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On a numerical math course I recently saw the following statement without a proof. How would one prove it?

Let $A$ be an $n$ x $n$ real matrix with singular values $s_1 \geq s_2 \geq \ldots \geq s_n$ and eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$, $|\lambda_i| \geq |\lambda_{i+1}|$. For all $k = 1, \ldots, n$, $|\lambda_1 \cdot \cdot \cdot \lambda_k| \leq s_1 \cdot \cdot \cdot s_k$.

The cases $k=n$ and $k=1$ are implied by parts (1a) and (2) of this question/answer, but what about the cases in between?

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Miikka
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1 Answers1

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Let $A=UTU^*$ be the Schur complement decomposition of $A$ such that $\mathrm{diag}(T)=[\lambda_1,\ldots,\lambda_n]$. Let $U_k$ be the matrix composed of the first $k$ columns of $U$. Then $$ |\lambda_1\cdots\lambda_k|=|\det (U_k^*AU_k)| $$ (note that $U_k^*AU_k$ is triangular with $\lambda_1,\ldots,\lambda_k$ on the diagonal). The remainder of the proof consists of two facts:

  • $|\det(X)|=s_1(X)\cdots s_k(X)$, where $s_i(X)$ are the singular values of the $k\times k$ matrix $X$.
  • Interlacing property: for $X\in\mathbb{C}^{m\times n}$, $V\in\mathbb{C}^{m\times k}$, $W\in\mathbb{C}^{n\times k}$, where $V$ and $W$ have orthonormal columns, and $k\leq\min\{m,n\}$, $s_i(V^*XW)\leq s_i(X)$ for $i=1,\ldots,k$.

The first fact relates $|\det (U_k^*AU_k)|$ to the product of the singular values of $U_k^*AU_k$, while the second bounds the singular values of $U_k^*AU_k$ in terms of the singular values of $A$.

Algebraic Pavel
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