2 - 2 = -4$.

    • Let $x = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ and $y = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$.             $x \cdot y = 0(1) + 0(0) + 1(0) = 0$.

• Orthogonality and Properties of the Dot Product:

  • Definition: Two vectors $x$ and $y$ are said to be orthogonal (perpendicular) if and only if $x \cdot y = 0$.

  • Properties:

    • 1. Positivity and Definiteness: $x \cdot x \ge 0$, and $x \cdot x = 0$ if and only if $x = 0$.

    • 2. Commutativity: $x \cdot y = y \cdot x$ for all vectors $x, y \in V$.

    • 3. Distributivity over Vector Addition: $x \cdot (y + z) = x \cdot y + x \cdot z$ for all $x, y, z \in V$.

    • 4. Homogeneity with Scalars: $k(x \cdot y) = (k x) \cdot y = x \cdot (k y)$ for all scalars $k \in \mathbb{R}$ and vectors $x, y \in V$.

• The Norm (Magnitude) of a Vector:

  • In $\mathbb{R}^n$, the norm (magnitude) of a vector $x$ is defined using the inner product:         $\|x\| = \sqrt{x \cdot x}$

  • Properties of the Norm:

    • (i) $\ travel{\|x\|} \ge 0$, and $\ travel{\|x\|} = 0$ if and only if $x = 0$.

    • (ii) $\ travel{\|k x\|} = |k| \|x\|$ for all scalars $k \in \mathbb{R}$ and vectors $x \in \mathbb{R}^n$.

    • (iii) Triangle Inequality: $\ travel{\|x + y\|} \le \|x\| + \|y\|$ for all $x, y \in \mathbb{R}^n$.

• Worked Problems on Dot Products and Norms:

  • Problem 1 (Norm Computation): Given $x = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$ and $y = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}$, evaluate $\ travel{\|2x + 3y\|}$.

    • Compute the combined linear vector:             $2x + 3y = 2\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} + 3\begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 0 \end{pmatrix} + \begin{pmatrix} 6 \\ 9 \\ -3 \end{pmatrix} = \begin{pmatrix} 8 \\ 7 \\ -3 \end{pmatrix}$.

    • Evaluate the inner product with itself:             $\begin{pmatrix} 8 \\ 7 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ 7 \\ -3 \end{pmatrix} = 8^2 + 7^2 + (-3)^2 = 64 + 49 + 9 = 122$.

    • Compute the root:             $\|2x + 3y\| = \sqrt{122} \approx 11.045$.

  • Problem 2 (Expansion with Orthogonality): Simplify the expression $(x + 2y) \cdot (3x - 2y)$ given that $x$ and $y$ are orthogonal vectors.

    • Expand using the distributive law:             $(x + 2y) \cdot (3x - 2y) = 3(x \cdot x) - 2(x \cdot y) + 6(y \cdot x) - 4(y \cdot y) = 3\|x\|^2 + 4(y \cdot x) - 4\|y\|^2$.

    • Since $x$ and $y$ are orthogonal, $x \cdot y = 0$:             $3\|x\|^2 + 4(0) - 4\|y\|^2 = 3\|x\|^2 - 4\|y\|^2$.

  • Problem 3 (System of Equations for Norms): Given that $x$ and $y$ are orthogonal, $\ travel{\|x + 2y\|^2} = 4$, and $\ travel{\|2x - 3y\|^2} = 9$, find $\ travel{\|x\|}$ and $\ travel{\|y\|}$.

    • Expand the first equation using orthogonality ($x \cdot y = 0$):             $\|x + 2y\|^2 = \|x\|^2 + 4(x \cdot y) + 4\|y\|^2 = \|x\|^2 + 4\|y\|^2 = 4$.

    • Expand the second equation:             $\|2x - 3y\|^2 = 4\|x\|^2 - 12(x \cdot y) + 9\|y\|^2 = 4\|x\|^2 + 9\|y\|^2 = 9$.

