# $AB$ is not invertible

Is it true that if $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix, with $m > n$ then $\det(AB)=0$?

• Hint: What rank could $AB$ maximally have? – Daniel Fischer Aug 31 '13 at 20:15
• This is true, even when $n=1$, as Daniel Fischer points out. Notice that $rank(AB)\le rank(A)\le n<m$, so AB can't be invertible. – user84413 Aug 31 '13 at 20:20
• Mapping from $\mathbb{R}^m$ to $\mathbb{R}^n$ inherently must lose information. You can't simply recover that with another map $\mathbb{R}^n$ to $\mathbb{R}^m$ -- the map $AB$ must lose at least as much information as $A$ does and so it is only intuitive it is singular. – oldrinb Aug 31 '13 at 20:25
• @Twink O.K., let $$A = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & 4 \end{bmatrix}.$$ Then $$AB = \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix},$$ so $\det(AB) = 3 \cdot 8 - 6 \cdot 4 = 24 - 24 = 0$. Not really a counterexample, is it? – Vedran Šego Aug 31 '13 at 20:33

If $$A=(a_{ij}) \in \mathbb{R}^{m\times n}, \ B=(b_{ij})\in \mathbb{R}^{n\times m},$$ with $m>n$, then $$C:=AB=(c_{ij}) \in \mathbb{R}^{m\times m},$$ with $$c_{ij}=\sum_{k=1}^na_{ik}b_{kj}.$$ For $j=1,\ldots,m$ let $$\mathbf{c}_j=Ce_j \in \mathbb{R}^m,$$ where $(e_1,\ldots, e_m)$ stands for the canonical basis of $\mathbb{R}^m$. Then we have $$\mathbf{c}_j=\left[ \begin{array}{c} \sum_{k=1}^na_{1k}b_{kj}\\ \vdots\\ \sum_{k=1}^na_{mk}b_{kj} \end{array} \right]=\sum_{k=1}^nb_{kj} \left[ \begin{array}{c} a_{1k}\\ \vdots\\ a_{mk} \end{array} \right]= \sum_{k=1}^nb_{kj}\mathbf{a}_k,$$ with $$\mathbf{a}_j=\left[ \begin{array}{c} a_{1j}\\ \vdots\\ a_{mj} \end{array} \right] \in \mathbb{R}^n.$$ We now have \begin{eqnarray} \det(C)&=&\det(\mathbf{c}_1,\ldots,\mathbf{c}_m)=\sum_{k_1=1}^n\sum_{k_2=1}^n\ldots\sum_{k_m=1}^nb_{k_1,1}b_{k_2,2}\ldots b_{k_m,m}\det(\mathbf{a}_{k_1},\ldots,\mathbf{a}_{k_m}) \end{eqnarray} Since there are exactly $n$ vectors $\mathbf{a}_j$, and the determinant has $m>n$ entries we have $$\det(\mathbf{a}_{k_1},\ldots,\mathbf{a}_{k_m})=0 \quad \forall k_1,\ldots,k_m \in \{1,\ldots,n\}$$ because at least two entries must be equal. Hence $$\det(C)=0.$$

Since $rank(AB)\le rank(A)\le \min\{m,n\}=n<m$ and $AB$ is an $m\times m$ matrix,

$AB$ is not invertible and therefore $\det(AB)=0$.

• Is the converse of this statement true? – imranfat Aug 31 '13 at 23:33
• Yes, an $m\times m$ matrix is invertible iff its rank is m. – user84413 Sep 1 '13 at 17:25

Since $m>n$, we must have that $\ker B$ is non-trivial (look at row canonical form, for example). Hence $Bv=0$ for some $v \neq 0$, and so $ABv=0$. It follows that $AB$ is singular and that $\det (AB) = 0$.

Twink, there is another way of looking at your statement. Take for example a 3x2 matrix in which you choose your numbers. Then take a 2x3 matrix in which you put letters for your entries, say a,b,c,d,e,f Now perform matrix multiplication. You will end up with a 3 by 3 matrix, but more interestingly, examine its rows. All your coefficients are the same when you compare the entries! In fact, no matter what you ;pick for your a,b,c,... You will have in each row the same value!. A simple row opration then shows that there is linear dependence. Now my approach is NOT a rigorous prove like colleagues above have done. It's just a different approach as to see how your statement "works"

Basically the same as cooper.hat's answer with lower technology.

Consider the system $$B\mathbf x = 0.$$ Since $B$ is an $n\times m$ matrix and $n\lt m$, this system has infinite solutions. So there is one non trivial solution say $\mathbf y$. But then $\mathbf y$ is also a nontrivial solution to the system $$(AB)\mathbf x = 0$$ and so $AB$ is not invertible.

So $\det(AB)=0$.