# A matrix equation involving eigendecomposition

Let $$p and

-$$H\in\mathbb{R}^{n\times n}$$ be symmetric with eigendecompoistion being $$H=U\Lambda U^{\text{T}}$$,

-$$A\in\mathbb{R}^{n\times p}$$,

-$$D\in\mathbb{R}^{p\times p}$$ be a diagonal matrix.

We have $$H=U\Lambda U^{\text{T}}=A D A^{\text{T}}$$. I want to find $$A$$ and $$D$$ as functions of $$U$$ and $$\Lambda$$.

For the above equation to be true only $$p$$ eigenvalues of $$H$$ in $$\Lambda$$ are nonzero (RHS is of rank $$p$$). Forming $$A$$ by the corresponding $$p$$ eigenvectors (columns of $$U$$) and $$D$$ with those nonezero eigenvalues is a solution.

My guess is that is the only solution but don't know how to prove it.

Suppose $$U\Lambda U^T = ADA'$$. Without loss of generality, we may assume that the eigenvalues in $$\Lambda$$ are arranged such that the first $$p$$ nonzero eigenvalues are first. Then partition $$U = [U_1, U_2]$$ and where $$U_1$$ corresponds to the nonzero eigenvalues of $$\Lambda$$. It is clear then that $$U \Lambda U^T = U_1 \Lambda' U_1^T$$ where $$\Lambda'$$ is the diagonal matrix consisting of just nonzero eigenvalues. Hence $$U_1\Lambda' U_1^T = ADA'$$.
Multiplying both equations by $$U_1^T$$ on the left and $$U_1$$ on the right, we see that $$\Lambda' = U_1^T A D A' U_1$$. This implies that $$U_1^T A$$ is at most a permutation matrix since both $$D$$ and $$\Lambda'$$ are diagonal.
• Thanks for the answer. One question though. If both $D$ and $\Lambda$ have some duplicated entries, how can we show $U_1^{\text{T}}A$ is at most a permutation matrix? Jan 25, 2019 at 1:36
• Note that $\Lambda' (A^T U_1)^T = U_1^T A D$; that is, $\Lambda'$ multiplied by an orthogonal matrix on the right is the same as multiplying an orthogonal matrix on the right. Since both $\Lambda'$ and $D$ are diagonal, they scale the columns and rows respectively. Hence, there is a contradiction if $U_1^TA$ is not a permutation matrix. Jan 25, 2019 at 16:18