1,455 views

Let $x$ and $y \in \mathbb{R}^{n}$ be non-zero column vectors, from the matrix $A=xy^{T}$, where $y^{T}$ is the transpose of $y$. Then the rank of $A$ is:

1. $2$
2. $0$
3. At least $n/2$
4. None of the above

Suppose  $\vec x = \begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}$, and $\vec y = \begin{bmatrix}y_1\\y_2\\\vdots\\y_n\end{bmatrix}$.

Then the column vectors of matrix $xy^T$ are: $$\begin{matrix}\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}\times y_1, && \begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix} \times y_2, && \ldots && \begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix} \times y_n \end{matrix}$$

So every column vector of matrix is a multiple of vector $x$, or we can say that every column vector of matrix is multiple of first column vector $y_1 \times \begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}$

Hence there is only one independent column vector in the matrix, and so rank is one.

So option (d) is correct.

### 1 comment

Thanks for the edit Pragy :)

1
495 views