What is the SVD of $ab^T + ba^T$ What is the SVD of $A = ab^T + ba^T$ when a and b are column vectors? It is also given that $a^Tb = 0$.
I've tried finding $AA^T$ and $A^TA$ to find the e-values and e-vectors for $U$ and $\Sigma$ to no avail. This method worked for finding the SVD of $A = ab^T$, but no luck with this case. Any pointers to the right direction would be greatly appreciated!
Edit: I saw this post, but I didn't quite get how they got the n-2 dimension null space. Moreover, it was confusing to me when they jumped from $(\alpha a+\beta b)=\lambda(\alpha a+\beta b)$ to the next equation, as I have no clue where they got ||a|| and ||b|| from.
 A: In fact, I just realized that the post referenced in your question was about the eigen-decomposition of $ab^T+b^Ta$, not about its Singular Values Decomposition (SVD). Therefore, I have chosen to postpone the answer to your questions to an Appendix.
Here is a simple way to address the SVD issue.
Let us consider for the sake of simplicity the case of dimension $3$ (the general case being a straightforward extension of this one).
Setting $$A=\frac{1}{\|a\|}a, B=\frac{1}{\|b\|}b,$$
one can obtain at once the SVD:
$$ab^t+ba^t=\|a\|\|b\|(AB^T+B^TA)$$
$$ab^t+ba^t=\underbrace{\begin{pmatrix}|&|&|\\A&B&W\\|&|&| \end{pmatrix}}_U\underbrace{\begin{pmatrix}\|a\|\|b\|&0&0\\0&\|a\|\|b\|&0\\0&0&0 \end{pmatrix}}_{\Sigma}\underbrace{\begin{pmatrix}-&B^T&-\\-&A^T&-\\-&W^T&- \end{pmatrix}}_{V^T}$$
where vector $W$ is completing $A,B$ into an ortho-normalized basis for $\mathbb{R^3}$.
In particular, we have a double non-zero singular value $\|a\|\|b\|$ (product of norms).
Appendix:
First question about the $n-2$ dimensions of the nullspace:
By (orthogonal) basis extension theorem one can add to $a,b$,  $n-2$ vectors $v_1,v_2,... v_{n-2}$ that constitute an orthogonal basis of $\mathbb{R^n}$. Then,
$$(ab^T+ba^T)v_k=a(b^Tv_k)+b(a^Tv_k)=a0+b0=0,$$
establishing that the kernel contains all the $v_k$s, therefore has dimension $\ge (n-2)$. As the dimension of the range is at least $2$ (it contains $a,b$), we deduce, using the rank-nullity theorem that the dimension of the kernel is exactly $(n-2)$ and the dimension of the range is $2$.
Second question:
$$(\alpha a+\beta b)=\lambda(\alpha a+\beta b)$$
$$(ab^T+ba^T)(\alpha a+\beta b)=\lambda(\alpha a+\beta b)$$
$$\alpha a\underbrace{(b^Ta)}_0+\beta a\underbrace{(b^Tb)}_{\|b\|^2}+\alpha b\underbrace{(a^Ta)}_{\|a\|^2}+\beta b\underbrace{(a^Tb)}_0=\lambda \alpha a+ \lambda \beta b$$
It remains to identify the coefficients of vectors $a$ and $b$.
A side remark: I had been working on the eigenvalues of a matrix of this form $ab^T+ba^T$ in a very interesting issue about average values of "products of dot products": see here.
