Lecture 32: Quiz 3 Review

Scope of Quiz 3: Chapter 6, eigenvalues. Chapter 7 (linear transformations) is saved for the final.

Topics Strang lists:

Finding eigenvalues and eigenvectors, including shortcuts that avoid $\det(A - \lambda I)$.
Differential equations $du/dt = Au$ and the matrix exponential $e^{At}$.
Symmetric matrices: real $\lambda$, orthogonal eigenvectors, $A = Q\Lambda Q^\top$.
Positive definite matrices.
Similar matrices, $B = M^{-1}AM$. Same eigenvalues, and $B^k = M^{-1}A^kM$, so powers behave alike.
SVD.

Problem 1: A differential equation with an antisymmetric matrix

\frac{du}{dt} = Au, \qquad A = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix}

General solution

u(t) = c_1e^{\lambda_1t}x_1 + c_2e^{\lambda_2t}x_2 + c_3e^{\lambda_3t}x_3

The three constants are fixed by the initial vector $u(0)$.

Eigenvalues without much computing

One eigenvalue is 0. Row 1 and row 3 are negatives of each other, so $A$ is singular.

The others are imaginary. $A^\top = -A$, so $A$ is antisymmetric (skew-symmetric). Antisymmetric matrices have purely imaginary eigenvalues.

Confirm with the determinant:

\det(A - \lambda I) = -\lambda^3 - 2\lambda = -\lambda(\lambda^2 + 2)

So $\lambda = 0,\ \pm i\sqrt2$.

Behavior of the solution

Neither decay nor blow-up.

$e^{0t} = 1$ stays constant.
$e^{\pm i\sqrt2\,t}$ has magnitude 1 and just circles the unit circle.

Periodic. The solution returns to its starting value when $\sqrt2\,T = 2\pi$:

T = \pi\sqrt2

Which matrices have orthogonal eigenvectors?

The eigenvectors here are orthogonal. This holds for every antisymmetric matrix, and not only for symmetric ones.

Exact condition (stated without proof):

\boxed{AA^\top = A^\top A}

Such matrices are called normal. For complex eigenvectors, "orthogonal" uses the complex inner product $x^Hy$.

Family	Why it satisfies $AA^\top = A^\top A$
Symmetric, $A^\top = A$	both sides equal $A^2$
Antisymmetric, $A^\top = -A$	both sides equal $-A^2$
Orthogonal, $Q^\top Q = I$	both sides equal $I$

Computing $e^{At}$

If $A$ is diagonalizable,

e^{At} = S\,e^{\Lambda t}\,S^{-1}, \qquad e^{\Lambda t} = \text{diag}(e^{\lambda_1t}, \dots, e^{\lambda_nt})

Then $u(t) = e^{At}u(0)$.

Problem 2: What can a matrix be, knowing only its eigen-data?

Given: a $3\times3$ matrix with eigenvalues $0,\ c,\ 2$, where $c$ is unknown, and three given eigenvectors. Those eigenvectors turn out to be mutually orthogonal.

Format. Each answer must give all values of $c$ that work.

Question	Answer	Reason
Diagonalizable?	All $c$	Three independent eigenvectors are given. Repeated eigenvalues don't matter when the eigenvectors are there.
Symmetric?	All real $c$	Orthogonal eigenvectors plus real eigenvalues give a symmetric matrix. Complex $c$ would not.
Positive definite?	Never	One eigenvalue is 0.
Positive semidefinite?	$c \geq 0$	The "consolation prize".
Markov matrix?	Never	A Markov matrix has no eigenvalue larger than 1, and this one has $\lambda = 2$.
Is $\tfrac12A$ a projection?	$c = 0$ or $c = 2$	A projection satisfies $P^2 = P$, so $\lambda^2 = \lambda$ and every eigenvalue is 0 or 1. Halving $0, c, 2$ must give only 0s and 1s.

The lesson. The orthogonality of the eigenvectors did most of the work. Without it, symmetric, positive definite, and projection would all have been ruled out immediately.

Problem 3: The SVD, including the sign correction from Lecture 29

A = U\Sigma V^\top

This holds for every matrix, rectangular included.

Computing it

Step 1: find $V$ and the $\sigma$'s from $A^\top A$.

A^\top A = V\,\Sigma^\top\Sigma\,V^\top

This is $Q\Lambda Q^\top$ for the symmetric matrix $A^\top A$.
$V$ = eigenvectors of $A^\top A$.
$\sigma_i = \sqrt{\lambda_i(A^\top A)}$, taking the positive square root.