    • From the first equation, substitute $\ travel{\|x\|^2} = 4 - 4\|y\|^2$ into the second:             $4(4 - 4\|y\|^2) + 9\|y\|^2 = 9$.             $16 - 16\|y\|^2 + 9\|y\|^2 = 9 \implies 16 - 7\|y\|^2 = 9 \implies 7\|y\|^2 = 7 \implies \|y\|^2 = 1$.             $\|y\| = 1$.

    • Solve for $\ travel{\|x\|}$.             $\|x\|^2 = 4 - 4(1) = 0 \implies \|x\| = 0$.

  • Problem 4 (Parallelogram Law): Prove that for all vectors $x, y \in \mathbb{R}^n$:         $\|x + y\|^2 + \|x - y\|^2 = 2\|x\|^2 + 2\|y\|^2$

    • Proof: Expand both squared norms using the dot product:             $\|x + y\|^2 = (x + y) \cdot (x + y) = \|x\|^2 + 2(x \cdot y) + \|y\|^2$.             $\|x - y\|^2 = (x - y) \cdot (x - y) = \|x\|^2 - 2(x \cdot y) + \|y\|^2$.

    • Summing the two equations eliminates the cross terms:             $\|x + y\|^2 + \|x - y\|^2 = \|x\|^2 + 2(x \cdot y) + \|y\|^2 + \|x\|^2 - 2(x \cdot y) + \|y\|^2 = 2\|x\|^2 + 2\|y\|^2$.

  • Problem 5 (Diagonals of a Rhombus): Prove that the diagonals of a rhombus intersect at right angles.

    • Proof: Let the sides of the rhombus be represented by vectors $a$ and $b$, with equal side lengths: $\ travel{\|a\|} = \|b\|$.

    • The two diagonals are given by the vector sum and difference: $d_1 = a + b$ and $d_2 = a - b$.

    • Take their dot product:             $d_1 \cdot d_2 = (a + b) \cdot (a - b) = a \cdot a - a \cdot b + b \cdot a - b \cdot b = \|a\|^2 - \|b\|^2$.

    • Since $\ travel{\|a\|} = \|b\|$, it follows that $\ travel{\|a\|^2} - \|b\|^2 = 0$.

    • Thus, $d_1 \cdot d_2 = 0$, proving the diagonals are orthogonal.

  • Problem 6 (Pythagorean Theorem): Prove that in a right-angled triangle with hypotenuse $c$, $\ travel{\|c\|^2} = \|a\|^2 + \|b\|^2$.

    • Proof: Let sides $a$ and $b$ be orthogonal, so $a \cdot b = 0$.

    • The hypotenuse is represented as $c = a + b$.

    • Taking the square of the norm:             $\|c\|^2 = (a + b) \cdot (a + b) = \|a\|^2 + 2(a \cdot b) + \|b\|^2$.

    • Substituting $a \cdot b = 0$ yields $\ travel{\|c\|^2} = \|a\|^2 + \|b\|^2$.

The Angle Between Vectors and Geometric Inequalities

• Derivation of the Angle Formula via the Law of Cosines:

• Consider triangle $\Delta OAB$ where sides are formed by vectors $a$ and $b$, with angle $\theta$ between them, and the third side being $b - a$.

  • Applying the Law of Cosines:         $\|b - a\|^2 = \|a\|^2 + \|b\|^2 - 2\|a\|\|b\|\cos(\theta)$

  • Expanding the squared norm using dot product properties:         $\|b - a\|^2 = (b - a) \cdot (b - a) = b \cdot b - b \cdot a - a \cdot b + a \cdot a = \|b\|^2 - 2(a \cdot b) + \|a\|^2$

  • Equating both representations:         $\|a\|^2 + \|b\|^2 - 2(a \cdot b) = \|a\|^2 + \|b\|^2 - 2\|a\|\|b\|\cos(\theta)$

