Lecture 05

[[lecture-data]]

2024-09-06

Readings

• a

1. Eigenvalues and Similarity

Last time, we talked about diagonalization $A=SD{S}^{-1}$ having the same eigenvalues as $D$ which are easy to find. Not only are the diagonal entries of $D$ the eigenvalues, but the columns of $S$ are the eigenvectors!

In particular, the change of basis from $A$ to $D$ is through eigenvectors. In the perspective of the eigenvectors $S$, $A$ is just a diagonal operator (it breaks individual coordinates into piece and scales along those dimensions).

We can call diagonalization an eigenvalue-eigenvector decomposition.

Suppose we have a system of differential equations like $y^{\prime }=Ay$ which might be tricky to solve if each derivative is in terms of the other functions. but if $A$ is diagonalizable, it may be easier to solve this system by changing to $z^{\prime }=Dz$ which is much easier to deal with.

$S_i$

Let $A\in M_n(\mathbb{C})$ have eigenvalues ${\lambda }_1,{\lambda }_2,..,{\lambda }_n$. For $i=1,\dots ,n$, define $S_i:={\sum }_{\text{multisubsets }U\text{ of }i\text{ eigenvalues}}{\prod }_{\lambda \in U}\lambda$

$S_2={\lambda }_1{\lambda }_2+\cdots +{\lambda }_1{\lambda }_n+\cdots +{\lambda }_{n-1}{\lambda }_n$ $S_3={\lambda }_1{\lambda }_2{\lambda }_3+{\lambda }_1{\lambda }_2{\lambda }_4+\cdots +{\lambda }_{n-2}{\lambda }_{n-1}{\lambda }_n$ Take all $\left(\genfrac{}{}{0ex}{}{n}{i}\right)$ ways to multiply the eigenvalues and then add them.

$E_i$ and Principal Submatrix

$E_i={\sum }_{\text{all }\left(\genfrac{}{}{0ex}{}{n}{i}\right)i\times i\text{ principle submatrices}M}\det M$

Submatrices: destroy some columns and rows. We call it a principle submatrix if we get rid of the same rows as columns (we delete each $i$th row and $i$th column).

$E_1=a_{11}+\cdots +a_{nn}=\mathrm{Tr}(A)$

Proposition

If $A\in M_n$ , then

P_{A} &= \lambda^n -S_{1}\lambda^{n-1}+S_{2}\lambda^{n-2}-\dots\pm S_{n}\lambda^0 \ &=\lambda^n -E_{1}\lambda^{n-1}+E_{2}\lambda^{n-2}-\dots\pm E_{n}\lambda^0 \end{aligned}$$

It’s easy to see why the first equality holds. If we factor the polynomial into $P_A=(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)\dots (\lambda -{\lambda }_n)$ and expand out, we get exactly $S_i$ as the coefficients. (When select $i$ of the terms to get a ${\lambda }^{n-i}$ term we get exactly $S_i$)

Exercise

Prove the second equality via induction with the Laplace expansion

Consequences: $\det A={\prod }_{\lambda \in \sigma (A)}\lambda$ For a diagonalizable matrix, this is easy to see. But this is useful to know for matrices that are not diagonalizable. Another perspective: when there is 0 eigenvalue, the determinant collapses to zero and the matrix is singular. When there is no 0 eigenvalue, then the matrix is invertible which we saw last time. (see determinant)

$\mathrm{Tr}A={\sum }_{i=1}^n{\lambda }_i$ This implies that the trace of similar matrices are the same! So this means that the trace is a property of the transformation - not necessarily obvious. (see trace)

Coming up:

• Seeing a relationship between $AB$ and $BA$ spectrum. They have the same nonzero eigenvalues! (for rectangular matrices $A$ and $B$)
• Commutativity of $A$ and $B$ if and only if simultaneously diagonalizable. To get there: some background

Multiplying partitioned matrices Suppose $A$ is partitioned, not necessarily regularly. Suppose $B$ is partitioned conformally (in the same way for the rows as the columns of $A$)

The $A_{ij}$th block is $m_i\times n_j$, a submatrix $B_{ij}$ is $n_i\times p_j$

$AB=C$ where $C_{ij}$ is $m_i\times p_j$ and $C_{ij}={\sum }_{k=1}^s{A}_{ik}{B}_{kj}$ This is suspiciously like multiplying regular matrices with single entries! And each matrix in the multiplication is of size $m_i\times p_j$. So why does this work?

Say I want to take a single row and a single column and compute their inner product like doing it for a “normal” matrix with single entries. We are doing the same thing but chunk by chunk.

Permutation Matrices

Permutation Matrix

A permutation matrix is a square matrix $P$ such that there is one $1$ in every row, one $1$ in every column, and the rest are zeroes.

(this reorders/rearranges the rows of $A$ if we multiply $PA$ and reorders the columns of $B$ if we multiply $B{P}^T$ according to the pattern of the rows or columns of $P$ respectively)

Note that $P{P}^T=I⟹P^T=P^{-1}$ and also $PD{P}^T$ reorders diagonals.

$P$ is also a similarity transformation.

(see permutation matrix)

Suppose $C$ is a matrix and $C$ is block traingular. (Blocks along diagonal are triagular in the same way) Then $\sigma (C)={\bigcup }_{i=1}^k\sigma (C_{ii})$

$\det (\lambda I-C)=\prod \det (I\lambda -C_{ii})$ which gives us the same characteristic polynomial.

Next time: $A$ and $B^T$ have the same dimensions. $AB$ and $BA$ have the same nonzero eigenvalues.