An routine exercise about matrix norm

Question

If $T_{n}\in M_{k(n)}(\mathbb{C})$ and $||T_{n}^{*}T_{n}-1_{k(n)}||\rightarrow0$, then $||T_{n}T_{n}^{*}-1_{k(n)}||\rightarrow 0$ too? (Here, $M_{k(n)}(\mathbb{C})$ denotes the $k(n) \times k(n)$ complex matrix.)

Answer 1

HINT:

We have $$||I - T^* T || = || I - T T^*||$$

Indeed for any self adjoint operator $S$ we have

$$||S|| = \sup \, \{ |\lambda| \ \mid \ \lambda\ \text{eigenvalue of } S \}= \rho(S)$$

Moreover, for any $U$, $V$ operators on a finite dimensional space we have

$$\sigma(I - UV) = \sigma( I - VU)$$

Alternatively, use the singular decomposition of $T$ and check that $I - T^* T$, $I - T T^*$ are unitarily conjugate self adjoint operators.