I was just wondering how many singular values a linear map $T: V \to W $ has, provided $\dim V=n$ and $\dim W=m$.

We only learnt about singular values of an operator $T \in \mathcal{L}(V)$, which we defined as the eigenvalues of $\sqrt{T^*T}$, repeated according to multiplicity.

I thought that it should be $n$ since the matrix of $\sqrt{T^*T}$ is $n \times n$, and hence has $n$ eigenvalues. It seems that the answer is $m$ however.

What is the correct number of singular values for the linear map $T$?


1 Answer 1


By your definition, you are correct: $T^*T$ is a $n\times n$ matrix and has $n$ (non-negative!) eigenvalues.

By the usual definition, a singular value decomposition of a $m\times n$ matrix $A$ is a factorization of the form $$A=U\Sigma V^*$$ where $U,V$ are unitary $m\times m$, $n\times n$ matrices, respectively, and $\Sigma$ is a $m\times n$ diagonal matrix with non-negative diagonal entries. The non-zero diagonal elements of $\Sigma$ are then called the singular values of $A$. Therefore, by this definition, there are at most $\min\{m,n\}$ singular values.

I believe in your definition, you should have the non-zero eigenvalues of $T^*T$? This will reconcile the difference, as then if $m<n$, at least $n-m$ of the diagonal entries of $\Sigma$ (and indeed $n-m$ eigenvalues of $T^*T$!) will necessarily be zero (the argument for when $m>n$ is similar).

  • $\begingroup$ Thanks for your answer! As an example, we are asked to find the singular values of the matrix: $\begin{pmatrix} 0 & 1 \\ 2 & 0 \\ 0 & 1 \end{pmatrix}$. The answer states that they are $0, 2$ and $4$, but I found that they were the square roots of the above. By your explanation above, should it just be $2$ and $\sqrt{2}$? $\endgroup$
    – MrMazgari
    Nov 14, 2015 at 1:11
  • $\begingroup$ Yes, the singular values should be $2$ and $\sqrt{2}$. Also, I'm not sure where they got the $0$? For your matrix, $A^*A=\left[\begin{array}{cc} 4 & 0\\ 0 & 2\end{array}\right]$? So their answer isn't even consistent with their own definition...? $\endgroup$ Nov 14, 2015 at 1:21

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