# Product of matrices of different order is not invertible

If $A$ is an $m \times n$ and $B$ is an $n \times m$ matrix with $n < m$ then prove that $C = AB$ is not invertible.

I am not getting any idea. May be I am missing vary basic point. $C = AB$ where both $A$ and $B$ are square matrices will be invertible iff both $A$ and $B$ be invertible. I know it and its proof.

May be this question has been discussed earlier, but I am not seeing the link. Thank you for your help.

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Hints: The rank of a $j\times k$ matrix is at most $\min\{j,k\}$. If the ranks of matrices $M,N$ are $s,t$ (respectively) and they are sized appropriately for $MN$ to make sense, then the rank of $MN$ is at most $\min\{s,t\}.$

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Thank you for your help. I get it. If we consider $A_{3 \times 2}$ and $B_{2 \times 3}$ then the rank of $A$ and $B$ will be at most 2 and the product matrix $AB$ will have rank at most $2$. For being invertible the matrix $AB$ should have rank 3. So. – Dutta Dec 26 '13 at 16:17
That's exactly right. – Cameron Buie Dec 26 '13 at 17:29

Hint: Suppose $C$ is an inverse of $AB$. Consider what the linear transformation $v\mapsto CAv$ does to the vectors $B\mathbf e_1,B\mathbf e_2,\ldots,B\mathbf e_m\in\mathbb R^n$. Are these vectors linearly dependent? Are their image after $CA$? Is it possible for a linear transformation to map a linearly dependent family of vectors to a linearly independent one?

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Thank you for your hints. It is giving more generalization. – Dutta Dec 26 '13 at 16:17

More hints in the form of words:

When you multiply two matrices, the rank can't increase.

Rank can't be more than the smaller of the number of rows and number of columns

For an invertible matrix, the rank should equal the size

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One way to see this is that if a composite function $f=g\circ h:X\to Y$ is to be bijective, it is necessary (though not sufficient) that $h$ be injective (if you would start mapping two elements of$~X$ to the same image, no further map could make their images distinct again and $f$ would not be injective), and also that $g$ be surjective (if some $y\in Y$ would fail to be in the image of $g$ it would also fail to be be in the image of$~f$, and $f$ would not be surjective). Now if $n<m$, you know that (the linear map corresponding to) a $n\times m$ matrix like $B$ cannot be injective, and also that a $m\times n$ matrix like $A$ cannot be surjective, which gives you two reasons why the composition $C=AB$ cannot be bijective (i.e., invertible).

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Thank you for a new answer. A matrix is nothing but a linear transformation. A bijective linear map is invertible and so the corresponding matrix is invertible. The linear transformation $T$ corresponding to the matrix $A_{m \times n}$ maps $\mathbb{R}^n$ to $\mathbb{R}^m$. If $m < n$, $T$ can not be injective as the basis of the range set contains lace number of vector than the basis of the domain set. Similarly the mapping $T$ can not be surjective when $m > n$. The composition mapping will not be bijective and so not invertible. – Dutta Dec 27 '13 at 2:15
Sir, the cardinality of both $\mathbb{R}^n$ and $\mathbb{R}^m$ will be same weather $m > n$ or $m < n$. So there is bjective mappings $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$. Can we say "there is no such $f$ which is linear"? – Dutta Dec 27 '13 at 2:19
@HopelessFool: Yes you are right: without requiring "linear" there do exist maps $\def\R{\Bbb R}\R^m\to\R^n\to\R^n$ whose composition is bijective (though such maps are not very easy to describe explicitly). So a mere cardinality argument is not possible here. With the requirement of linear maps the notion of dimension can replace cardinality, so that such simple arguments can again be usefully exploited. – Marc van Leeuwen Dec 27 '13 at 6:15
Thank you sir.. – Dutta Dec 27 '13 at 12:10