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Recall that a matrix A is said to be invertible if there exists a matrix B such that A⋅B=I and B⋅A=I, where a priori these identity matrices may have different sizes. In order for this to make sense, if A is an n×m matrix, then B must be an m×n matrix, so we could write this as
A⋅B=In
B⋅A=Im
As is often the case with linear algebra concepts, it is better to think of matrix inversion in terms of linear transformations instead of matrices. In the language of linear transformations, the fact that only square matrices are invertible is a consequence of the fact that "dimension of vector spaces is invariant under invertible linear transformations."
Suppose we have two finite-dimensional vector spaces VV and WW and that f:V→Wf:V→W is a linear transformation. Then the following three things are equivalent:
- ff is one-to-one and onto: for every vector w∈Ww∈W there exists a unique v∈Vv∈V with f(v)=wf(v)=w.
- There exists an inverse linear transformation g:W→Vg:W→V to ff, i.e. g∘f=idVg∘f=idV and f∘g=idWf∘g=idW.
- After choosing bases for VV and WW, the matrix representing ff is invertible.
These equivalences are easy to see. Condition 1 implies condition 2 since we can define gg to be the inverse function f−1f−1, and then it turns out that gg is actually linear. In the opposite direction, Condition 2 implies condition 1 since g∘f=idVg∘f=idV implies ff is one-to-one and f∘g=idWf∘g=idW implies ff is onto. Condition 2 implies condition 3 since the matrix corresponding to ff has the matrix corresponding to gg as its inverse; conversely, if the matrix corresponding to ff has an inverse matrix, then the linear transformation corresponding to that matrix will be an inverse linear transformation to ff.
A linear transformation satisfying any of the above 3 equivalent conditions is often called an isomorphism.
Coming back down to earth a bit, suppose we have an n×mn×m matrix AA that admits an inverse. The corresponding linear transformation f:Rm→Rnf:Rm→Rn is one-to-one and onto.
Claim: If f:V→Wf:V→W is a one-to-one and onto linear transformation, then dimV=dimWdimV=dimW.
In other words, dimension is an isomorphism invariant.
Since the dimension of RnRn is nn, this will imply that n=mn=m whenever AA is an invertible n×mn×m matrix.
Proof of claim: If {e1,…,em}{e1,…,em} is a basis for VV then {f(e1),…,f(em)}{f(e1),…,f(em)} is a basis for WW.
Indeed, the set {f(e1),…,f(em)}{f(e1),…,f(em)} is linearly independent since ff is one-to-one--any relation
a1f(e1)+a2f(e2)+⋯+amf(em)=0a1f(e1)+a2f(e2)+⋯+amf(em)=0
can be rearranged to
f(a1e1+a2e2+⋯+amem)=0=f(0)f(a1e1+a2e2+⋯+amem)=0=f(0)
by linearity, which then forces
a1e1+⋯+amem=0,a1e1+⋯+amem=0,
since ff is one-to-one, giving ai=0ai=0 for all ii since e1,…,eme1,…,em are a basis for VV.
On the other hand {f(e1),…,f(em)}{f(e1),…,f(em)} spans WW. Since ff is onto, for any w∈Ww∈W we can find some v∈Vv∈V with f(v)=w.f(v)=w. Then we can write
v=a1e1+⋯+amemv=a1e1+⋯+amem
and hence
w=f(v)=a1f(e1)+⋯+amf(em)w=f(v)=a1f(e1)+⋯+amf(em)
by linearity.
Thus both VV and WW are mm-dimensional. □
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