I have a $2n\times n$ real matrix $A$ which has full rank $n$. I would like to zero value eigenvectors of $P=AA^+$, i.e. the orthogonal projector onto the range of $A$.
$A^+$ is the Moore-Penrose pseudoinverse.
I am guessing the way to do this is using singular value decomposition. I can break $A$ into
$$A=U\Sigma V^T$$
and then calculate the Moore-Penrose pseudoinverse as
$$A^+=V\Sigma^+ U^T$$
which gives
$$AA^+=U\Sigma V^TV\Sigma^+U^T=U\Sigma \Sigma^+U^T$$
Since $A$ has rank $n$ then the matrix $\Sigma \Sigma^+$ is going to be a $2n\times 2n$ matrix of the form
$$\Sigma \Sigma^+=\left(\begin{matrix}I&0\\0&0\end{matrix}\right)$$
This means that the eigenvectors of $P$ with eigenvalue $0$ will be those corresponding to the columns of $U$.
I have a conjecture that the eigenvectors of $P$ should be of the form
$$\mathbf b=\left(\begin{matrix}b_1\\\vdots\\b_n\\0\\\vdots\\0\end{matrix}\right)$$
where $b_1,\dots,b_n$ can be any real number or does this only hold if $U$ is an identity matrix?
Best Answer
You want the eigenvectors for $UBU^T$. Eigenvalues in your case are $0$ and $1$, each with multiplicity $n$.
If $$UBU^Tw=\lambda w,$$ this is the same as $$BU^Tw=\lambda U^Tw.$$ That is, $U^Tw$ has to be an eigenvector for $B$. In your case the eigenvectors for $B$ are $e_1,\ldots,e_n$ for $1$ and $e_{n+1},\ldots,e_{2n}$ for $0$. Thus the eigenvectors for $AA^+$ are $$ U^Te_1,\ldots,U^Te_{2n}. $$ These are the rows of $U$. The first $n$ rows form a basis for the range of $P$, while the last $n$ rows form a basis for the kernel of $P$.
For the eigenvector in the range of $P$ to be of the form you want is to ask for $U$ to be of the form $$U=\begin{bmatrix}X&0\\ Y&Z\end{bmatrix}.$$ Being a unitary, this means that $$U=\begin{bmatrix}X&0\\ 0&Z\end{bmatrix},$$ with $X,Z$ unitaries. This in turn implies that $$ P=U\Sigma\Sigma^+U^T=\begin{bmatrix}X&0\\ 0&Z\end{bmatrix}\begin{bmatrix}I&0\\ 0&0\end{bmatrix}\begin{bmatrix}X^T&0\\ 0&Z^T\end{bmatrix}=\begin{bmatrix}I&0\\ 0&0\end{bmatrix}=\Sigma\Sigma^+. $$ Since $A=AA^+A$, writing $A=\begin{bmatrix}A_0\\ A_1\end{bmatrix}$ this gives $$ A=\begin{bmatrix}I&0\\ 0&0\end{bmatrix}\begin{bmatrix}A_0\\ A_1\end{bmatrix}=\begin{bmatrix}A_0\\ 0\end{bmatrix}. $$