# Using SVD to write an eigendecomposition

Let $$A\in\mathbb{R}^{n\times n}$$. Use the SVD of A to write down an explicit eigendecomposition of $$H = \begin{bmatrix}0 &A^{T}\\A & 0\end{bmatrix}.$$ Hint: If $$\sigma$$ is a singular value of $$A$$ then $$±\sigma$$ are eigenvalues of $$H$$.
I do not understand how to proceed. My idea is to set $$U=V=I$$ so that $$H = \begin{bmatrix}0 &\Sigma\\\Sigma & 0\end{bmatrix}$$ And try to find a matrix $$X$$ such that $$\begin{bmatrix}0 &\Sigma\\\Sigma & 0\end{bmatrix}=X\Sigma X^{-1}$$ Is there a more straighforward way to solve it?

It is possible to find a matrix $$X$$ such that $$X^{-1}HX = \pmatrix{0 & \Sigma\\ \Sigma & 0}.$$ In particular, if $$A = U \Sigma V^T$$ is the singular value decomposition of $$A$$, we can take $$X = \pmatrix{V & 0\\0 & U}.$$ On the other hand, the matrix $$Y = \frac{1}{\sqrt{2}} \pmatrix{I & I\\I & -I}$$ Is such that $$Y^{-1}\pmatrix{0 & \Sigma\\\Sigma & 0}Y = \pmatrix{\Sigma&0\\0& -\Sigma}.$$ Put these together to get the eigendecomposition of $$H$$.

• Can you explain how I can use it to write an eigendecomposition of $H$? Commented Mar 22, 2022 at 16:44
• See my latest edit. Commented Mar 22, 2022 at 17:08

The characteristic polynomial of $$H$$ is given by

$$p_H(\lambda)=\det(\lambda I-H)=\det(\lambda I)\det(\lambda I-\lambda^{-1}A^TA)=\det(\lambda^2I-A^TA)$$

where the second equality comes from the Schur complement (see also https://en.wikipedia.org/wiki/Block_matrix#Block_matrix_determinant).

So, we can see that $$\lambda^2$$ must be an eigenvalue of $$A^TA$$ for $$p_H(\lambda)$$ to be zero. This means that the eigenvalues of $$H$$ consist of the singular values of $$A$$ and their additive inverse.

This means that the matrix $$H$$ can be expressed in diagonal form modulo some suitable basis change with diagonal matrix

$$D=\begin{bmatrix}\Sigma & 0\\0 & -\Sigma\end{bmatrix}.$$

The basis change $$P$$ obeys $$HP=PD$$, so let

$$P=\begin{bmatrix}P_1 & P_2\\ P_3 & P_4\end{bmatrix}.$$

This yields

$$\begin{bmatrix}0 & V\Sigma U^T\\U\Sigma V^T & 0 \end{bmatrix}\begin{bmatrix}P_1 & P_2\\ P_3 & P_4\end{bmatrix}=\begin{bmatrix}P_1 & P_2\\ P_3 & P_4\end{bmatrix}\begin{bmatrix}\Sigma & 0\\0 & -\Sigma\end{bmatrix}.$$

Solving for this system of equations yields

$$P=\begin{bmatrix}V & -V\\ U & U\end{bmatrix}\ \mathrm{and}\ P^{-1}=\dfrac{1}{2}\begin{bmatrix}V^T & U^T\\ -V^T & U^T\end{bmatrix}$$

where we have used the fact that $$U^TU=UU^T=I$$ and $$V^TV=VV^T=I$$.

We can then easily verify that $$PHP^{-1}=D$$.

• Yes, that was given by a hint. My question is how to use SVD of A to write the eigendecomposition of H Commented Mar 22, 2022 at 16:32
• @sebab2101 I have updated my answer.
– KBS
Commented Mar 22, 2022 at 17:05
• I don't understand what you mean by P^-1. If it is supposed to be the inverse of P, then the first block of the matrix product P P^-1 is V U^T + V U^T which is not necessarily equal to I_n. Commented Mar 18, 2023 at 12:24
• You can use the orthogonality of U and V to show that the inverse of 1/sqrt(2) P is its transpose, but I don't believe your expression for P^-1 is correct as-is. Commented Mar 18, 2023 at 12:25
• @user14464173 Sorry, that was a typo, and I have corrected it.
– KBS
Commented Mar 22, 2023 at 7:42