Let $p<n$ 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.


1 Answer 1


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.

Hence, you are correct up to permutation of duplicated eigenvalues and their corresponding eigenvectors.

  • $\begingroup$ 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? $\endgroup$
    – RozaTh
    Jan 25, 2019 at 1:36
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 25, 2019 at 16:18

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