  • Subtracting $\ travel{\|a\|^2} + \|b\|^2$ and dividing by $-2$ yields:         $a \cdot b = \|a\|\|b\|\cos(\theta)$

  • Rearranging gives the formula for the cosine of the angle:         $\cos(\theta) = \frac{a \cdot b}{\|a\| \|b\|}$

• Special Orientations:

  • Parallel Vectors: If $a$ and $b$ are parallel, then $b = k a$ for some scalar $k$.         $\cos(\theta) = \frac{a \cdot (k a)}{\|a\| \|k a\|} = \frac{k \|a\|^2}{|k| \|a\|^2} = \pm 1$

    • $\cos(\theta) = 1 \implies \theta = 0^\circ$ (same direction).

    • $\cos(\theta) = -1 \implies \theta = 180^\circ$ (opposite direction).

  • Perpendicular Vectors: If $a$ and $b$ are orthogonal, then $a \cdot b = 0$, so:         $\cos(\theta) = \frac{0}{\|a\|\|b\|} = 0 \implies \theta = 90^\circ$

• Worked Examples for Angles:

  • Example (a): Find the acute angle between $a = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $b = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$.

    • Compute the dot product: $a \cdot b = 1(-3) + 2(4) = -3 + 8 = 5$.

    • Compute magnitudes:             $\|a\| = \sqrt{1^2 + 2^2} = \sqrt{5}$.             $\|b\| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

    • Substitute into the formula:             $\cos(\theta) = \frac{5}{\sqrt{5} \times 5} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$.

    • Calculate angle:             $\theta = \arccos\left(\frac{\sqrt{5}}{5}\right) \approx 63.435^\circ$.

  • Example (b): Find the acute angle between $a = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}$ and $b = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}$.

    • Compute the dot product: $a \cdot b = (-1)(-1) + 2(0) + 0(2) = 1 + 0 + 0 = 1$.

    • Compute magnitudes:             $\|a\| = \sqrt{(-1)^2 + 2^2 + 0^2} = \sqrt{1 + 4 + 0} = \sqrt{5}$.             $\|b\| = \sqrt{(-1)^2 + 0^2 + 2^2} = \sqrt{1 + 0 + 4} = \sqrt{5}$.

    • Substitute into the formula:             $\cos(\theta) = \frac{1}{\sqrt{5} \times \sqrt{5}} = \frac{1}{5}$.

    • Calculate angle:             $\theta = \arccos\left(\frac{1}{5}\right) \approx 78.463^\circ$.

• Proof of the Triangle Inequality:

  • Theorem: For all vectors $x, y \in \mathbb{R}^n$, $\ travel{\|x + y\|} \le \|x\| + \|y\|$.

  • Proof: Consider the squared norm of the sum:         $\|x + y\|^2 = (x + y) \cdot (x + y) = \|x\|^2 + 2(x \cdot y) + \|y\|^2$.

  • Substitute $x \cdot y = \|x\|\|y\|\cos(\theta)$:         $\|x + y\|^2 = \|x\|^2 + 2\|x\|\|y\|\cos(\theta) + \|y\|^2$.

  • Since $\cos(\theta) \le 1$ for all $\theta$:         $\|x + y\|^2 \le \|x\|^2 + 2\|x\|\|y\|(1) + \|y\|^2 = (\|x\| + \|y\|)^2$.

  • Taking the non-negative square root of both sides establishes the inequality:         $\|x + y\| \le \|x\| + \|y\|$.

  • Geometric Meaning: In any triangle, the length of the longest side cannot exceed the sum of the lengths of the other two sides.

Lines and Planes in Three Dimensions

• Equation of a Straight Line:

  • The vector equation of a line passing through point $A$ with position vector $a$ and directed toward point $B$ with position vector $b$ is:         $OP = a + t(b - a)$         where $t \in \mathbb{R}$ is a scalar parameter and $b - a$ represents the direction vector.

