# Some basic questions regarding rank-1 matrices

If an $n\times n$ matrix $B$ has rank 1, and A is another $n\times n$ matrix, then why does $AB$ also have rank 1? This showed up in a solution that I read through, but it doesn't seem like an obvious fact.

And one more thing that came up in this solution: it says that since this matrix has rank 1, then it must have $(n-1)$ eigenvalues that are all zero, and only one non-zero eigenvalue. I don't see how this has to be true either.

Any ideas are welcome.

Thanks,

## 4 Answers

Okay, we can show this in different ways, depending on how we define rank. I'll try to show these using both these definitions:

• $$\operatorname{rank}(A)$$ is the dimension of the range of $$A$$.
• $$\operatorname{rank}(A)$$ is the least number $$r$$ of rank 1 matrices $$A_1, A_2, \dots, A_r$$ so that $$A = A_1 + A_2 + \dots + A_r$$ and a rank one matrix $$A_1$$ is defined as a matrix that can be written as an outer product: $$A_1 = xy^T$$ for vectors $$x, y$$.

These two definitions are equivalent. The proof of this is left as an exercise to the reader.

## If $$B$$ has rank 1, then $$AB$$ has at most rank one

Fixed the formulation for you on this one. $$AB$$ can have rank zero.

## Definition 1

$$B$$ has rank one, so its range is one dimensional. It follows that its nullspace is $$n-1$$-dimensional. Consider the range of $$AB$$. If $$x$$ is in the nullspace of $$B$$, then $$ABx = A0 = 0$$, so $$AB$$'s nullspace is also at least $$n-1$$-dimensional, so its range is at most 1-dimensional.

Note that for $$AB$$ to have rank 1 you must have that $$Ax \neq 0$$ for some $$x$$ in the range of $$B$$. A sufficient, but not necessary, condition for this is that $$A$$ is invertible.

## Definition 2

$$B$$ has rank one, so it can be written $$B = xy^T$$. Then $$AB = Axy^T = (Ax)y^T$$, so $$AB$$ can be written as one outer product (but we don't know if it can be written using zero outer products, which would be the case if $$Ax = 0$$).

## Definition 1

As said before, if $$B$$ has rank 1, then its nullspace is $$n-1$$ dimensional. Pick a basis $$v_1, \dots, v_{n-1}$$ for the nullspace of $$B$$. These are eigenvectors of $$B$$ with eigenvalue zero.

## Definition 2

Say $$B = xy^T$$. Then if you multiply a vector $$v$$ with $$B$$ you get $$xy^Tv$$. Note that $$y^Tv$$ is the inner product of $$y$$ and $$v$$. The orthogonal complement $$\mathcal U$$ to the subspace spanned by $$y$$ has dimension $$n-1$$. If $$u \in \mathcal U$$ then $$y^Tu = 0$$ and hence $$Bu = xy^Tu = x \cdot 0 = 0$$ so if you pick a basis for $$\mathcal U$$, this basis will be eigenvectors to $$B$$ with eigenvalue 0.

## Is there a non-zero eigenvalue?

In short, it is not guaranteed that there will be a non-zero eigenvalue.

This rank one nilpotent matrix: $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ has zero as eigenvalue with algebraic multiplicity 2 and geometric multiplicity 1. It therefore has $$n-1$$ zero eigenvalues, but it does not have 1 non-zero eigenvalue. In other words, it is not guaranteed that you will have a non-zero eigenvalue for rank one matrices.

• @User001, yes, it would, I mentioned this under "Definition 1" under the section dealing with $AB$ having rank at most one. Dec 12 '15 at 15:18
• @User001, since $B$ has rank one, there exists an $x$ such that $Bx = y \neq 0$. If $A$ is invertible we know that $Ay \neq 0$, so $AB$ has at least rank one, but we know that $AB$ has at most rank one, so $1 \leq \operatorname{rank}(AB) \leq 1$ so the rank is one. If $A$ would not be invertible, we would not know that $Ay \neq 0$. Dec 12 '15 at 15:40
• Sorry, I meant $AB$. :) Dec 12 '15 at 15:43
• Ok, got it. Thanks so much for your truly awesome solution, additional comments, and especially making many of us on this page aware of our mistake of assuming that a rank-1 matrix had to have a non-zero eigenvalue. I really appreciate it. Have a great night @Calle :-) Dec 12 '15 at 15:53
• @User001, Thank you for asking a question I enjoyed answering. :) Dec 12 '15 at 15:54

