GATE 2023 EC Q.38: solution and answer

Official answer: ASource: Official answer key, GATE 2023 EC (IIT Kanpur)

1. Write $\mathbf{x}={\left[\begin{array}{cccc}{x}_1& x_2& \cdots & x_n\end{array}\right]}^T$ with real entries. The outer product $P=\mathbf{x}{\mathbf{x}}^T$ is an $n\times n$ matrix whose entries are $P_{ij}=x_i{x}_j$.

2. The trace is the sum of the diagonal entries: $\mathrm{tr}(P)={\sum }_{i=1}^n{P}_{ii}={\sum }_{i=1}^n{x}_i^2$.

3. The inner product is also ${\mathbf{x}}^T\mathbf{x}={\sum }_{i=1}^n{x}_i^2$, so $\mathrm{tr}(\mathbf{x}{\mathbf{x}}^T)={\mathbf{x}}^T\mathbf{x}$. The cyclic property $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ gives this directly: $\mathrm{tr}(\mathbf{x}{\mathbf{x}}^T)=\mathrm{tr}({\mathbf{x}}^T\mathbf{x})$, and ${\mathbf{x}}^T\mathbf{x}$ is a $1\times 1$ matrix, so its trace is the number itself.

4. The length is defined as $l=\sqrt{{\mathbf{x}}^T\mathbf{x}}$, so ${\mathbf{x}}^T\mathbf{x}=l^2$.

5. Therefore $$\mathrm{tr}(P)=l^2,$$ which is option (A).

6. Check: take $\mathbf{x}=\left[\begin{array}{c}3\\ 4\end{array}\right]$, so $l=\sqrt{9+16}=5$. Then $P=\left[\begin{array}{cc}9& 12\\ 12& 16\end{array}\right]$ and $\mathrm{tr}(P)=9+16=25=l^2$.