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Let $\lambda_{i}(M)$ denote the $i$th eigenvalue of the square matrix $M$, and $T$ denote the matrix transpose.

Is it true that $|\lambda_{i}(A^{T}A)|=|\lambda_{i}(A)|^{2}$ for every square matrix $A$?

Thank you very much!

Davide Giraudo
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Anna Smirnova
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2 Answers2

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No. Take $A=\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}$. The spectral radius is zero, but $A^TA=\begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}$ has an eigenvalue at one.

copper.hat
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The square roots of the eigenvalues of $A^TA$ are also called "singular values" of $A$, and have no relationship to the eigenvalues of $A$, in general. The only exception is when $A$ is symmetric (or more general a normal matrix). Then your relation holds. When $A$ is even symmetric (or Hermitian) and positive definite the eigenvalues and the singular values coincide.

Elmar Zander
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  • There are some relationships that hold in general, $\sigma_1(A) \ge \rho(A)$, and $\sigma_n(A) = 0$ iff $0$ is in the spectrum of $A$. – copper.hat Mar 12 '13 at 16:06
  • Yes, you're right. I had only direct algebraic relations like in the OP's question in mind, so I forgot about that. Thanks! – Elmar Zander Mar 12 '13 at 19:00