Can anyone give a very simple proof (or explanation) as to why only square matrix can possibly be invertible?
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5 Answers
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Basically, an $\,n\times m\,$ matrix represents a linear map between linear spaces over some field of dimensions $\,m\,,\,n\,$ .
That a matrix is invertible means the map it represents is invertible, which means it is an isomorphism between linear spaces, and we know this is possible iff the linear spaces' dimensions are the same, and from here $\,n=m\,$ and the matrix is a square one.
answered Mar 30, 2013 at 11:31
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A product of two matrices of order $m\times n$ and $n\times p$ is a matrix $m\times p$.
But when a matrix $A$ has an inverse $B$, it has a two sided inverse, that is $AB=BA=I$.
The only possibility is $m=n=p$.
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5$\begingroup$ Well, just purely as function, we don't say a function has an inverse if $f\circ g = g\circ f$. Rather, we say $f:X\to Y$ has an inverse if there is a $g:Y\to X$ such that $g\circ f = \operatorname{id}_X$ and $f\circ g = \operatorname{id}_Y$. So we'd want $p=m$ and $AB=I_m$ and $BA=I_n$. $\endgroup$ Commented Mar 30, 2013 at 11:31
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$\begingroup$ Thanks. Another quick question: You can find the inverse using both the determinant AND gauss jordan elimation, right? $\endgroup$ Commented Mar 30, 2013 at 13:43
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An inverse of a square matrix $A$ is $B$ such that $A B = I$. You can also find a $m \times n$ matrix $A$ and $n \times m$ matrix $B$ such that $A B = I$, and call $B$ inverse of $A$. However such inverse need not be unique, nor does it endow any subset of matrices with a group structure under multiplication. Therefore it isn't as interesting.
answered Mar 30, 2013 at 11:32
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1$\begingroup$ Actually, an inverse is $B$ such that $AB=I$ and $BA=I$. What you are describing is a 1-sided inverse. $\endgroup$ Commented Mar 30, 2013 at 11:33
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An $m\times n$ matrix $A$ is invertible if there exists an $n\times m$ matrix $B$ such that the products $AB=I_m$ and $BA=I_n$ are identity matrices. I'll show that this implies $m=n$. I'll work over an arbitrary commutative ring with $1\neq0$ (e.g. fields, the integers, all integral domains, etc.).
I just need the following basic properties of the determinant of square matrices:
- $\det(CD)=\det(C)\det(D)$ for any two square matrices $C$ and $D$.
- $\det(I)=1$ for $I$ the identity matrix of any size.
- $\det(C)=0$ if $C$ has any zero row or column.
We argue by contradiction. Assume that $m>n$ (one can do assume without loss of generality, since the roles of $A$ and $B$ can be exchanged). We extend $A$ and $B$ to $m\times m$ square matrices by adding zeroes, $$C=(A|0),\qquad D=\left(\frac{B}{0}\right).$$ Then $$CB=AB=I.$$ By 1, 2, 3, above $$1=\det(I)=\det(CD)=\det(C)\det(D)=0,$$ reaching the desired contradiction.
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Say $A_{m \times n}$ with $m\ne n$.
If $A_{m \times n}\times B_{n \times m} = I_{m\times m}$ and $C_{n \times m}\times A_{m \times n} = I_{n\times n}$
Of course, the 2 identity matrices are of different size.But despite the size problem, the larger identity matrix is actually impossible to get.
This is because $dim(C(A)) = dim(C(A^T)) = rank(A)\le min\{m,n\} \lt max\{m,n\}$ which is the dimension of the larger indentity matrix.
