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January 22nd, 2014, 03:33 AM   #1
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square of invertible matrix

Hi,

I have to show for an invertible matrix A, its square AČ is also invertible.


I'm pretty sure this is correct, or did I miss something?
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January 22nd, 2014, 09:19 AM   #2
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Re: square of invertible matrix

We can use







We now have exactly what we are looking for on the RHS and the left hand side is gauranteed to be meaningful since both exist , are nxn and certainly commute with each other and their inverses under matrix multiplication.

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January 23rd, 2014, 01:26 AM   #3
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Re: square of invertible matrix

That's a good idea, thank you.
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January 23rd, 2014, 01:33 AM   #4
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Re: square of invertible matrix

Quote:
Originally Posted by PrototypePHX
Hi,

I have to show for an invertible matrix A, its square AČ is also invertible.


I'm pretty sure this is correct, or did I miss something?
This is not bad, but it's better to avoid relying on for a matrix. Instead:



This shows that the matrix is the inverse of A^2.

You really need to prove this before you can start cancelling powers of matrices. Not the other way round.
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January 25th, 2014, 06:45 AM   #5
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Re: square of invertible matrix

Matrix is invertible if and only if its determinant is different from 0. Observe also that so if then also . This is a way to verify that is invertible if is.
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January 25th, 2014, 04:43 PM   #6
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Re: square of invertible matrix

Quote:
Originally Posted by Pero
Quote:
Originally Posted by PrototypePHX
Hi,

I have to show for an invertible matrix A, its square AČ is also invertible.


I'm pretty sure this is correct, or did I miss something?
This is not bad, but it's better to avoid relying on for a matrix. Instead:



This shows that the matrix is the inverse of A^2.

You really need to prove this before you can start cancelling powers of matrices. Not the other way round.
Yes, but: the set of invertible nxn matrices is closed under matrix multiplication, since

is demonstrably an inverse for AB (on both sides).

Also, since , it is clear the inverse of an invertible matrix is invertible. Thus we clearly have a group structure on the set of invertible matrices.

As such, we can talk about the subgroup generated by the (invertible) matrix A, and for this subgroup the laws of (integer) exponents work exactly as expected, if we set:

. We then have:

for ANY integers k and m, as well as:

(including the special case k =2, m = -1 under discussion here).

So, in my humble opinion, your concern over this misuse of exponent rules here is unfounded.
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January 25th, 2014, 11:36 PM   #7
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Re: square of invertible matrix

Quote:
Originally Posted by Deveno

So, in my humble opinion, your concern over this misuse of exponent rules here is unfounded.
The original question,if you recall, was to prove that A^2 is invertible. If you use the exponent rules for matrices, you are assuming that -ve powers of matrices exist, which is actually what you're trying to prove.

The exponent rules were not being misused, but used prematurely given the original question.
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