Find $\alpha$ that minimizes $\|(I-\alpha H)^2 A\|$ Suppose $A,H$ are positive definite matrices. How do I find $\alpha$ which minimizes the following?
$$\|(I-\alpha H) A(I-\alpha H)^T\|_\text{op}$$
It can be written as minimizing linear function with semidefinite cone constraint, however this constraint has an $\alpha^2$ entry, so SDP solvers won't take it.
$$\begin{align}
\text{minimize}_{\alpha,t}\ & t \\
\text{subject to } & 
A-\alpha AH - \alpha HA + \alpha^2 HAH \prec t I
\end{align}$$
For matrices below I can use visual inspection to check that $\alpha\approx 3.2$
$$\text{H=}\left(
\begin{array}{cc}
 4 & 0 \\
 0 & 1 \\
\end{array}
\right)\\
\text{A=}\left(
\begin{array}{cc}
 11 & 9 \\
 9 & 11 \\
\end{array}
\right)
$$

 A: This has already been noted in the comments above but for completeness here is the SDP solving the above problem
$$
\begin{aligned}
\min_{t,\alpha} & \quad t \\
\mathrm{s.t.} & \quad \begin{pmatrix} t\, \mathbb{I} & (\mathbb{I} - \alpha H) \\
(\mathbb{I} - \alpha H) & A^{-1} \end{pmatrix} \succeq 0
\end{aligned}
$$
This SDP representation can be derived in two steps. Firstly note that the SDP
$$
\begin{aligned}
\min_{t} & \quad t \\
\mathrm{s.t.} & \quad t\, \mathbb{I} - X \succeq 0
\end{aligned}
$$
is equal to $\|X\|_{\mathrm{op}}$ (largest eigenvalue of $X$) for any positive semidefinite $X$. Secondly we have by the Schur complement form of block PSD matrices
$$
\begin{pmatrix}
A & B \\
B^T & C
\end{pmatrix} \succeq 0 \quad \iff \quad C \succeq 0, \quad A - B C^{-1} B^T \succeq 0, \quad (\mathbb{I} - C C^{-1} ) B^T = 0\,.
$$
Using this result we see that the block positive semidefinite constraint in the first SDP is equivalent to the constraint
$$
t\,\mathbb{I} - (\mathbb{I}- \alpha H) A (\mathbb{I} - \alpha H) \succeq 0
$$
and hence the first SDP is computing $\min_{\alpha} \|(\mathbb{I}- \alpha H) A (\mathbb{I} - \alpha H)\|_{\mathrm{op}}$.
