Jtdylktuy

Given $A in mathbb{R}^{n times m}$, consider the symmetric matrix

$M = begin{pmatrix} 0 & A \ A^{t} & 0 end{pmatrix} in mathbb{R}^{(n+m) times (n+m)}$.

Show that a simple relationship exists between the singular vectors of $A$ and the eingenvectors of $M$. Show how to build an orthbogonal basis of $mathbb{R}^{n+m}$ consisting of eigenvectors of $M$, given the singular vectors of $A$.

My attempt: Consider the singular decomposition $A=USigma V^{t}$, where $U_{ntimes n} $, $V_{m times m}$ are orthogonal and $Sigma$ is rectangular diagonal matrix of singular values.

$displaystyle det M = det((-lambda I)(-lambda I) - AA^{t}) = det(lambda^2I - AA^{t}) = det(lambda^2I - USigma V^{t}VSigma^{t}U^{t})=det(lambda^2I - USigma^{2}U^{t}) = det begin{pmatrix}(lambda^2 - sigma_1^2 & 0 & dots & dots &0 \0 & lambda^2 - sigma_2^2 & 0 & dots & 0 \ vdots & ddots & ddots & ddots& vdots \ 0 & dots & dots & 0 & lambda^2 - sigma_n^2 end{pmatrix}=prod_{k=1}^{n} (lambda^2 - sigma_k^2)=0 implies lambda = pm sigma_k$

for some $k in{1,...,n}$

Let $v_k$ be the eigenvector associated to $lambda$.

$Mv_k = lambda v_k = pm sigma_k v_k $, and $sigma_kv_k$ is equal to either $Av_k, A^{t}v_k$ or $0$ (By a theorem in the book), depending on which index $k$.

I'm insecure, specially at the last part, it doesn't feel right. Also, I'd like to some orientation on how to build this basis of orthogonal eigenvectors. I know Gram-schmidt, but not when singular vectors are involved...

Please verify what I did and show me how to improve, or maybe start all over...

Thanks.

I would start the same way you did: let $A = U Sigma V^t$ be the SVD of $A$. Then, the SVD of $A^t$ is given by $A^t = V Sigma^t U^t$. Inserting into $M$:

$$M = begin{bmatrix} 0_{n times n} & U Sigma V^t \ V Sigma^t U^t & 0_{m times m} end{bmatrix}.$$

Now, let us try the eigenvectors $q_k = [u_k^t, pm v_k^t]^t$. Inserting into $M$ yields

$$M cdot q_k = begin{bmatrix} U Sigma V^t v_k \ V Sigma^t U^t u_k end{bmatrix} =begin{bmatrix} sigma_k u_k \ pm sigma_k v_k end{bmatrix} = sigma_k q_k.$$

This shows that $q_k$ are eigenvectors. Note that for $nneq m$ you will also get some zero eigenvalues. This is due to the fact that then either $A$ or $A^t$ will have a non-empty kernel and thus we can find nonzero vectors that give $M cdot q = 0$.