  • Worked Example 1: Line joining $(0, 1, -1)$ and $(3, 0, 2)$.

    • Direction vector: $\begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}$.

    • Vector equation:             $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}$.

    • Parametric equations:             $x = 3t$             $y = 1 - t$             $z = -1 + 3t$

  • Worked Example 2: Line joining $A(2, -1, 3)$ and $B(0, 5, -1)$.

    • Direction vector: $\begin{pmatrix} 0 \\ 5 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \\ -4 \end{pmatrix}$.

    • Vector equation:             $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} + t \begin{pmatrix} -2 \\ 6 \\ -4 \end{pmatrix}$.

    • Parametric equations:             $x = 2 - 2t$             $y = -1 + 6t$             $z = 3 - 4t$

• Equation of a Plane in $\mathbb{R}^3$:

  • The general Cartesian equation of a plane is:         $a x + b y + c z = k$         where the vector $n = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ represents the normal vector perpendicular to the plane.

  • Derivation: If $P_0(x_0, y_0, z_0)$ is a known fixed point on the plane and $P(x, y, z)$ is a general point on the plane, the vector $P - P_0$ lies within the plane and must be perpendicular to the normal vector $n$:         $\begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} x - x_0 \\ y - y_0 \\ z - z_0 \end{pmatrix} = 0$         $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$         $a x + b y + c z = a x_0 + b y_0 + c z_0 = k$

  • Special Geometric Orientations:

    • When $k = 0$, the equation $a x + b y + c z = 0$ describes a plane passing through the origin $(0,0,0)$.

    • When $a = b = 0$, the plane $c z = k$ is parallel to the $x-y$ coordinate plane.

    • When $b = c = 0$, the plane $a x = k$ is parallel to the $y-z$ coordinate plane.

    • When $a = c = 0$, the plane $b y = k$ is parallel to the $x-z$ coordinate plane.

• Intersection of a Line and a Plane:

  • Find the point of intersection between the plane $2x - 3y + 5z = 15$ and the line joining $(0, -1, 1)$ and $(3, -1, 0)$.

    • Direction vector of the line: $\begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ -1 \end{pmatrix}$.

    • Parametric equations of the line:             $x = 0 + 3t = 3t$             $y = -1 + 0t = -1$             $z = 1 - t$

    • Substitute the parametric coordinates into the plane equation:             $2(3t) - 3(-1) + 5(1 - t) = 15$             $6t + 3 + 5 - 5t = 15 \implies t + 8 = 15 \implies t = 7$.

    • Compute the intersection coordinates:             $x = 3(7) = 21$             $y = -1$             $z = 1 - 7 = -6$

    • The line intersects the plane at the point $(21, -1, -6)$.

• Acute Angle Between Two Planes:

  • The angle between two planes equals the angle between their respective normal vectors.

  • Find the acute angle between the planes $x - 2y + 5z = 3$ and $2x + y - 3z = 15$.

    • Normal vectors: $n_1 = \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix}$ and $n_2 = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}$.

    • Inner product: $n_1 \cdot n_2 = 1(2) + (-2)(1) + 5(-3) = 2 - 2 - 15 = -15$.

    • Magnitudes:             $\|n_1\| = \sqrt{1^2 + (-2)^2 + 5^2} = \sqrt{1 + 4 + 25} = \sqrt{30}$.             $\|n_2\| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$.

    • Cosine of the acute angle $\theta$:             $\cos(\theta) = \frac{|n_1 \cdot n_2|}{\|n_1\| \|n_2\|} = \frac{|-15|}{\sqrt{30} \sqrt{14}} = \frac{15}{\sqrt{420}} = \frac{15}{2\sqrt{105}}$.

• Distance Formulas for Planes:

  • Distance of a Plane from the Origin:

    • Let $A(\alpha, \beta, \gamma)$ be the nearest point on the plane $a x + b y + c z = k$ to the origin.

