Understanding the proof of the sum of eigenvalues and singular values

Question

I am trying to Understand this proof . So far I understand everything but the part where the author says the following:

"Due to Schur decomposition, there exist a unitary matrix $U$ and an upper triangular matrix $T$, such that $A=U T U^{*}$. Note that $A$ and $T$ share the same eigenvalues and singular values, so we may assume that $A$ is upper triangular."

I don't understand why the fact that $A$ and $T$ share the same eigenvalues and singular values enables us to assume that $A$ is upper triangular.

I know that since $A=U T U^{*}$ means that $A$ and $T$ are unitarily equivalent ($T$ being upper triangular), but i don't see how this enables us to assume that $A$ is upper triangular.

Please help.

Answer 1

You are trying to prove that $$\sum_{i=1}^n \sigma_i(A) \geq\sum_{i=1}^n \left |\lambda_i(A) \right |$$ where the $\lambda_i$ are the eigenvalues and the $\sigma_j$ are the singular values.

If you can solve the problem for upper triangular matrices, and $A$ is arbitrary, then let $T$ be upper triangular such that $A=UTU^*$. Then you know the result is true for $T$, so $$\sum_{i=1}^n \sigma_i(T) \geq\sum_{i=1}^n \left |\lambda_i(T) \right |$$ and since the singular values of $T$ agree with those of $A$, and the eigenvalues of $T$ agree with those of $A$, you have $$\sum_{i=1}^n \sigma_i(A) = \sum_{i=1}^n \sigma_i(T) \geq\sum_{i=1}^n \left |\lambda_i(T) \right | = \sum_{i=1}^n\left|\lambda_i(A)\right|$$ which gives you the result.

That is, it is sufficient to prove the result for upper triangular matrices. And of course, if the result is true, then it is true for upper triangular matrices, so you'll need to prove it for upper triangular matrices. So you may as well just assume that the matrix $A$ that you have is upper triangular and prove it for those matrices.