Proving that a right (or left) inverse of a square matrix is unique using only basic matrix operations

Proving that a right (or left) inverse of a square matrix is unique using only basic matrix operations

-- i.e. without any reference to higher-order matters like rank, vector spaces or whatever ( :)).

More precisely, armed with the knowledge only of:

• rules of matrix equality check, addition, multiplication, distributive law and friends
• Gauss-Jordan elimination and appropriate equation system solution cases for the reduced row-echelon form

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The idea I've got is that since elementary transforms of I will never give a zero row, any matrix A which is not row-equivalent to I, will produce an unsolvable set of equations for at least one column of B in A*B = I. Hence, for that matrix equation to hold, A must be row-equvalent to I and thus there's only one B sufficing the equation. –  mlvljr Feb 17 '12 at 13:56
Or is it really that exotic? –  mlvljr Feb 17 '12 at 14:50
"...and thus there's only one B sufficing the equation." How did you conclude that? Why do you need to show that A is row equivalent to I? Focus on the question: uniqueness. –  osa Feb 17 '12 at 15:37
Whatever row operations we apply to A to get it into reduced row-echelon form, if the same row operations are applied to I, the resulting matrix will have no zero rows. Then, if and only if A is row-equivalent to I, there will be a single set of solutions (i.e. B's column values) for every equation A_<j>*B=I_<j>, where A_ and I_ are transformed A and I and <j> denotes j column. If there's a zero row i in A_, there will be at least one such j that I_<j> has a non-zero element in row i, and thus the system of equations has no solution. –  mlvljr Feb 17 '12 at 15:56
@mlvljr: If you can solve the problem for inverses on one side, taking transposes solves the problem for inverses on the other side: $AB=I$ if and only if $B^TA^T=I$, and if you know that right inverses are unique when they exist, then it follows that left inverses are unique when they exist, and vice-versa. –  Arturo Magidin Feb 17 '12 at 17:30

So, let us suppose that $A$ is a square matrix, and that $B$ is a matrix such that $BA=I$. You want to show that $B$ is the unique left inverse of $A$ (that is).

Note that a system $A\mathbf{x}=\mathbf{b}$ has at most one solution, namely $B\mathbf{b}$: if $A\mathbf{x}=\mathbf{b}$, then $$\mathbf{x} = I\mathbf{x} = BA\mathbf{x} = B\mathbf{b}.$$

If $CA=I$, then again a system $A\mathbf{x}=\mathbf{b}$ has at most one solution, namely $C\mathbf{b}$. Thus, $B\mathbf{b}=C\mathbf{b}$ for any $\mathbf{b}$ for which the system has a solution.

If we can show that $A\mathbf{x}=\mathbf{e}_i$ has a solution for each $i$, where $\mathbf{e}_i$ is the $i$th standard basis vector ($1$ in the $i$th entry, $0$s elsewhere) this will show that $B=C$, since they have the same columns.

Because $A\mathbf{x}=\mathbf{0}$ has a solution, that solution must be $B\mathbf{0}=\mathbf{0}$. That means that the reduced row-echelon form of $A$ is $I$. Because the reduced row-echelon form of $A$ is $I$, performing row reduction on the augmented coefficient matrix $[A|\mathbf{e}_i]$ yields the matrix $[I|\mathbf{y}]$ for some $\mathbf{y}$, with $\mathbf{y}$ being the solution to $A\mathbf{x}=\mathbf{e}_i$. Since this vector is equal to both $\mathbf{b}_i=B\mathbf{e}_i$ (the $i$th column of $B$) and to $\mathbf{c}_i=C\mathbf{e}_i$, as noted above, then the $i$th columns of $B$ and $C$ are equal; thus, $B=C$, and the matrix has a unique left inverse.

Now, let us suppose that $A$ is a square matrix and has a right inverse, $AB=I$. We want to show that $B$ is the unique right inverse of $A$. Taking transposes, we get $I = I^T = (AB)^T = B^TA^T$. By what was proven above, $B^T$ is the unique left inverse of $A^T$. If $AC=I$, then $C^TA^T=I^T = I$, so $C^T=B^T$, hence $C=B$. Thus, $B$ is the unique right inverse of $A$.

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Suppose you have square matrices $A$ and $B$ with $BA = I$. To show that $B$ is the unique left inverse of $A$ it suffices to show that $A$ is invertible, because then $B$ is the unique inverse. (If $C$ is an inverse for $A$ we have $B = BAC = C$.)

The equation $Ax = 0$ has a solution, namely $x = 0$, and for every vector $v$ with $Av = 0$ we have $v = BAv = 0$, hence there is only the trivial solution to the equation $Ax = 0$. This shows that applying Gauss-Jordan elimination to $A$ gives you the identity matrix, hence $A$ is the product of elementary matrices and these are all invertible. This shows that $A$ is invertible and we are done.

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HINT Show that $I_m = AA^{-1} = A^{-1}A = I_n$; if $m \neq n$ then you have different dimensions for $I_A$...this means they can't commute!!

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Huh, that's nice! :) –  mlvljr Nov 25 '13 at 23:01