Can an upperbound constraint on the squared Frobenius norm of a matrix be expressed as a linear matrix inequality? Is it possible to express an inequality constraint on the squared Frobenius norm of a matrix $X$:
$$\|X\|_F^2 = \mathop{tr}( X^T X ) \le t$$
as a linear matrix inequality?
I want to say that it's:
$$\left[\begin{array}{cc} I & X \\ X^T  & tI\end{array}\right] \succcurlyeq  0$$
but I've lost confidence that this is correct since the Schur complement would be $tI - X^TX \succcurlyeq 0 $ and I can't figure how the trace gets in there.
 A: There might be a compacter and more elegant way, but one way you can represent it as LMI's is by using intermediate values for each diagonal term of $X^\top X$. These can be calculated using
$$
Y_i = X\,e_i
$$
with $e_i$ a vector with the $i$th element equal to one and the rest zeros (so $Y_i$ is the $i$th column of $X$), such that $Y_i^\top Y_i$ is the $i$th diagonal term of $X^\top X$. Then using the Schur complement you can write for every diagonal term an LMI for $Y_i^\top Y_i \leq \alpha_i$, namely
$$
\begin{bmatrix}
I & Y_i \\ Y_i^\top & \alpha_i
\end{bmatrix} \succeq 0,
$$
with $\alpha_i \in \mathbb{R}$. Now a bound for $\|X\|_F^2$ can be found by summing all $\alpha_i$, which should be smaller or equal to $t$
$$
\sum \alpha_i \leq t,
$$
which is also a linear inequality.

By using an intermediate LMI for $X^\top X$ you might also be able to write $X^\top X \preceq M$, with $M = M^\top$, as
$$
\begin{bmatrix}
I & X \\ X^\top & M
\end{bmatrix} \succeq 0.
$$
An upper bound for $\|X\|_F^2$ would then be $\text{Tr}(M)$, so adding the linear inequality $\text{Tr}(M) \leq t$ would make this system of LMI's equivalent to your problem. To show that $X^\top X \preceq M$ also implies that $\text{Tr}(X^\top X) \leq \text{Tr}(M)$ you can use that the trace of a matrix is equal to the sum of all its eigenvalues. Namely $X^\top X \preceq M$ is equivalent to $M - X^\top X \succeq 0$, thus $M - X^\top X$ can only have non-negative eigenvalues and therefore $\text{Tr}(M - X^\top X)$ is the sum of these non-negative eigenvalues, which is also non-negative. The trace inequality $\text{Tr}(X^\top X) \leq \text{Tr}(M)$ is equivalent to $\text{Tr}(M - X^\top X) \geq 0$ and in the previous sentence it was shown that it holds when $X^\top X \preceq M$, thus $\text{Tr}(X^\top X) \leq \text{Tr}(M)$ should hold in that case as well.
It can be noted that $M$ might add more degrees of freedom than all $\alpha_i$ so might or might not be an attractive alternative of writing the problem as LMI's.
