Express $\|y\|_2 \leq t$ as a linear matrix inequality (LMI) For $y \in \mathbf{R}^n$ and $t \in \mathbf{R}$, show that:
$$\|y\|_2 \leq t ~~\iff~~ F(y)
\succeq 0$$
Where $\mathrm{I}$ is the $n \times n$ identity matrix, and
$$F(y) = \begin{pmatrix}
t      & y_1 & ...    & y_n \\ 
y_1    & t   &        & \\ 
\vdots &     & \ddots & \\ 
y_n    &     &        & t 
\end{pmatrix} = \begin{pmatrix}
t & y^T \\
y & t~\mathrm{I}
\end{pmatrix}$$
I came across this problem on a few sites/books, but couldn't readily find a solution. I've posted my answer below in the hopes that it might be useful to someone (assuming it is correct).
 A: From $\| \mathrm y \|_2 \leq t$, we obtain $\| \mathrm y \|_2^2 \leq t^2$, or, 
$$t^2 - \mathrm y^\top \mathrm y \geq 0$$
Dividing both sides by $t > 0$, we get
$$t - \mathrm y^\top \left( t \,\mathrm I_n \right)^{-1} \mathrm y \geq 0$$
and, using the Schur complement, we finally get the following LMI
$$\begin{bmatrix} t \, \mathrm I_n & \mathrm y\\ \mathrm y^\top & t \end{bmatrix} \succeq \mathrm O_{n+1}$$
A: Applying the Schur Complement Lemma, we see that $F(y)$ is positive semi-definite if and only if, first, $t \geq 0$ (which is implied from the original inequality since $||y||_2 \geq 0$) and second:
$$t~\text{I} - y \left ( \frac{1}{t}~\text{I} \right ) y^T \succeq 0 \quad~~ \text{or equivalently} ~~\quad t~\text{I} - \frac{y ~ y^T}{t} \succeq 0$$
Now note that $y~y^T$ is a rank-one matrix with the only eigenvalue $y^T y$ or $||y||_2^2$:
$$(y~y^T)y = y(y^Ty)= ||y||_2^2~y$$
Also note that scaling this matrix by $1/t$ scales the eigenvalue by the same amount, and adding $t \text{I}$ shifts all eigenvalues by $t$. Thus, the matrix $t~\text{I} - \left ( 1/t \right ) \left (y ~ y^T \right )$ has eigenvalues:
$$\lambda_1 = t - \frac{y^T y}{t}$$
$$\lambda_{2,...,n} = t $$
Thus, $F(y)$ is positive semi-definite when (in addition to $t \geq 0$):
$$ t - \frac{y^Ty}{t} \geq 0 $$
$$ t^2 \geq y^Ty $$
$$ ||y||_2 \leq t $$
Where we have used the fact that $t \geq 0$ to multiply both sides of the inequality without flipping the sign.
