# Multiplying quadratic form using block matrix multiplication

My textbook gives the following example:

The second-order cone is the norm cone for the Euclidean norm, i.e.,

\begin{align} C &= \{ (x, t) \in \mathbb{R}^{n + 1} \mid \left\lVert x \right\rVert_2 \le t \} \\ &= \left\{ \begin{bmatrix} x \\ t \end{bmatrix} \;\middle|\; \begin{bmatrix} x \\ t \end{bmatrix}^T \begin{bmatrix} I & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ t \end{bmatrix} \le 0, t \ge 0 \right\} \end{align}

The second-order cone is also known by several other names. It is called the quadratic cone, since it is defined by a quadratic inequality. It is also called the Lorentz cone or ice-cream cone.

I'm wondering how to multiply out $$\begin{bmatrix} x \\ t \end{bmatrix}^T \begin{bmatrix} I & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ t \end{bmatrix}$$

It seems to me that I need to use block matrix multiplication (assuming $$I$$ is the identity matrix), but even after reviewing examples (see this and Wikipedia), I keep getting the matrix $$\begin{bmatrix} x^2 I - t^2 \end{bmatrix}$$ as my final result, which doesn't make sense, since $$x^2 I$$ is an invalid matrix multiplication.

I would greatly appreciate it if people could please take the time to demonstrate this.

$$\newcommand{\R}{\mathbb{R}}$$Let $$\mathbf{0}\in\R^n$$ be a column of all $$0$$'s, so $$\mathbf{0}^T$$ denotes a row of $$n$$ zeros. Note that by block matrix multiplication, we have $$\begin{bmatrix} I & \mathbf{0} \\ \mathbf{0}^T & -1\end{bmatrix}\begin{bmatrix} \mathbf{x} \\ t\end{bmatrix} = \begin{bmatrix}I\mathbf{x} + \mathbf{0} t\\ \mathbf{0}^T \mathbf{x} +(-1)t\end{bmatrix} = \begin{bmatrix}\mathbf{x} \\ -t\end{bmatrix}\in \R^{n+1}.$$ Also, $$\begin{bmatrix}\mathbf{x} \\ t\end{bmatrix}^T = \begin{bmatrix}\mathbf{x}^T & t\end{bmatrix}\in\R^{1\times (n+1)}$$ and so

$$\color{blue}{\begin{bmatrix}\mathbf{x} \\ t\end{bmatrix}^T \begin{bmatrix}\mathbf{x} \\ -t\end{bmatrix} = \begin{bmatrix}\mathbf{x}^T & t\end{bmatrix} \begin{bmatrix}\mathbf{x} \\ -t\end{bmatrix} = \mathbf{x}^T \mathbf{x} - t^2}.$$

(It looks like your error was maybe in the above step.)

Combining these facts tells us that $$\boxed{\begin{bmatrix}\mathbf{x} \\ t\end{bmatrix}^T \begin{bmatrix} I & \mathbf{0} \\ \mathbf{0}^T & -1\end{bmatrix}\begin{bmatrix} \mathbf{x} \\ t\end{bmatrix} = \mathbf{x}^T \mathbf{x} - t^2}.$$

Note that this means the inequality $$\begin{bmatrix}\mathbf{x} \\ t\end{bmatrix}^T \begin{bmatrix} I & \mathbf{0} \\ \mathbf{0}^T & -1\end{bmatrix}\begin{bmatrix} \mathbf{x} \\ t\end{bmatrix} \le 0$$ becomes $$\mathbf{x}^T \mathbf{x} - t^2\le 0,$$ which is equivalent to $$\left\|\mathbf{x}\right\|_2 \le t$$ (for $$t\ge 0$$), as was claimed.

• Thanks for the answer. Can you please explain why we require that $\mathbf{0}^T$ denote a row of $n$ zeroes? – The Pointer Apr 5 at 23:31
• That is the meaning of the lower-left "0" in your original post's matrix (becausethe overall matrix is $(n+1)\times (n+1)$. Also the other (top-right) 0 in that matrix refers to a column of $n$ zeros. It's a good idea to keep track of the dimensions of things in these block matrix problems. – Minus One-Twelfth Apr 5 at 23:51
• Ahh, I didn’t realize that. So the author has already partitioned the matrix correctly to allow for easy computation? – The Pointer Apr 5 at 23:52
• Correct! $\ddot{\smile}$ – Minus One-Twelfth Apr 5 at 23:53