I have the matrix $A^2$$^0$$^1$$^5$ and have to use the associativity of matrix multiplication to prove that if that matrix is invertible, then so is A. I know how to find the inverse of a matrix (check the determinant isn't 0, and then use elementary row operations) but any hints on this proof as I have no idea how to start or where to go.

For part 2, the question says "Consider an m × n matrix A. Show that A has a left inverse if and only if $A^T$ has a right inverse" Again no idea how to start so any tips/full proofs would be great

  • $\begingroup$ Hi @Lauren, in fact, if $AB$ is invertible for any matrix $B$, then $A$ must be invertible. Can you see why this more general statement is true? What definitions of "invertible" do you know (you can use the "determinant non-zero" definition, but any definition will work)? $\endgroup$ – Amitesh Datta Mar 12 '15 at 14:24
  • $\begingroup$ Here's an idea which might help. Can you prove that if $A^2$ is invertible then $A$ is as well? It's the same technique. $\endgroup$ – Cameron Williams Mar 12 '15 at 14:28
  • $\begingroup$ @AmiteshDatta You should probably state explicitly that $A$ is square before saying that $AB$ is invertible implies $A$ invertible for arbitrary $B$ :). $\endgroup$ – Erick Wong Mar 12 '15 at 18:11

Let $I_n$ be the unit matrix of order $n$. We know:

$$A^{2015} \times (A^{2015})^{-1} = I_n$$

We can write:

$$A \times (A^{2014})(A^{2015})^{-1} = I_n$$

Let $B = (A^{2014})(A^{2015})^{-1}$


$$A \times B = I_n$$

We get that $A$ is invertible by the definition of invertibility.

  • $\begingroup$ Did you see where associativity was used? $\endgroup$ – user207710 Mar 12 '15 at 14:49

If you are only using the associativity, try considering $A^{2015}$=$AA^{2014}$ and use the fact that $A^{2015}$ is invertible. More than proving that $A$ is invertible, this process will actually give you an expression of the inverse of $A$ in terms of the inverse of $A^{2015}$.

  • $\begingroup$ If I'm understanding your answer correctly, you would have to assume that $A^{-1}$ exists before you could apply this approach directly. $\endgroup$ – rnrstopstraffic Mar 12 '15 at 14:36
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    $\begingroup$ Not really, the existence of such matrix would follow from the existence of $\big(A^{2015}\big)^{-1}$ plus the associativity of matrix product. However, I cannot go further since the next step would already be the solution of the problem :) $\endgroup$ – Alberto Debernardi Mar 12 '15 at 14:38
  • $\begingroup$ You are entirely correct. I misunderstood you to be pointing at the $(AB)^{-1}=B^{-1}A^{-1}$ property of inverses of products. $\endgroup$ – rnrstopstraffic Mar 12 '15 at 16:04

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