    • $A$ lies along the normal line through the origin, so $\begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix} = t \begin{pmatrix} a \\ b \\ c \end{pmatrix}$.

    • Substitute into the plane equation:             $a(t a) + b(t b) + c(t c) = k \implies t(a^2 + b^2 + c^2) = k \implies t = \frac{k}{a^2 + b^2 + c^2}$.

    • The vector from the origin to $A$ is:             $O A = \frac{k}{a^2 + b^2 + c^2} \begin{pmatrix} a \\ b \\ c \end{pmatrix}$.

    • The shortest distance $d = \|O A\|$ is given by:             $d = \frac{|k|}{\sqrt{a^2 + b^2 + c^2}}$

  • Distance of a Plane from an Arbitrary Point:

    • Let $Q(x_1, y_1, z_1)$ be a given point and $A(x, y, z)$ be the nearest point on the plane $a x + b y + c z = k$.

    • Translate the origin to $Q$ by applying the coordinate transformation:             $X = x - x_1$             $Y = y - y_1$             $Z = z - z_1$

    • The plane equation becomes:             $a(X + x_1) + b(Y + y_1) + c(Z + z_1) = k$             $a X + b Y + c Z = k - a x_1 - b y_1 - c z_1$

    • The shortest distance from $Q$ to the plane is given by the origin distance formula in the translated coordinate system:             $d = \frac{|k - a x_1 - b y_1 - c z_1|}{\sqrt{a^2 + b^2 + c^2}}$

Vector Subspaces and Linear Combinations

• Definition of a Vector Subspace:

  • Given a vector space $V$, a non-empty subset $W \subseteq V$ is called a vector subspace of $V$ if $W$ is itself a vector space with respect to the vector addition and scalar multiplication defined on $V$.

• The Two-Step Subspace Criterion Theorem:

  • Theorem: A non-empty subset $W$ of a vector space $V$ is a subspace of $V$ if and only if:

    • (i) $a + b \in W$ for all $a, b \in W$ (closure under vector addition).

    • (ii) $k a \in W$ for all $a \in W$ and $k \in \mathbb{R}$ (closure under scalar multiplication).

  • Proof:

    • Necessity: If $W$ is a subspace, it is a vector space in its own right, so properties (i) and (ii) hold by definition.

    • Sufficiency: Assume (i) and (ii) hold. Since $W \ne \emptyset$, choose any $a \in W$.

    • By (ii), setting $k = 0$ gives $0 a = 0 \in W$, guaranteeing that the zero vector is in $W$.

    • By (ii), setting $k = -1$ gives $(-1)a = -a \in W$, guaranteeing additive inverses.

    • The remaining six vector space axioms (associativity, commutativity, distributivity, scalar identity) hold automatically in $W$ because they hold for all elements in the parent space $V$.

    • Hence, $W$ is a vector space, and thus a subspace of $V$.

• Examples and Non-Examples of Subspaces:

  • Trivial Subspaces: For any vector space $V$, the singleton zero set $W = \{0\}$ is a subspace (the trivial subspace), and $W = V$ is a subspace (the improper subspace).

  • Example in $\mathbb{R}^2$ ($a + 2b = 0$):

    • Let $W = \left\{ \begin{pmatrix} a \\ b \end{pmatrix} \in \mathbb{R}^2 \;\middle|\; a + 2b = 0 \right\}$.

    • Closure under addition: Let $u = \begin{pmatrix} a \\ b \end{pmatrix} \in W$ and $v = \begin{pmatrix} x \\ y \end{pmatrix} \in W$, so $a + 2b = 0$ and $x + 2y = 0$.             $u + v = \begin{pmatrix} a + x \\ b + y \end{pmatrix}$.             $(a + x) + 2(b + y) = (a + x) + 2b + 2y = (a + 2b) + (x + 2y) = 0 + 0 = 0$.             Therefore, $u + v \in W$.

    • Closure under scalar multiplication: For any scalar $k \in \mathbb{R}$,             $k u = \begin{pmatrix} k a \\ k b \end{pmatrix}$.             $(k a) + 2(k b) = k(a + 2b) = k(0) = 0$.             Therefore, $k u \in W$.

    • Conclusion: $W$ is a subspace of $\mathbb{R}^2$.

  • Non-Subspace Example 1 ($x + y = 1$):

    • Let $W = \left\{ \begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^2 \;\middle|\; x + y = 1 \right\}$.

    • The zero vector is $0 = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.

    • Testing the condition: $0 + 0 = 0 \ne 1$.

• Since $0 \notin W$, $W$ cannot be a vector subspace of $\mathbb{R}^2$.

• Non-Subspace Example 2 ($y = x^2$):

  • Let $W = \left\{ \begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^2 \;\middle|\; y = x^2 \right\}$.

  • The vector $\begin{pmatrix} 1 \\ 1 \end{pmatrix} \in W$ since $1 = 1^2$.

  • Adding this vector to itself:         $\begin{pmatrix} 1 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}$.

  • Testing membership: $x = 2, y = 2$, but $x^2 = 2^2 = 4 \ne 2$.

• Since $\begin{pmatrix} 2 \\ 2 \end{pmatrix} \notin W$, $W$ is not closed under addition and is not a subspace.

• Linear Combinations:

  • Definition: A vector $x$ is a linear combination of vectors $x_1, x_2, \dots, x_n \in V$ if there exist scalars $c_1, c_2, \dots, c_n \in \mathbb{R}$ such that:         $x = c_1 x_1 + c_2 x_2 + \dots + c_n x_n$

  • Numerical Verification: Let $x_1 = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $x_2 = \begin{pmatrix} 1 \\ -5 \end{pmatrix}$.         $x = 3 x_1 + 5 x_2 = 3\begin{pmatrix} 2 \\ -1 \end{pmatrix} + 5\begin{pmatrix} 1 \\ -5 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \end{pmatrix} + \begin{pmatrix} 5 \\ -25 \end{pmatrix} = \begin{pmatrix} 11 \\ -28 \end{pmatrix}$.         Thus, $\begin{pmatrix} 11 \\ -28 \end{pmatrix}$ is a linear combination of $x_1$ and $x_2$.

• Finding Scalars Using Orthogonality:

  • If the vectors $x_1, x_2, \dots, x_n$ are mutually orthogonal ($x_i \cdot x_j = 0$ for $i \ne j$), the expansion scalars can be isolated directly by taking the dot product with each basis vector:         $x_1 \cdot x = x_1 \cdot (c_1 x_1 + c_2 x_2 + \dots + c_n x_n) = c_1(x_1 \cdot x_1) + c_2(x_1 \cdot x_2) + \dots + c_n(x_1 \cdot x_n)$.         $x_1 \cdot x = c_1 \|x_1\|^2 + 0 + \dots + 0 = c_1 \|x_1\|^2$.

• For $x_i \ne 0$, the scalar $c_i$ is computed via:         $c_i = \frac{x_i \cdot x}{\|x_i\|^2}, \quad i = 1, 2, \dots, n$

  • Worked Numerical Example 1: Write $x = \begin{pmatrix} 7 \\ -5 \end{pmatrix}$ as a linear combination of $x_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $x_2 = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$.

    • Verify orthogonality:             $x_1 \cdot x_2 = 1(2) + 2(-1) = 2 - 2 = 0$.

    • Compute $c_1$:             $c_1 = \frac{x_1 \cdot x}{\|x_1\|^2} = \frac{1(7) + 2(-5)}{1^2 + 2^2} = \frac{7 - 10}{5} = -\frac{3}{5}$.

    • Compute $c_2$:             $c_2 = \frac{x_2 \cdot x}{\|x_2\|^2} = \frac{2(7) + (-1)(-5)}{2^2 + (-1)^2} = \frac{14 + 5}{5} = \frac{19}{5}$.