Step 2: find $U$ from $V$, not separately.

$AA^\top = U\,\Sigma\Sigma^\top\,U^\top$ shows that $U$ holds the eigenvectors of $AA^\top$. But eigenvectors are determined only up to sign. That is exactly what went wrong in Lecture 29: Strang picked a sign at random and was wrong on one of them.

The correct rule. Once the $v_i$ are chosen, compute

u_i = \frac{Av_i}{\sigma_i}

This forces the right signs, so that $AV = U\Sigma$. It is the same fix I gave in the Lecture 29 summary.

Reading the SVD

$\Sigma = \text{diag}(3, 2)$. Both singular values are nonzero, so $A$ is invertible.

$\Sigma = \text{diag}(3, -5)$. This is not an SVD. Singular values are never negative.

$\Sigma = \text{diag}(3, 0)$. Then:

$A$ is singular, with rank 1.
$\dim N(A) = 1$, and $v_2$ spans the nullspace, since $Av_2 = 0 \cdot u_2 = 0$.
By the same reasoning (my addition): $u_1$ spans $C(A)$, $u_2$ spans $N(A^\top)$, and $v_1$ spans $C(A^\top)$.

Strang's point. The SVD takes more work to compute, but it displays everything: rank, all four subspaces, and orthonormal bases for each.

Problem 4: A matrix that is both symmetric and orthogonal

Possible eigenvalues

Symmetric means $\lambda$ is real.

Orthogonal means $|\lambda| = 1$, because $Q$ preserves length:

\|Qx\| = \|x\| \;\Rightarrow\; |\lambda|\,\|x\| = \|x\| \;\Rightarrow\; |\lambda| = 1

Together: $\lambda = \pm1$.

True or false

Statement	Answer	Reason
Sure to be positive definite	False	It can have $\lambda = -1$.
Has no repeated eigenvalues	False	For $n \geq 3$, the only two possible values must repeat.
Diagonalizable	True	Symmetric and orthogonal matrices are always diagonalizable, with orthogonal eigenvectors.
Nonsingular	True	Orthogonal matrices are always invertible, and no eigenvalue is 0.
$\tfrac12(A + I)$ is a projection	True	See below.

Proof that $P = \tfrac12(A + I)$ is a projection

Key fact. $A = A^\top = A^{-1}$, so $A^2 = I$.

Method 1: check the two properties directly.

$P$ is symmetric, because $A$ is.
Squaring: $P^2 = \tfrac14(A^2 + 2A + I) = \tfrac14(2I + 2A) = \tfrac12(A + I) = P$

Method 2: use eigenvalues.

$A$ has eigenvalues $\pm1$.
So $A + I$ has eigenvalues $2$ and $0$.
So $\tfrac12(A + I)$ has eigenvalues $1$ and $0$.
A symmetric matrix whose eigenvalues are all 0 or 1 is a projection.

Review checklist for Chapter 6

Spot eigenvalues fast.
- Singular means $\lambda = 0$.
- Trace and determinant give the sum and product.
- Triangular matrices show them on the diagonal.
Structure determines the spectrum:

Matrix type	Eigenvalues
Symmetric	real
Antisymmetric	imaginary
Orthogonal	$\lvert\lambda\rvert = 1$
Projection	0 or 1
Markov	$\lambda = 1$ appears, the rest $\lvert\lambda\rvert \leq 1$
Positive definite	$\lambda > 0$

Orthogonal eigenvectors exactly when $AA^\top = A^\top A$ (normal). This covers symmetric, antisymmetric, and orthogonal matrices.
$du/dt = Au$: the solution is $u = e^{At}u(0)$ with $e^{At} = Se^{\Lambda t}S^{-1}$. $\text{Re}(\lambda)$ controls growth; imaginary $\lambda$ means oscillation.
SVD: get $V$ and $\sigma$ from $A^\top A$, then $u_i = Av_i/\sigma_i$.
Similar matrices have the same eigenvalues, and their powers satisfy $B^k = M^{-1}A^kM$.

Family	Why it satisfies \(AA^\top = A^\top A\)
Symmetric, \(A^\top = A\)	both sides equal \(A^2\)
Antisymmetric, \(A^\top = -A\)	both sides equal \(-A^2\)
Orthogonal, \(Q^\top Q = I\)	both sides equal \(I\)