Is it true that product of rank $n$ matrices is at most rank $n$?

Question

The context is when doing $A^TA$ for a matrix $A \in M_{m,n}(\mathcal{R})$ say. Then $A^TA$ is invertible if column of $A$ are independent (I'm not sure about this fact either)?

$A$ is invertible if and only if $A^T$ is. The product $A^TA$ is then a product of invertible matrices, and so is invertible. — David P, Mar 15 '20 at 15:29
@DavidPeterson It doesn't make sense to say that $A$ is invertible when $A$ is not a square matrix — Ben Grossmann, Mar 15 '20 at 15:45

score 0 · Accepted Answer · answered Mar 15 '20 at 16:43

Yes. Since the rank of a matrix $A$ is the dimension of the image of the linear map $m_A: v \mapsto Av$, and matrix multiplication corresponds to composition of linear maps, we have, for $A$ a matrix of rank $n$ and $B$ a matrix of any rank,

\begin{align*}\mathrm{rank}(AB) &= \mathrm{dim}\,\mathrm{Im}(m_{AB}) \\&= \mathrm{dim}\,\mathrm{Im}(m_Am_B)\\&\leq \mathrm{dim}\,\mathrm{Im}(m_A)\\&= \mathrm{rank}(A)\\&=n.\end{align*}

In your particular case, if $A_1,\ldots,A_n$ are matrices of rank $n$, apply the above with $A = A_1$ and $B = \prod\limits_{i=2}^nA_i$ to see that $\mathrm{rank}\left(\prod_iA_i\right)\leq \mathrm{rank}(A_1) = n$.

The same result applies for each other $A_i$ in place of $A_1$.

score 0 · Answer 2 · answered Mar 15 '20 at 23:07

There are straightforward formal proofs of this, but I think its worth developing some intuition about what’s going on by thinking about the situation informally.

The rank of a matrix is equal to the dimension of its column space, which is the image of the linear transformation that it represents. If you apply another linear transformation, can the image get any bigger (i.e., increase in dimension)? For that to happen, the kernel would have to shrink, but that means that the additional transformation would have to map the zero vector to something nonzero, but that violates linearity, among other things.

So, when you compose linear transformations, the kernel can’t be any smaller than the kernel of the first transformation, which also means that the image can’t be any bigger than the image of the first transformation, and thus the rank of a product of two matrices is at most the rank of the first one. It’s not too hard to convince yourself that the rank of the product is also no greater than the rank of the second matrix, i.e., $\operatorname{rank}(AB)\le\min\{\operatorname{rank}(A),\operatorname{rank}(B)\}$. It’s a worthwhile exercise to work out when the rank of the product is less than the rank of both matrices (hint: consider the intersection of their null spaces).

Is it true that product of rank $n$ matrices is at most rank $n$?

2 Answers2