    • Linear combination:             $\begin{pmatrix} 7 \\ -5 \end{pmatrix} = -\frac{3}{5}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + \frac{19}{5}\begin{pmatrix} 2 \\ -1 \end{pmatrix}$.

  • Worked Non-Example 2: Show that $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ is not a linear combination of $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}$.

    • Set up the vector equation:             $c_1\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + c_2\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

    • Equating components:             $\begin{pmatrix} c_1 \\ c_1 + 2c_2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

    • The third row yields $0 = 1$, which is impossible.

    • Thus, $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ cannot be expressed as a linear combination of the two vectors.

  • Worked Numerical Example 3: Express $x = \begin{pmatrix} 4 \\ -2 \\ 3 \end{pmatrix}$ as a linear combination of $x_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$, $x_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$, and $x_3 = \begin{pmatrix} 2 \\ -2 \\ 0 \end{pmatrix}$.

    • Verify mutual orthogonality:             $x_1 \cdot x_2 = 0(1) + 0(1) + 1(0) = 0$.             $x_1 \cdot x_3 = 0(2) + 0(-2) + 1(0) = 0$.             $x_2 \cdot x_3 = 1(2) + 1(-2) + 0(0) = 2 - 2 + 0 = 0$.

    • Compute coefficients:             $c_1 = \frac{x_1 \cdot x}{\|x_1\|^2} = \frac{0(4) + 0(-2) + 1(3)}{0^2 + 0^2 + 1^2} = \frac{3}{1} = 3$.             $c_2 = \frac{x_2 \cdot x}{\|x_2\|^2} = \frac{1(4) + 1(-2) + 0(3)}{1^2 + 1^2 + 0^2} = \frac{4 - 2}{2} = \frac{2}{2} = 1$.             $c_3 = \frac{x_3 \cdot x}{\|x_3\|^2} = \frac{2(4) + (-2)(-2) + 0(3)}{2^2 + (-2)^2 + 0^2} = \frac{8 + 4}{4 + 4} = \frac{12}{8} = \frac{3}{2}$.

    • Linear combination:             $\begin{pmatrix} 4 \\ -2 \\ 3 \end{pmatrix} = 3\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} + 1\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \frac{3}{2}\begin{pmatrix} 2 \\ -2 \\ 0 \end{pmatrix}$.

• The Linear Span:

  • Definition: The set of all linear combinations of a finite set of vectors $\{x_1, x_2, \dots, x_n\}$ in a vector space $V$ is called the linear span (or subspace generated by the set), denoted:         $W = \operatorname{span}\{x_1, x_2, \dots, x_n\}$

  • Theorem: The linear span of any set of vectors $\{x_1, x_2, \dots, x_n\}$ is a vector subspace of $V$.

    • Proof: Let $u, v \in \operatorname{span}\{x_1, \dots, x_n\}$.

    • Then $u = c_1 x_1 + \dots + c_n x_n$ and $v = c'_1 x_1 + \dots + c'_n x_n$.

    • Addition:             $u + v = (c_1 x_1 + \dots + c_n x_n) + (c'_1 x_1 + \dots + c'_n x_n) = (c_1 + c'_1)x_1 + \dots + (c_n + c'_n)x_n$.             Since $(c_i + c'_i) \in \mathbb{R}$, $u + v \in \operatorname{span}\{x_1, \dots, x_n\}$.

    • Scalar Multiplication:             $k u = k(c_1 x_1 + \dots + c_n x_n) = (k c_1)x_1 + \dots + (k c_n)x_n$.             Since $(k c_i) \in \mathbb{R}$, $k u \in \operatorname{span}\{x_1, \dots, x_n\}$.

    • Both closure conditions are verified, so the span is a vector subspace.

  • Remarks:

    • $\operatorname{span}\{0\} = \{0\}$.

    • $\operatorname{span}(V) = V$.