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Suppose that $A$ and $B$ are self-adjoint bounded linear operators on a Hilbert space and $\lambda \in \mathbb{C}$. It turns out that if $\lambda \notin \{-1, 1\}$ then $AB=\lambda BA \implies AB = B …

Fix $r \geq 1$ and let $A_n$ be the $n \times n$ identity matrix with the bottom $n-r$ rows replaced with rows of zeros. Then $||A_n|| = 1$ for all $n$ and $||T_n \circ A_n|| = ||A_n||$ so we have th …

The Hilbert-Schmidt norm, $||A||_F= (\sum_{i,j=1}^n a_{i,j}^2)^{1/2}$ is clearly always larger than $||A||_{max}$ and is also submultiplicative. Hence, $||AB||_{max} \leq ||AB||_F \leq ||A||_F ||B||_ …

This may be a partial solution to your problem: I claim that if there exists a shared eigenvector, $x$ of $A$ and $B$ with common eigenvalue of $1$ then $\det(AB - BA) = \det[A,B] = 0$. Proof: Supp …

I wrote my dissertation on a problem in matrix analysis and I found that I had to read from several different sources to understand the material. I don't know which is the best book on the subject bu …