# Subtracting orthogonal projection matrix to find second largest eigenvalue

Let α = λ/(v'v), with λ being A's largest eigenvalue and v its corresponding eigenvector, how do I prove that the largest absolute eigenvalue of A − αvv' equals the second largest eigenvalue of A by use of the power method? I think that αvv' is an ortogonal projection matrix but what is the use of multiplying it by λ and subtracting from A?

• Try looking at what this does in the case that $A$ is diagonal, and I think it should then be clear why this works in general – Ben Grossmann Oct 31 '18 at 18:03
• Do you assume $A$ is square and diagonalizable? – abolfazl Oct 31 '18 at 19:07

I assume $$A$$ is an $$n\times n$$ square matrix having $$n$$ eigenvectors. Therefore, $$A = VDV^{-1}$$ where I assume columns of $$V$$ are orthogonal to each other but, they are not normalized. We can write $$A$$ equivalently as $$A = \lambda_1v_1v_1^\top+\sum_{i=2}^n\lambda_iv_iv_i^\top$$. Let $$B = A-\alpha v_1v_1^\top$$ where $$\alpha = \frac{\lambda_1}{\|v_1\|_2^2}$$. Then, $$B = \lambda_1(1-\frac{1}{\|v_1\|_2^2})v_1v_1^\top+\sum_{i=2}^n\lambda_iv_iv_i^\top$$. So, if $$\lambda_1(1-\frac{1}{\|v_1\|_2^2}) > \lambda_2,$$ the largest absolute eigenvalue of $$B = A-\alpha v_1v_1^\top$$ DOES NOT equal the second largest eigenvalue of $$A$$. You can easily construct many counterexamples. If you have MATLAB, the following code generates a random counterexample:

clear; clc;
Q = randn(3); Q = orth(Q)*diag([3,2,1])
sum(Q.^2)
D = diag([200,1.5,1])
A = Q*D*Q^(-1)
eig(A)
a = 2/norm(Q(:,1))^2
B = A-a*Q(:,1)*Q(:,1)'
eig(B)


If the eigenvectors where normalized, $$\lambda_1(1-\frac{1}{\|v_1\|_2^2}) = 0$$ and you could easily see $$\lambda_2$$ is the largest eigenvalue of $$B$$.

• That’s great, thank you! How could I, without using matlab, show that the last term is not greater than lambda2, so that the power method would indeed return the second greatest? – Walter Nap Oct 31 '18 at 20:05