It comes from the associavity of matrix multiplication. If $B$ has rank-1 then it can be written in the form $B = u v^T$ for some vectors $u$ and $v$. So, $$AB = A (uv^T) = (Au) v^T$$ but $Au$ is a vector itself, so now we have a rank-1 expression for $AB$. Maybe it will be helpful to see a picture of the shapes of the matrices during this process:

$$\underbrace{\begin{bmatrix}\cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot\end{bmatrix}}_{A} \left(\underbrace{\begin{bmatrix}\cdot \\ \cdot \\ \cdot\end{bmatrix}}_{u}\underbrace{\begin{bmatrix}\cdot & \cdot & \cdot\end{bmatrix}}_{v^T}\right)= \left(\underbrace{\begin{bmatrix}\cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot\end{bmatrix}}_{A}\underbrace{\begin{bmatrix}\cdot \\ \cdot \\ \cdot\end{bmatrix}}_{u}\right)\underbrace{\begin{bmatrix}\cdot & \cdot & \cdot\end{bmatrix}}_{v^T} = \underbrace{\begin{bmatrix}\cdot \\ \cdot \\ \cdot\end{bmatrix}}_{Au}\underbrace{\begin{bmatrix}\cdot & \cdot & \cdot\end{bmatrix}}_{v^T}$$

(Of course, note that if $Au=0$, then the result is the zero matrix so the rank would be zero instead of 1)

• What if $Au = 0$? Dec 12 '15 at 22:31
• Then the rank is zero and the statement in the original question is false, obviously... Dec 12 '15 at 22:33
• I was pointing out that your approach seems to indicate that the rank of $AB$ is always 1, or at least that is how I read it. Dec 12 '15 at 22:47
• @JacobMaibach Ok, I added a sentence to explicitly mention this. Dec 13 '15 at 1:19
• Hi @NickAlger, your answer is awesome. So simple and intuitive to understand. Thanks so much :-) Dec 13 '15 at 5:29

Concerning the first point, remember $A$ and $B$ represent endomorphisms $f$ and $g$ in some basis, and $AB$ represent $f\circ g$.

Now $\;\DeclareMathOperator\rk{rank}\DeclareMathOperator\img{Im}\rk B=1$ means $\;\dim\img g=1$. Then we know $\;\dim\img(f\circ g)\le\dim\img g$, which means $\;\rk AB\le\rk B$. So $\rk AB=0$ or $1$.

For the second point, if $AB$ has indeed rank $1$, its kernel has dimension $n-1$. As the kernel is the eigenspace for the eigenvalue $0$, and has dimension at most equal to the multiplicity of this eigenvalues, it implies the eigenvalue $0$ has multiplicity at least $n-1$, and it can't have multiplicity $n$ as it would imply $\rk AB=0$.

• Hi @Bernard, I apologize for probably withholding a key piece of information that I just found from rereading that solution. A is invertible. Does that now make $AB$ have rank 1, for certain? Thanks, Dec 12 '15 at 14:41
• Yes: if $A$ is invertible, $\operatorname{rank}AB=\operatorname{rank}B$, and similarly , $\operatorname{rank}BA=\operatorname{rank}B$. Dec 12 '15 at 19:16

The rank of a matrix is the dimension of its image ${\rm rank} \ B = \dim \{Bx: x \in \mathbb R^n\},$ or equivalently the dimension of the space spanned by its columns.

Since applying $AB$ on $\mathbb R^n$ is the same as applying $A$ on the one-dimensional space $B\mathbb R^n$, we see that $AB$ can only have at most rank 1. Note that rank 0 is possible when the columns of $B$ are orthogonal to the rows of $A$.

We have the rank-nullity theorem $n = {\rm rank} \ B + \dim \ker B$ (i.e. what doesn't contribute to the dimensions of the image must land in the kernel.

If ${\rm rank} \ B = 1$, then the dimension of the kernel must be $n-1$, i.e. the eigenspace for the eigenvalue is $n-1$. We can deduce that there is also another nonzero eigenvalue since otherwise we'd have ${\rm rank} \ AB = 0$.

• Hi @Roland, the matrix $A$ is actually assumed to be invertible. I reread the solution and noticed that. Does the invertibility of $A$ make $AB$ have rank 1, for certain? Thanks so much, Dec 12 '15 at 14:58
• "We can deduce that there is also another nonzero eigenvalue since otherwise we'd have rank B = 0." This is not true. Dec 12 '15 at 15:14
• @User001: Yes, because then $A^{-1}(AB) = B$ has rank 1, and so $AB$ must have rank at least 1. Dec 12 '15 at 15:23
• HI @IlmariKaronen, how does your equation imply that $AB$ must have rank at least 1? It seems like you are using the invertibility of $A^{-1}$, but I'm not sure how...thanks, Dec 12 '15 at 15:28
• @User001: By the same argument given by Roland above, the product of two matrices cannot have a higher rank than either of the matrices alone. Since we know that $A^{-1}(AB) = (A^{-1}A)B = B$ has rank 1, this means that both $A^{-1}$ and $AB$ must have rank at least 1. (Of course, for $A^{-1}$, we actually know more; namely that, as invertible matrices, both $A$ and $A^{-1}$ in fact have full rank. In fact, we can prove this simply by replacing $B$ in the previous argument with, say, the identity matrix. But for the proof that $AB$ has rank 1, we don't really need this extra information.) Dec 12 '15 at 16:04