Given a square matrix A, the characteristic polynomial of A is given by
p(λ)=detThe zeros of the characteristic polynomial are called the eigenvalues of A. Given an eigenvalue \lambda, a non-zero vector v such that
Av = \lambda vis called an eigenvector for the eigenvalue \lambda.
Note: If v is an eigenvector, the \alpha v is also for any \alpha \in \mathbb R. Given any norm \|\cdot\|, you can always choose v so that \|v\| = 1.
A = [1,2;2,-1];
eigs(A)
[vs,lambdas] = eigs(A)
Find the eigenvalues of
A = \begin{bmatrix} 2 & 0 & 0 \\ 1 & -1 & 2 \\ -1 & 1 & -1 \end{bmatrix}.The spectral radius of a square matrix A is given by
\rho(A) = \max_i \left|\lambda_i\right|where the maximum is taken overall eigenvalues.
Recall that if \lambda = \alpha + i \beta is a complex number then |\lambda| = \sqrt{\alpha^2 + \beta^2}.
If A is an n \times n symmetric matrix A = A^T then all the eigenvalues of A are real and there exists a matrix U with U^T = U^{-1} such that
A = U \begin{bmatrix} \lambda_1 \\ & \lambda_2 \\ && \ddots \\ &&& \lambda_n \end{bmatrix} U^T.Recall that in the previous lecture we came up with explicit formulae for \|A\|_1 and \|A\|_\infty but we mentioned nothing about \|A\|_2. One needs to use eigenvalues to characterize it:
If A is an n\times n matrix, then
Because A^TA is a symmetric matrix, A^T A = U \Lambda U^T for a diagonal matrix \Lambda that contains the eigenvalues of A^TA.
For such a matrix U, note that
\|x\|_2^2 = x^T x = x U^T U x = \|Ux\|_2^2\|x\|_2^2 = x^T x = x U U^T x = \|U^Tx\|_2^2.So \|x\|_2 = 1 if and only if \|Ux\|_2= 1. Let v be an eigenvector corresponding to an eigenvalue \lambda:
A^TA v = \lambda v.
We note that for any vector y, \|y\|_2^2 = y^T y. We apply x^T to the above equation:
v^TA^T A v = \lambda v^T v \Leftrightarrow \|Av\|_2^2 = \lambda \|x\|^2.This implies that \lambda \geq 0. So \rho(A^TA) = \max_i \lambda_i. Going back to the definition of the norm
\begin{align} \|A\|^2_2 &= \max_{\|x\|_2 = 1} \|Ax\|^2_2 = \max_{\|x\|_2 = 1} x^T A^TAx \\ & = \max_{\|x\|_2 = 1} x^T U \Lambda U^T x = \max_{\|U^T x\|_2 = 1} x^T U \Lambda U^T x \\ & = \max_{\|y\|_2 = 1} y^T \Lambda y = \max_{\|y\|_2 = 1} y^T \Lambda^{1/2} \Lambda^{1/2} y \\ & = \max_{\|y\|_2 = 1} \| \Lambda^{1/2} y \|_2^2 = \|\Lambda\|_2. \end{align}Assume \Lambda = \mathrm{diag}(\lambda_1,\lambda_2,\ldots,\lambda_n) where \lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_n. If we choose x = e_n = [0,0,\ldots,0,1]^T then \|x\|_2 =1 and
\|\Lambda x\| = \max_i \lambda_i = \rho(\Lambda).This shows that \|A\|_2 \geq [\rho(\Lambda)]^{1/2}. Now, for any vector x \in \mathbb R^n, \|x\|_2 = 1
\|\Lambda x\|_2^2 = \sum_{i=1}^n \lambda^2_i x^2_i \leq \|x\|_2^2 \max_i \lambda_i^2 = \max_i \lambda_i^2.This last inequality shows that \|\Lambda\|_2 \leq \rho(\Lambda) and hence \|A\|_2 \leq [\rho(\Lambda)]^{1/2}. And so, we find that
\|A\|_2 = [\rho(\Lambda)]^{1/2} = [\rho(A^T A)]^{1/2},because \Lambda and A^TA have the same eigenvalues.
Let \|\cdot\| be any induced matrix norm:
\|A\| = \max_{\|x\|=1} \|Ax\|.Let v be a eigenvector, \|v\| = 1, associated to the largest eigenvalue \lambda, \rho(A) = |\lambda| = \max_i |\lambda_i. Then
\|Av\| = \|\lambda v\| = |\lambda| \|v\| = \rho(A).
For any specific choice of vector \|v\|=1, \|A\| \geq \|Av\|, so \|A\| \geq \rho(A).
hold on
%A = [10 2 3 1; 1 -3 1 -1; -1 -3 -1 4; 2 2 2 5];
%A = rand(10)-1/2; A = A + 1j*(rand(10)-1/2);
A = [4,1,1;0,2,1;-2,0,9];
for i = 1:length(A)
a = A(i,i); r = norm(A(i,:),1)-abs(a);
theta = linspace(0,2*pi,100);
x = r*cos(theta); y = r*sin(theta);
plot(x+real(a),y + imag(a),'k')
end
lambda = eigs(A);
plot(real(lambda),imag(lambda),'*')
Estimate the spectral radius of A using the Gershgorin Circle Theorem
A = \begin{bmatrix} 4 & 1 & 2 \\ 0 & 2 &1 \\ 2 & 1 & 9 \end{bmatrix}.