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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?

Chav Likit
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  • Did you mean a and b are column vectors? I don't think $Xa=b$ makes sense if a and b are row vectors and X is an $n \times n$ matrix. Also check out https://math.stackexchange.com/questions/1151491/under-what-conditions-is-aat-invertible – Theo C. Aug 01 '17 at 02:42
  • Sorry, you're right. I have fixed the post. Thanks. – Chav Likit Aug 01 '17 at 02:44
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    AS @SiongThyeGoh notes, the matrix $A$ has rank $1$, hence is not invertible except in the case $n = 1$. On the other hand, the equation $Xa =b$ generally doesn't have a solution, either, so ... – John Hughes Aug 01 '17 at 02:48

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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.

Siong Thye Goh
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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.

Anders Beta
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