Nonsingular Matrix Proof.

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February 13th 2011, 03:01 PM
tommiegirl13

Nonsingular Matrix Proof.

The Proof:
a. Show that if A(nxn) is a nonsingular matrix, so is A^2.
b. Generalize to n: Show that if A is a nonsingular matrix, so is A^n.

I understand that a nonsingular matrix is one that has an inverse (in our course we have not yet talked about determinants so please do not use those in responses), but I cannot think how I would generally any matrix A to always be nonsingular. If I could figure this out, I think I could prove how A^2 or A^n would follow as being nonsingular as well. Any help on this proof you could offer me is greatly appreciated! Thanks!
February 13th 2011, 03:14 PM
FernandoRevilla

Perhaps (perhaps) the definition you have covered is: $A\in \mathbb{K}^{n\times n}$ is non singular iff $Ax=0$ implies $x=0\;\;(x\in\mathbb{K}^{n\times 1})$ .

If so, suppose $A$ is non singular then,

$A^2x=0\Rightarrow A(Ax)=0\Rightarrow Ax=0\Rightarrow x=0$

which implies $A^2$ is non singular.

Try the second part.

Fernando Revilla
February 13th 2011, 03:21 PM
tommiegirl13

I'm not sure I'm understanding your definition of nonsingularity. My professor taught us that if you can find an inverse of a matrix, then it is nonsingular. This is not necessarily when talking about systems of equations which is what I'm getting from your definition, talking about x=0. Why would A or x have to equal 0 to be nonsingular?
February 13th 2011, 03:31 PM
FernandoRevilla

Quote:

Originally Posted by tommiegirl13

My professor taught us that if you can find an inverse of a matrix, then it is nonsingular.

For that reason I said perhaps . So, you have covered: $A$ non singular iff there exists $B$ such that $AB=BA=I$ . Suppose $A$ non singular then,

$A^2B^2=A(AB)B=AIB=AB=I$

$B^2A^2=B(BA)A=BIA=BA=I$

which implies $A^2$ is non singular.

Fernando Revilla
February 13th 2011, 03:40 PM
tommiegirl13

Thank you so much!! I get it now. You really made it click for me!