I am asked to prove following proposition:
Proposition 1. If an invertible matrix $\mathbf A$ has a left inverse $\mathbf{B}$ and a right inverse $\mathbf{C}$, then $\mathbf{B} = \mathbf {C}$
My attempt:
"$\mathbf{B}$ is the left inverse of $\mathbf{A}$" implies:
$$\mathbf{BA} = \mathbf I$$
And "$\mathbf{C}$ is the right inverse of $\mathbf{A}$" implies:
$$\mathbf{AC} = \mathbf I$$
Hence
$$\tag 1\mathbf{AC} = \mathbf{BA}$$
Premultiply $(1)$ by $\mathbf A^{-1}$:
$$ \mathbf A^{-1}\mathbf{AC} = \mathbf A^{-1}\mathbf{BA} \implies $$
$$\mathbf{C} = \mathbf A^{-1} $$
Postmultiply $(1)$ by $\mathbf A^{-1}$:
$$\mathbf{AC}\mathbf A^{-1} = \mathbf{BA}\mathbf A^{-1} \implies$$
$$\mathbf A^{-1} = \mathbf{B}$$
Hence $\mathbf B = \mathbf C.$$\Box$
Is it correct?
I have one more question:
Assume we already proved following theorem:
The inverse of an invertible matrix is unique
Knowing this, do we really need to prove the proposition 1? My reasoning is, if we know $\mathbf{B}$ is the left inverse of $\mathbf{A}$, and we also know theorem above, then we can infer that $\mathbf{B}$ will also be right inverse of $\mathbf{A}$. Or am I missing something?
Best Answer
Assuming that $\mathbf A^{-1}$ is meaningful seems premature until you have proved the assertion that $\mathbf B = \mathbf C$
Instead use something like $\mathbf{B(AC)} = \mathbf{(BA)C}$ since matrix multiplication is associative