# Finding the Eigenvalues of Special Block Matrix

I'm trying to find the eigenvalues of the block matrix

$$\begin{bmatrix} 0 & A \\\ A^T & A^T A \end{bmatrix}$$

in terms of the eigenvalues or singular values of $$A$$. My plan was to calculate the determinant of

$$\begin{bmatrix} \lambda I & -A \\\ -A^T & \lambda I - A^T A \end{bmatrix}$$

For this, I use an identity about the determinant of block matrices to get

$$\det(\lambda I) \det(\lambda I - A^T A - \frac{1}{\lambda}AA^T )$$

However, the second factor doesn't quite look the characteristic polynomial of a matrix related to $$A^TA$$ or $$A$$ yet. I'd really appreciate any help about where to go from here!

• @amsmath I was a bit too fast there. How about $[I A]^T [I A] - [I 0]^T [I 0]$ ? Oct 30, 2019 at 14:39
• This should allow simplification with Kronecker products and $\text{eig}(A)$. Oct 30, 2019 at 14:41
• I found by using Schur complements that the eigenvalues of your matrix must be within $\sigma(A^TA)\cup\{\frac{\mu}2\pm\sqrt{\frac{\mu^2}2+\mu} : \mu\in\sigma(A^TA)\}$. Is $A$ a square matrix? Oct 30, 2019 at 14:44
• @amsmath Not necessarily square.
– anon
Oct 30, 2019 at 15:05
• @rhacksby Now, I have a complete answer. Please have a look. Oct 30, 2019 at 16:27

Let me call your matrix $$S$$. The set of eigenvalues of a square matrix $$X$$ will be denoted by $$\sigma(X)$$. I will show that
$$\sigma(S)=\left\{\frac{\mu}2\pm\sqrt{\frac{\mu^2}4+\mu} \,:\, \mu\in\sigma(A^TA)\cup\sigma(AA^T)\right\}.$$Or, equivalently, $$\lambda\in\sigma(S)\,\Longleftrightarrow\,\frac{\lambda^2}{1+\lambda}\in\sigma(A^TA)\cup\sigma(AA^T).$$
Note that $$\sigma(AA^T)$$ and $$\sigma(A^TA)$$ coincide up to zero. If $$A$$ is not square, zero will be in at least one of them. If $$A$$ is square, then the two sets coincide.
Proof. Set $$L := S+I$$. For $$\lambda\neq 1$$ you can easily check that $$L-\lambda I = \begin{pmatrix}I&0\\(1-\lambda)^{-1}A^T&I\end{pmatrix}\begin{pmatrix}(1-\lambda)I&0\\0&T(\lambda)\end{pmatrix}\begin{pmatrix}I&(1-\lambda)^{-1}A\\0&I\end{pmatrix},$$ where $$T(\lambda) = -\frac{\lambda}{1-\lambda}A^TA + (1 - \lambda)I = -\frac{\lambda}{1-\lambda}\left(A^TA - \frac{(1-\lambda)^2}{\lambda}I\right).$$ Since the two matrices enclosing the diagonal matrix are invertible, we see that $$L-\lambda I$$ is not invertible iff $$T(\lambda)$$ is not invertible, i.e., iff $$\frac{(1-\lambda)^2}{\lambda}\in\sigma(A^TA)$$. Let us consider $$\lambda = 1$$. Then $$L-\lambda = S$$ is easily seen to be non-invertible iff $$0\in\sigma(A^TA)\cup\sigma(AA^T)$$. Hence, we get that $$\lambda\in\sigma(L)\,\Longleftrightarrow\,\frac{(1-\lambda)^2}{\lambda}\in\sigma(A^TA)\cup\sigma(AA^T).$$ The claim now follows from $$\lambda\in\sigma(S) \Longleftrightarrow \lambda+1\in\sigma(L)$$.