Computing Eigenvectors from eigenvalues by making Symmteric Matrix from $X$

Question

I need to find out the eigenvectors of the matrix

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

Here it is given that $X$ is a matrix of shape m x n. The eigenvectors of the matrix need to be represented as singular vectors of X.

I tried by doing it using individual elements but it did not really work and the expression has become very complicated. Also, I don't know how to multiply block matrices

Answer 1

Posting as an aswer due to length.

Let $M=\begin{bmatrix}\mathbf{0}&\mathbf{X}\\\mathbf{X}^T&\mathbf{0}\end{bmatrix}$. Note that $M^TM = \begin{bmatrix}\mathbf{XX}^T&\mathbf{0}\\\mathbf{0}&\mathbf{X}^T\mathbf{X}\end{bmatrix}$ and eigenvalues/eigenvectors of $M^TM$ are constructed by eigenvalues/eigenvectors of $\mathbf{XX}^T$ and $\mathbf{X}^T\mathbf{X}$, which will yield singular values/vectors of $\mathbf{X}$.

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Other form is based on the definition, and using properties of the determinant of block matrices. Eigenvalues of $M$ are the solutions to $\det(M-\lambda I_{m+n})=0$. As the identity is invertible by definition, we can write $$\begin{align}\det\begin{bmatrix}-\lambda\mathbf{I}&\mathbf{X}\\\mathbf{X}^T&-\lambda\mathbf{I}\end{bmatrix}&=\det(-\lambda\mathbf{I}_n)\det\Big(-\lambda\mathbf{I}_n+\frac{1}{\lambda}\mathbf{X}^T\mathbf{I}_m\mathbf{X}\Big)\\ &=(-\lambda)^n\det\Big(-\lambda\mathbf{I}_n+\frac{1}{\lambda}\mathbf{X}^T\mathbf{X}\Big)\\ &=\det\Big(\lambda^2\mathbf{I}_n-\mathbf{X}^T\mathbf{X}\Big)=0\\ \end{align}$$ and you'll also get $\det\Big(\lambda^2\mathbf{I}_m-\mathbf{X}\mathbf{X}^T\Big)=0$

Answer 2

Let $X = USV^T$ be an SVD, then $$ A = \begin{pmatrix}0 & X\\X^T & 0\end{pmatrix} = \begin{pmatrix}0 & USV^T\\VS^TU^T & 0\end{pmatrix} = \begin{pmatrix}U&0\\0&V\end{pmatrix}\begin{pmatrix}0 & S\\S^T & 0\end{pmatrix}\begin{pmatrix}U^T&0\\0&V^T\end{pmatrix} = \tilde{U}\tilde{S}\tilde{U}^T. $$ Can you finish from here? (Hint: diagonalize $\tilde{S}$ using a block matrix of the form $\frac{1}{\sqrt{2}}\begin{pmatrix}I_m & I_{m,n}\\-I_{n,m} & I_n\end{pmatrix}$, where $I_{m,n}$ is an m-by-n matrix with ones on its main diagonal and the rest zeros, and $I_m$ is the identity of size m).