Question

Let $\mathbf{x}$ be an $n \times 1$ real column vector with length $l=\sqrt{\mathbf{x}^T\mathbf{x}}$. The trace of the matrix $P=\mathbf{x}\mathbf{x}^T$ is _____________

• A. $l^2$
• B. $\dfrac{l^2}{4}$
• C. $l$
• D. $\dfrac{l^2}{2}$

Hint:

For any vectors $a\in\mathbb{R}^{n\times 1}$ and $b\in\mathbb{R}^{n\times 1}$, $\operatorname{tr}(ab^T)=b^Ta$. In particular, $\operatorname{tr}(\mathbf{x}\mathbf{x}^T)=\mathbf{x}^T\mathbf{x}=\|\mathbf{x}\|^2$.

Answer 1

The Correct Option is A

Step 1: Use trace cyclicity. $\operatorname{tr}(P)=\operatorname{tr}(\mathbf{x}\mathbf{x}^T)=\operatorname{tr}(\mathbf{x}^T\mathbf{x})$.
Step 2: Reduce to a scalar. $\mathbf{x}^T\mathbf{x}$ is the sum of squares of the components of $\mathbf{x}$, i.e., $\|\mathbf{x}\|_2^2=l^2$.
Therefore, $\operatorname{tr}(P)=\mathbf{x}^T\mathbf{x}=l^2$.
\[ \boxed{l^2} \]