# 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,

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