Lecture 05

[[lecture-data]]

2024-09-06

Readings

1. Eigenvalues and Similarity

Last time, we talked about diagonalization $A = S D S^{- 1}$ 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 $y^{'} = A y$ 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 $z^{'} = D z$ which is much easier to deal with.

S_{i}

Let $A \in M_{n} (C)$ have eigenvalues $λ_{1}, λ_{2}, . ., λ_{n}$ . For $i = 1, \dots, n$ , define

S_{i} := \sum_{multisubsets U of i eigenvalues} \prod_{λ \in U} λ

Example

$S_{2} = λ_{1} λ_{2} + \dots + λ_{1} λ_{n} + \dots + λ_{n - 1} λ_{n}$
$S_{3} = λ_{1} λ_{2} λ_{3} + λ_{1} λ_{2} λ_{4} + \dots + λ_{n - 2} λ_{n - 1} λ_{n}$
Take all $(\binom{n}{i})$ ways to multiply the eigenvalues and then add them.

E_{i}

and Principal Submatrix

$E_{i} = \sum_{all (\binom{n}{i}) i \times i principle submatrices M} det M$

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).

$E_{1} = a_{11} + \dots + a_{n n} = Tr (A)$

Proposition

If $A \in M_{n}$ , then

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

It's easy to see why the first equality holds. If we factor the polynomial into $P_{A} = (λ - λ_{1}) (λ - λ_{2}) \dots (λ - λ_{n})$ and expand out, we get exactly $S_{i}$ as the coefficients. (When select $i$ of the terms to get a $λ^{n - i}$ term we get exactly $S_{i}$ )

Exercise

Prove the second equality via induction with the Laplace expansion

Consequences:
$det A = \prod_{λ \in σ (A)} λ$
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)

$Tr A = \sum_{i = 1}^{n} λ_{i}$
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 $A B$ and $B A$ 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 $A_{i j}$ th block is $m_{i} \times n_{j}$ , a submatrix
$B_{i j}$ is $n_{i} \times p_{j}$

$A B = C$ where $C_{i j}$ is $m_{i} \times p_{j}$ and $C_{i j} = \sum_{k = 1}^{s} A_{i k} B_{k j}$
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 $P A$ 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 $P D P^{T}$ 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 $σ (C) = ⋃_{i = 1}^{k} σ (C_{i i})$

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

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