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

  1. $l^{2}$
  2. $\frac{l^{2}}{4}$
  3. $l$
  4. $\frac{l^{2}}{2}$
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Let $\mathbf{A} = \begin{bmatrix} A_{11} \\ A_{21} \\ A_{31} \\ \vdots \\ A_{N1} \end{bmatrix}$ be a column vector.

The product $\mathbf{A}^\top \mathbf{A}$ (transpose of the column vector multiplied by the column vector) results in:

\[
\mathbf{A}^\top \mathbf{A} = \begin{bmatrix} A_{11} & A_{21} & A_{31} & \cdots & A_{N1} \end{bmatrix} \begin{bmatrix} A_{11} \\ A_{21} \\ A_{31} \\ \vdots \\ A_{N1} \end{bmatrix} = \begin{bmatrix}  A_{11}^2 + A_{21}^2 + A_{31}^2 + \cdots + A_{N1}^2\end{bmatrix}
\]

hence $l=\sqrt{A_{11}^2 + A_{21}^2 + A_{31}^2 + \cdots + A_{N1}^2}$

The product $\mathbf{A} \mathbf{A}^\top$ (column vector multiplied by its transpose) results in the matrix:

\[
\mathbf{A} \mathbf{A}^\top = \begin{bmatrix} A_{11} \\ A_{21} \\ A_{31} \\ \vdots \\ A_{N1} \end{bmatrix} \begin{bmatrix} A_{11} & A_{21} & A_{31} & \cdots & A_{N1} \end{bmatrix} =
\begin{bmatrix}
    A_{11} \cdot A_{11} & A_{11} \cdot A_{21} & A_{11} \cdot A_{31} & \cdots & A_{11} \cdot A_{N1} \\
    A_{21} \cdot A_{11} & A_{21} \cdot A_{21} & A_{21} \cdot A_{31} & \cdots & A_{21} \cdot A_{N1} \\
    A_{31} \cdot A_{11} & A_{31} \cdot A_{21} & A_{31} \cdot A_{31} & \cdots & A_{31} \cdot A_{N1} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    A_{N1} \cdot A_{11} & A_{N1} \cdot A_{21} & A_{N1} \cdot A_{31} & \cdots & A_{N1} \cdot A_{N1} \\
\end{bmatrix}
\]

Trace is the sum of diagonal elements $P=A_{11}^2 + A_{21}^2 + A_{31}^2 + \cdots + A_{N1}^2$

$P=l^2$

Answer: