Rank of the given Matrix

**Chandru1** · January 21st 2010, 11:55 PM

Let $\bf{x}$ and $\bf{y} \in \mathbb{R}^{n}$ be 2 non zero column vectors. Let $\bf{y}^{T}$ denote the transpose of $\bf{y}$ . Let $A=\bf{x} \bf{y}^{T}$ . What is the rank of A?

**Roam** · January 22nd 2010, 01:16 AM

What do you mean by $y^T$ , because we don't know what "y" is! I think maybe you're referring to the case where x=y. And I think in this case $xx^T$ will have the full column rank if $det(x^Tx) \neq 0$ , and it will have the full row rank if $det(xx^T) \neq 0$ .

Since $x^Tx$ is square, and it is an nxn matrix, it follows that $x^Tx$ is invertible if and only if $x^Tx$ has rank n. However $x^Tx$ has the same rank as x itself, so $x^Tx$ is invertible if rank(x)=n (this means if x has full column rank).

Again I'm assuming x=y.

**HallsofIvy** · January 22nd 2010, 04:13 AM

I see no reason to assume that x= y. The problem makes sense without that. For example, if $x= \begin{bmatrix}a \\ b\end{bmatrix}$ and $y= \begin{bmatrix}c \\ d\end{bmatrix}$ , then $y^T= \begin{bmatrix}c & d\end{bmatrix}$ and
$xy^T= \begin{bmatrix}a \\ b\end{bmatrix}\begin{bmatrix}c & d\end{bmatrix}=$ $\begin{bmatrix}ac & ad \\ bc & bd\end{bmatrix}$ . The determinant of that is is acbd- adbc= 0 so it does NOT have rank 2. It is not, unless at least one of x and y is the 0 vector, all "0"s so it does not have rank 0. It has rank 1.

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