Better upper bound involving linear operator

Question

Let $\mathcal{A} \colon \mathbb{R}^{n} \to \mathbb{R}^{m}$ be a linear operator. For any given $x,y \in \mathbb{R}^{n}$ and $t > 0$. Denote: $$ z = t \left( x - y \right) + \mathcal{A}^{*} \left( \mathcal{A} y - \mathcal{A} x \right) . $$ Then it can be seen that \begin{equation} \left\lVert z \right\rVert \leq t \left\lVert x - y \right\rVert + \left\lVert \mathcal{A}^{*} \left( \mathcal{A} y - \mathcal{A} x \right) \right\rVert \leq \left( t + \left\lVert \mathcal{A} \right\rVert ^{2} \right) \left\lVert x - y \right\rVert . \end{equation} However notice that $$ z = \left( t \mathrm{Id} - \mathcal{A}^{*} \mathcal{A} \right) \left( x - y \right) $$ and thus we can deduce $$ \left\lVert z \right\rVert = \left\lVert t \mathrm{Id} - \mathcal{A}^{*} \mathcal{A} \right\rVert \left\lVert x - y \right\rVert . $$ I guess that if I can find another upper bound for $t \mathrm{Id} - \mathcal{A}^{*} \mathcal{A}$ then I can improve the etimate for $\left\lVert z \right\rVert$ but I don't know how.

Edit

The main point is to express the formula $\left\lVert t \mathrm{Id} - \mathcal{A}^{*} \mathcal{A} \right\rVert$ in term of $t$ and $\left\lVert \mathcal{A}^{*} \mathcal{A} \right\rVert$.

Update (Thanks to @mechanodroid)

One idea is to try to prove $\left\lVert t \mathrm{Id} - \mathcal{A}^{*} \mathcal{A} \right\rVert = \left\lvert t - \left\lVert \mathcal{A} \right\rVert ^{2} \right\rvert$. Suppose that $t \geq \left\lVert \mathcal{A} \right\rVert ^{2}$ then he proved that $t - \left\lVert \mathcal{A} \right\rVert ^{2} \leq \left\lvert t \left\lVert x \right\rVert ^{2} - \left\lVert \mathcal{A}x \right\rVert ^{2} \right\rvert$. If we can prove that $\left\lvert t \left\lVert x \right\rVert ^{2} - \left\lVert \mathcal{A}x \right\rVert ^{2} \right\rvert \leq t - \left\lVert \mathcal{A} \right\rVert ^{2}$ then the claim is obtained.

Answer 1

Elaborating on the comment of @zwim:

The following proposition will be useful.

Let $T : X \to X$ be a self-adjoint operator on a inner product space $X$. Then we have:

$$\|T\| = \sup\big\{\left|\left\langle Tx, x\right\rangle\right| : x \in X, \|x\| =1\big\}$$

Proof for finite-dimensional $X$:

For any $x\in X, \|x\| =1$ we have $$|\langle Tx, x\rangle| \le \|Tx\|\|x\| \le \|T\|\|x\|^2 =\|T\|$$

Thus $\sup\big\{|\langle Tx, x\rangle| : x \in X, \|x\| =1\big\} \le \|T\|$.

For the reverse inequality, let $\{e_1, \ldots, e_n\}$ be an orthonormal basis for $X$ in which $T$ is diagonalized, that is $Te_i = \lambda_ie_i$ for some $\lambda_i \in \mathbb{R}$. Denote $M = \max\{|\lambda_1|, \ldots, |\lambda_n|\}$.

We have $M \le \sup\big\{|\langle Tx, x\rangle| : x \in X, \|x\| =1\big\}$ since $\left|\langle Te_i, e_i\rangle\right| = |\lambda_i|$.

For any $x\in X, \|x\| =1$ we have: \begin{align}\|Tx\|^2 &= \left\langle \sum_{i=1}^n \langle x, e_i\rangle Te_i,\sum_{i=1}^n \langle x, e_i\rangle Te_i\right\rangle\\ &= \left\langle \sum_{i=1}^n \lambda_i \langle x, e_i\rangle e_i,\sum_{i=1}^n \lambda_i \langle x, e_i\rangle e_i\right\rangle\\ &= \sum_{i=1}^n |\lambda_i|^2 |\langle x, e_i\rangle|^2\\ &\le M^2 \sum_{i=1}^n |\langle x, e_i\rangle|^2\\ &= M^2 \|x\|^2 \end{align}

Thus, $\|T\| \le M$.

We have:

$$\sup\big\{|\langle Tx, x\rangle| : x \in X, \|x\| =1\big\} \le \|T\| \le M \le \sup\big\{|\langle Tx, x\rangle| : x \in X, \|x\| =1\big\}$$

So $\|T\| = \sup\big\{|\langle Tx, x\rangle| : x \in X, \|x\| =1\big\} = M$.

Now in our case, notice that $t\cdot I - A^*A$ is a self-adjoint operator on $\mathbb{R}^n$.

Therefore $$\|t\cdot I - A^*A\| = \max\big\{\left|t - \lambda_1\right|, \ldots, \left|t - \lambda_n\right|\big\}$$

where $\lambda_1, \ldots, \lambda_n$ are eigenvalues of $A^*A$, which are real nonnegative numbers.

As you can see here, there seems to be no simple relationship between the eigenvalues of $A^*A$ and $A$.