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I found a proof of the fact that the largest singular value is always greater or equal to $|\lambda|_{\max}$.

In what cases equality holds and in what cases it is strictly greater?

Thank you.

Paul R
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  • One case would be a symmetric matrix with only negative eigenvalues. – Calle Nov 11 '23 at 15:59
  • I read that equality holds only for symmetric matrices. Any proof? – Paul R Nov 11 '23 at 16:02
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    What do you mean “I found a proof when it is greater or equal”? Can you share the statement with us? – Lorenzo Pompili Nov 11 '23 at 16:09
  • Also, with $\lambda_{max}$ do you mean the maximum of the absolute value of all eigenvalues? If so, then I don’t think strict inequality always holds for all non-symmetric matrices – Lorenzo Pompili Nov 11 '23 at 16:12
  • @LorenzoPompili https://math.stackexchange.com/questions/2034958/show-that-the-largest-singular-value-dominates-all-eigenvaluesany-matrix-a http://math.mit.edu/classes/18.095/2016IAP/lec2/SVD_Notes.pdf page 389, (9th page) – Paul R Nov 11 '23 at 16:13
  • @LorenzoPompili, yes the absolute value. – Paul R Nov 11 '23 at 16:15
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    I don’t see anywhere written that strict inequality holds for all non-symmetric matrices. Can you point me to the relevant part? – Lorenzo Pompili Nov 11 '23 at 16:17
  • What do you mean by $\lambda_\max$ when the eigenvalues of a matrix may not be real numbers? – user1551 Nov 11 '23 at 16:20
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    equality holds iff a $A$ is unitarily similar to $\pmatrix{a_1 & \mathbf 0^H\\mathbf0& A'}$ where $\big \Vert A'\big \Vert_2 \leq \sigma_1$ and $\vert a_1\vert =\sigma_1$ . One direction is obvious, for the other see here: https://math.stackexchange.com/questions/3496086/proving-that-left-and-right-eigenvector-are-equal-using-singular-values/ – user8675309 Nov 11 '23 at 17:09
  • @eMathHelp, you changed your question without stating that you changed it. Quite rude. – Calle Nov 12 '23 at 15:02
  • @Calle, sorry for that. – Paul R Nov 12 '23 at 15:28

1 Answers1

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Equality holds for symmetric matrices (it is a consequence of the spectral theorem).

Example of strict inequality: consider the matrix $$\left[\begin{array}{}0&1\\0&0\end{array}\right].$$ Its singular values are $0$, $1$, but all eigenvalues are zero.

There are some non symmetric matrices whose singular values coincide with the largest (in absolute value) eigenvalue though. For instance $$\left[\begin{array}{}0&1\\-1&0\end{array}\right]$$ (the singular values are all $1$, and the eigenvalues are $\pm i$).

I don’t know a general criterion for when $\sigma_1$ and $|\lambda|_{\max}$ coincide.

Lorenzo Pompili
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    A proof of what? – Lorenzo Pompili Nov 11 '23 at 16:05
  • I mean is there more general class of matrices? You gave some examples. They are great! So for non symmetric matrices could be both cases. What about symmetric? Symmetric positive semidefinite have this property: https://math.stackexchange.com/questions/867652/why-do-positive-definite-symmetric-matrices-have-the-same-singular-values-as-eig Can you give an example of symmetric matrix whose largest eigenvalue is strictly less then the largest singular value? – Paul R Nov 11 '23 at 16:27
  • I thought this was already clear to you, this is why I did not write that. For symmetric matrices equality always holds: by the spectral theorem, the matrix is diagonal in some orthonormal basis, and for diagonal matrices equality always holds – Lorenzo Pompili Nov 11 '23 at 16:31
  • Is there any rigorous proof for this? – Paul R Nov 11 '23 at 16:44
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    Sure. Take a symmetric (or self-adjoint in the complex case) matrix A: it is diagonalizable via an orthogonal change of basis, so there exists $U$ orthogonal (or maybe it should be called unitary) such that $U^AU=D$, with $D$ diagonal. The singular values of $A$ are the square roots of the eigenvalues of $A^A$. But $A^A=A^2=UD^2U^$, so $A^*A$ is diagonalizable and the eigenvalues are precisely those in the diagonal of $D^2$, i.e., the squares of the eigenvalues of $A$. So all singular values are the absolute values of the eigenvalues of $A$. – Lorenzo Pompili Nov 11 '23 at 16:55
  • Thank you very much! – Paul R Nov 11 '23 at 17:02