Determinants, Eigenvalues and Eigenvectors

What is the Laplace Expansion Theorem?

For any $n\times n$ matrix M and any natural numbers $1\le i\le n$ and $1\le j\le n$ the following hold:

• $$detM=\sum_{k=1}^{n}\left(-1\right)^{i+k}M_{ik}\det\overline{M_{ik}}$$

• $$detM=\sum_{k=1}^{n}\left(-1\right)^{k+j}M_{kj}\det\overline{M_{kj}}$$

Remark: The value $\det\overline{M_{ij}}$ is called the $(i,j)$ minor of M, while the value $(-1)^{i+j}det\overline{M_{ij}}$ is the $(i,j)$ cofactor of $M$.

If M is a $n\times n$ upper triangular matrix, what is $detM$?

$$detM=\prod_{i=1}^{n}M_{ii}$$

Let $A$ and $B$ be square matrices, what is the determinant of B equal to if A is transformed into B using each of the elementary row operations?

$A\underrightarrow{r_{i}\to\alpha r_{i}}B$ then $detB=\alpha detA$

$A\underrightarrow{r_{i}\to r_{i}+\lambda r_{j}}B$ then $detB=detA$

$A\underrightarrow{r_{i}\leftrightarrow r_{j}}B$ then $detB=-detA$

Proof that a matrix A is only invertible if and only if $detA\ne 0$.

Let R=RREF(A). No elementary row operations changes whether the determinant is 0.

Since R is obtained from A via a sequence of elementary row operations, $detA\ne 0 \iff detR\ne 0$. However, R is $n\times n$ and is in RREF, so it follows either $R=I_n$ or $R$ contains a row of zeros. It follows that $R=I_n$ iff $detR\ne 0$.

Hence $A$ is invertible $\iff R=I_n$

$$\iff detR\ne 0$$

$\iff detA \ne 0$.

What rules are the determinants of elementary matrices given by?

$$detE_{r_{i}\to\alpha r_{i}}=\alpha$$

$$detE_{r_i→r_i+\lambda r_j}=1$$

$$detE_{r_i\leftrightarrow r_j}=-1$$

What’s the adjugate of A, where A is a square matrix, and what are its properties?

let B be the matrix given by $B_{ij}=(-1)^{i+j}det\overline{A_{ji}}$

B is the adjugated of A.

$$AB=detAI_{n}=BA$$

in particular if $detA\ne 0$ then A is invertible and$A^{-1}=\frac{1}{detA} B$

What does $Det(AB)=?$ and the proof of this?

Split into two cases.

If $A$ is singular, then $AB$ is singular, because if $AB$ was invertible, $A$ would be invertible, so $\det(A)=0$ and $\det(AB)=0$, hence $\det(AB)=\det(A)\det(B)$.

If $A$ is invertible, write $A$ as a product of elementary matrices: $A=E_k\cdots E_1$. For each elementary matrix $E$, we know $\det(EB)=\det(E)\det(B)$.

Applying this repeatedly gives $\det(AB)=\det(E_k)\cdots\det(E_1)\det(B)$. But $\det(A)=\det(E_k)\cdots\det(E_1)$, so $\det(AB)=\det(A)\det(B)$.

How can we work out the determinant if the matrix is in LU decomposition.

If $PA=LU$, where $P$ is a permutation matrix, $L$ has diagonal entries $1$, and $U$ is upper triangular, then $\det(A)=\pm\prod_{i=1}^{n}U_{ii}$.

The sign depends on whether the number of row swaps in $P$ is even or odd (+ if even, - if odd).

What’s the definition of an eigenvector with eigenvalue $\lambda?$

Let $V$ be a v.s over $K$ and $T:V→V$ be a linear transformation.

A vector $v\in V$ is an eigenvector of $T$ with eigen value $\lambda \in K$ if $v\ne 0$ and $T(v)=\lambda v$.

If $A\in M_n(K)$, then eigenvalues and eigenvectors of $A$ are the eigenvalues and eigenvectors of the linear map $T_A:V→V$, $Av=\lambda v$.

Remark: The zero vector is never an eigenvector.

How do we find eigenvalues and eigenvectors?

To find eigenvalues:

• evaluate $det(A-\lambda I_n)=0$.

To find eigenvectors:

• $v$ is a eigenvector if it is a non-trivial solution to the system $(A-\lambda I_n)x=0$.

What is an eigenspace?

The $\lambda$-eigenspace of linear map $T:V→V$ is the subspace $E_{\lambda}=\{v\in V|T(v)=\lambda v\}$

or with matrices

the $\lambda$-eigenspace of A is the $\lambda$-eigenspace of the linear map $T_A(v)=AV$.

In other words the collection of all eigenvectors together with the zero vector.

What is the geometric multiplicity of an eigenvalue?

The dimension of the corresponding $\lambda$-eigenspace $E_\lambda$.

Why are eigenvectors corresponding to distinct eigenvalues linearly independent?

Assume for contradiction that the eigenvectors $v_1,\ldots,v_k$ are linearly dependent. Choose the first dependent one, so $\{v_1,\ldots,v_{j-1}\}$ is linearly independent but $v_j=\alpha_1v_1+\cdots+\alpha_{j-1}v_{j-1}$, with not all $\alpha_i=0$.

Applying $T$ gives $\lambda_jv_j=\alpha_1\lambda_1v_1+\cdots+\alpha_{j-1}\lambda_{j-1}v_{j-1}$.

But multiplying the original equation by $\lambda_j$ gives $\lambda_jv_j=\alpha_1\lambda_jv_1+\cdots+\alpha_{j-1}\lambda_jv_{j-1}$. Subtracting gives $0=\sum_{i=1}^{j-1}(\lambda_i-\lambda_j)\alpha_iv_i$. Since $v_1,\ldots,v_{j-1}$ are linearly independent, each $(\lambda_i-\lambda_j)\alpha_i=0$. The eigenvalues are distinct, so $\lambda_i-\lambda_j\neq0$, hence all $\alpha_i=0$, a contradiction. Therefore $v_1,\ldots,v_k$ are linearly independent.

What does it mean for a field to be algebraically closed?

If every polynomial $p(x)=a_0+a_1x+…+a_nx^n$ where $a_i\in K$ and $a_{i}\ne0$ can be factorised in the form $p(x)=b(x-c_1)…(x-c_n)$ for some $b,c_1,…,c_n\in K$.

What is the algebraic multiplicity of an eigenvalue $\lambda$?

How many times the eigenvalue appears as an root of the characteristic polynomial.

What is the trace of A?

$$tr(a)=A_{11}+\ldots+A_{nn}=\sum_{i=1}^{n}A_{ii}$$

If K is an algebraically closed field and $\lambda_1,..,\lambda_n$ are the eigenvalues of $A\in M_n(K)$, counted with multiplicity, what is the trace and determinant?

$$tr(A)=\lambda_1+…+\lambda_n$$

$$det(A)=\lambda_1…\lambda_n$$