if $\|T\|\leq 1$ then $I-TT^{*}\geq 0$ If $T\in\mathcal{B}(\mathcal{H})$ and $\|T\|\leq 1$ then $I-TT^{*}\geq 0.$ Is this statement true?
And why if, the matrix $\begin{pmatrix}
I & T^{*}\\
T & I
\end{pmatrix}\geq 0$ then $\|T\|\leq 1?$
Do we use $\|T\|=\|T^{*}\|$ here?
Any help would be appreciated.
 A: Let $S = I - TT^*$. Since $S$ self-adjoint, $S \geq 0$ iff $\langle Sv , v\rangle \geq 0$, which is equivalent to
$$ 0 \leq \langle v, v \rangle - \langle TT^*v, v \rangle = \langle v, v \rangle - \langle T^*v, T^*v \rangle.$$
And this inequality is equivalent to saying that $\|T^*\|\leq 1$. Now use that $\|T\| = \|T^*\|$ to conclude the reult.
A: Assuming you are using the induced 2-norm for a matrix, $||T||_2=\sigma_1$ where $\sigma_1$ is the largest singular value. Let $T=U\Sigma V^H$ be its singular decomposition. Thus $I-TT^H=U(I-\Sigma^2)U^H$. Thus, eigenvalues of the Hermitian matrix $I-TT^H$ are $(1-\sigma_i^2)$. Since $\sigma_1\leq$1, we have that $I-TT^H$ have all non-negative eigenvalues. Thus, it is positive-semidefinite.
A: $\langle TT^*v, v \rangle=\langle T^*v, T^*v \rangle=||T^*v||^2 \le ||T^*||^2||v||^2 =||T||^2 ||v||^2\le ||v||^2=\langle v, v \rangle$.
Can you proceed ?
A: If $\lVert T \rVert \leqslant 1$, then
$$
\langle x,(I-TT^*)x\rangle = \lVert x \rVert^2 - \lVert T^* x\rVert^2 \geqslant \lVert x \rVert^2(1-\lVert T^*\rVert^2) = \lVert x \rVert^2(1-\lVert T\rVert^2) \geqslant 0.
$$
If $\lVert T \rVert > 1$, then there are $x,y\in\mathcal{H}$ such that $\lVert x \rVert=\lVert y \rVert=1$ and $\langle y,T x\rangle<-1$. It follows that
$$
\left\langle\begin{bmatrix}
x \\
y
\end{bmatrix},
\begin{bmatrix}
I & T^* \\
T & I
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
\right\rangle = \lVert x \rVert^2 + \lVert y \rVert^2 + 2\mathrm{Re}\langle y,Tx\rangle = 2 + 2\langle y,Tx\rangle < 0.
$$
