Inverse of the matrix $aa^T$, where $a$ is a column vector?

Question

I have the following matrix equation $$Xa=b$$ where $X$ is an $n\times n$ matrix, and $a$ and $b$ column vectors. We can multiply both sides by $a^T$ to get $$XA = ba^T$$ where $A = aa^T$. Now, I'm wondering whether the matrix $A$ has an inverse. If so, how to prove it?

Answer 1

Unless $n=1$ and $a \neq 0$, it does not have an inverse, the rank of matrix $aa^T$ is rank $1$ if $a \neq 0$.

To see this clearly suppose $a= \begin{bmatrix} a_1 \\ \vdots \\a_n\end{bmatrix}$, $aa^T=\begin{bmatrix} a_1 a & \ldots & a_n a\end{bmatrix}.$ We can see that the columns are multiples of $a$.

Remark:

Now suppose you want to solve $Xa=b$ given $a$ and $b$. if $a=0$ and $b \neq 0$, then there is no solution. if both $a=0$ and $b=0$, then $X$ can be anything.

Suppose $a_i \neq 0$, let $X_i$ be the $i$-th column of $X$, then let $X_i = \frac{b}{a_i}$ and $X_j = 0, \forall j \neq i$, would be a feasible solution.

Answer 2

Assuming that you are working in $\mathbb R^n$ and $a\ne0$, the matrix $aa^T$ is the projector to the subspace generated by the vector $a$. Any vector orthogonal to $a$ will be an eigenvector of $aa^T$ with eigenvalue zero. So unless you are working in one dimension the matrix is not invertible.