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!

  • $\begingroup$ @amsmath I was a bit too fast there. How about $[I A]^T [I A] - [I 0]^T [I 0]$ ? $\endgroup$ Oct 30, 2019 at 14:39
  • $\begingroup$ This should allow simplification with Kronecker products and $\text{eig}(A)$. $\endgroup$ Oct 30, 2019 at 14:41
  • $\begingroup$ 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? $\endgroup$
    – amsmath
    Oct 30, 2019 at 14:44
  • $\begingroup$ @amsmath Not necessarily square. $\endgroup$
    – anon
    Oct 30, 2019 at 15:05
  • $\begingroup$ @rhacksby Now, I have a complete answer. Please have a look. $\endgroup$
    – amsmath
    Oct 30, 2019 at 16:27

1 Answer 1


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 $\lambda\in\sigma(S) \Longleftrightarrow \lambda+1\in\sigma(L)$.


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