Upper bounding the largest singular value of a matrix $X$ via LMI — is it correct?

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For $z \in \Bbb C$ and $\delta > 0$, the inequality $|z| < \delta$ is equivalent to the matrix inequality

$$\begin{bmatrix} -\delta & z\\ z^* & -\delta \end{bmatrix} \prec 0$$

(Source: https://onlinelibrary.wiley.com/doi/pdf/10.1002/9781118577325.app1)

and I believe this because

\begin{align} |z| < \delta &\Leftrightarrow \frac{|z|^2}{\delta^2} < 1 \\ &\Leftrightarrow 1 - \frac{zz^*}{\delta^2} > 0 \\ &\Leftrightarrow -\delta - z(-\delta)^{-1}z^* < 0 \quad (\text{multiplying $-\delta$)} \\ &\Leftrightarrow \begin{bmatrix} -\delta & z\\ z^* & -\delta \end{bmatrix} \prec 0 \quad (\text{by Schur complement formula}). \end{align}

Also, for a matrix $X \in \mathbb{C}^{p\times q}$, we have $\|X\| < 1$ equivalent to

$$\begin{bmatrix} I & X\\ X^* & I\end{bmatrix} \succ 0$$


Question

I want to know if the analogous result for a matrix is true, that is, is the following true?

$$\|X\| < \delta \Leftrightarrow \begin{bmatrix} -\delta I & X\\ X^* & -\delta I \end{bmatrix} \prec 0$$


Attempt

\begin{align} \|X\| < \delta &\Leftrightarrow \frac{\sigma_{max}(X)}{\delta} < 1 \\ &\Leftrightarrow \lambda_{max}\big(\frac{XX^{*}}{\delta^2}\big) < 1 \\ &\Leftrightarrow \delta^2 - \lambda_{max}\big({XX^{*}}\big) > 0 \\ &\Leftrightarrow \lambda_{min}\big(\delta^2 I - XX^{*}\big) > 0\\ &\Leftrightarrow \delta^2 I - XI^{-1}X^{*} \succ 0 \tag{1} \\ &\Leftrightarrow -\delta I - X(-\delta I)^{-1}X^{*} \prec 0 \quad (\text{multiplying $-\frac{1}{\delta}$}) \\ &\Leftrightarrow \begin{bmatrix} -\delta I & X\\ X^* & -\delta I \end{bmatrix} \prec 0 \quad \tag{2}(\text{by Schur complement formula}). \end{align}

However, from page 184 of S. Boyd's Convex Optimization text, (https://web.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf#page184), we can see that the condition $(1)$ above is equivalent to $\begin{bmatrix} \delta I & X\\ X^* & \delta I \end{bmatrix} \succ 0$.

Multiplying by $-1$ from this statement, we have $\begin{bmatrix} -\delta I & -X \\ -X^* & -\delta I \end{bmatrix} \prec 0. \tag{3}$

Is there anything wrong with my derivation? Or can it be that both $(2)$ and $(3)$ are equivalent?