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Let $V,W$ be finite-dimensional inner product spaces over $\mathbb{K}$.

Let $T:V\rightarrow W$ be a linear transformation.

How do I prove that the eigenvalues of $TT^*$ are the eigenvalues of $T^*T$?

I remember that it can be proven by using singular value decomposition, but I cannot remember..

EDIT:

I posted a proof using SVD as an answer.

However, I'm curious to know whether there is another way to prove this not invoking SVD.

Rubertos
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2 Answers2

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This is a special case of Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$? (Citation needed!).

Community
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PhoemueX
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Note that square roots of nonzero eigenvalues of $TT^*$ are the singular values of $T$. (Since $TT^*$ is self-adjoint, those eigenvalues are positive) Consider $T$ as a matrix and consider the singular value decomposition. Then, $T=U\sum V^* \Rightarrow T^* = V \sum^t U^*$. Since singular values are unique, the diagonal terms of $\sum^t$ are exactly the square roots of eigenvalues of $T^*T$.

Consequently, $TT^*$ and $T^*T$ have the same nonzero eigenvalues. Since $rank(TT^*)=rank(T^* T)$, they have the same eigenvalues.

Rubertos
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