Prove that $\bigg\| \begin{bmatrix} X \\ A\end{bmatrix} \bigg\|_2 \leq \sigma \iff X^* X + A^* A \preceq \sigma^2 I$

Question

Prove that

$$\bigg\| \begin{bmatrix} X \\ A\end{bmatrix} \bigg\|_2 \leq \sigma \iff X^* X + A^* A \preceq \sigma^2 I$$

Here, * denotes the conjugate transpose. This norm is the $2$-induced matrix norm. This equivalence is a part of other proof and I don't know how to start proving this equivalence.

obs: No need for a formal proof, the simply understanding of this inequality is enough for me.

The context from which I took this question is:

Answer 1

Short Answer: \begin{align} \left\|\begin{bmatrix} X\\A \end{bmatrix} \right\|^2_2 := \lambda_{max}\left(\begin{bmatrix} X\\A \end{bmatrix}\begin{bmatrix} X^\mathrm{*}&A^\mathrm{*} \end{bmatrix} \right) = \lambda_{max}\left(\begin{bmatrix}X^\mathrm{*} & A^\mathrm{*} \end{bmatrix}\begin{bmatrix} X\\A \end{bmatrix} \right) = \lambda_{max}\left(X^{*}X + A^*A \right) \end{align} Now notice $\lambda_{max}\left(X^{*}X + A^*A \right) \leq \sigma^2 \Leftrightarrow \left(X^{*}X + A^*A \right)-\sigma^2 I \preceq 0 \Leftrightarrow X^{*}X + A^*A \preceq \sigma^2 I$

Longer Answer: \begin{align} \left\|\begin{bmatrix} X\\A \end{bmatrix} \right\|^2_2 := \lambda_{max}\left(\begin{bmatrix} X\\A \end{bmatrix}\begin{bmatrix} X^\mathrm{*}&A^\mathrm{*} \end{bmatrix} \right) = \lambda_{max}\left(\begin{bmatrix}X^\mathrm{*} & A^\mathrm{*} \end{bmatrix}\begin{bmatrix} X\\A \end{bmatrix} \right) = \lambda_{max}\left(X^{*}X + A^*A \right) \end{align} Where $\lambda_{max}\left(AA^* \right) = \lambda_{max}\left(A^*A \right)$ can be verified by looking at the SVD of $A=U\Sigma V^{*}$ and noting that $\lambda_i(AA^*) = \lambda_i(U\Sigma^2 U^*) = \lambda_i(V\Sigma^2 V^*) = \lambda_i(A^*A)$

Now notice: \begin{align} \lambda_{max}\left(X^{*}X + A^*A \right) \leq \sigma^2 &\Leftrightarrow \lambda_{max}\left(X^{*}X + A^*A \right)- \sigma^2 \leq 0 \\ &\Leftrightarrow \lambda_{max}\left(X^{*}X + A^*A -\sigma^2 I\right) \leq 0 \\ & \Leftrightarrow \left(X^{*}X + A^*A \right)-\sigma^2 I \preceq 0 \\ & \Leftrightarrow X^{*}X + A^*A \preceq \sigma^2 I \end{align}

Answer 2

Note that \begin{align*} \gamma & \geq\left\Vert \left[\begin{array}{c} X\\ A \end{array}\right]\right\Vert \\ & =\sup_{\left\Vert x\right\Vert =1}\left\Vert \left[\begin{array}{c} X\\ A \end{array}\right]x\right\Vert \\ & =\sup_{\left\Vert x\right\Vert =1}\sqrt{\left(\left[\begin{array}{c} X\\ A \end{array}\right]x\right)^{*}\left[\begin{array}{c} X\\ A \end{array}\right]x}\\ & =\sup_{\left\Vert x\right\Vert =1}\sqrt{x^{*}\left[\begin{array}{cc} X^{*} & A^{*}\end{array}\right]\left[\begin{array}{c} X\\ A \end{array}\right]x}\\ & =\sup_{\left\Vert x\right\Vert =1}\sqrt{x^{*}X^{*}Xx+x^{*}A^{*}Ax}.\\ & =\sup_{\left\Vert x\right\Vert =1}\sqrt{x^{*}\left(X^{*}X+A^{*}A\right)x}. \end{align*} Therefore, $$ \gamma\geq\sqrt{x^{*}\left(X^{*}X+A^{*}A\right)x}\text{ for all }\left\Vert x\right\Vert =1.\tag{*} $$ Squaring both sides, $$ \gamma^{2}\geq x^{*}\left(A^{*}A+X^{*}X\right)x\text{ for all }\left\Vert x\right\Vert =1. $$ Now, if $x$ is not a unit vector, we can normalize to get $$ \gamma^{2}x^{*}Ix=\left\Vert x\right\Vert ^{2}\gamma^{2}\geq x^{*}\left(A^{*}A+X^{*}X\right)x. $$ Thus, we have proven $\gamma^{2}I\geq A^{*}A+X^{*}X$. The reverse direction is essentially the same (start with inequality $(*)$ and take supremums of both sides).