My Math Forum square of invertible matrix

 Linear Algebra Linear Algebra Math Forum

 January 22nd, 2014, 03:33 AM #1 Newbie   Joined: Sep 2012 Posts: 17 Thanks: 0 square of invertible matrix Hi, I have to show for an invertible matrix A, its square A² is also invertible. $\mathbf{A}^2 * \mathbf{A}^{-1}* \mathbf{A}^{-1}= \mathbf{A}^{2} * \mathbf{A}^{-2} = \mathbf{A}^{0} = \mathbf{I}$ $\mathbf{A}^{-1}* \mathbf{A}^{-1} * \mathbf{A}^2= \mathbf{A}^{-2} * \mathbf{A}^{2} = \mathbf{A}^{0} = \mathbf{I}$ I'm pretty sure this is correct, or did I miss something?
 January 22nd, 2014, 09:19 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 Re: square of invertible matrix We can use $B^{-1} \cdot A^{-1} \= \ (A \cdot B)^{-1}$ $A^{-1} \cdot A^{-1} \= \ (A \cdot A)^{-1}$ $A^{-1} \cdot A^{-1} \= \ (A^2)^{-1}$ We now have exactly what we are looking for on the RHS and the left hand side is gauranteed to be meaningful since both $\ \ A \ , \ A^{-1} \ \$ exist , are nxn and certainly commute with each other and their inverses under matrix multiplication.
 January 23rd, 2014, 01:26 AM #3 Newbie   Joined: Sep 2012 Posts: 17 Thanks: 0 Re: square of invertible matrix That's a good idea, thank you.
January 23rd, 2014, 01:33 AM   #4
Senior Member

Joined: Jun 2013
From: London, England

Posts: 1,316
Thanks: 116

Re: square of invertible matrix

Quote:
 Originally Posted by PrototypePHX Hi, I have to show for an invertible matrix A, its square A² is also invertible. $\mathbf{A}^2 * \mathbf{A}^{-1}* \mathbf{A}^{-1}= \mathbf{A}^{2} * \mathbf{A}^{-2} = \mathbf{A}^{0} = \mathbf{I}$ $\mathbf{A}^{-1}* \mathbf{A}^{-1} * \mathbf{A}^2= \mathbf{A}^{-2} * \mathbf{A}^{2} = \mathbf{A}^{0} = \mathbf{I}$ I'm pretty sure this is correct, or did I miss something?
This is not bad, but it's better to avoid relying on $A^2 * A^{-2}= A^0$ for a matrix. Instead:

$A^2*(A^{-1}*A^{-1})= A*(A*A^{-1})*A^{-1} = A*I*A^{-1} = A*A^{-1} = I$

This shows that the matrix $A^{-1}*A^{-1}$ 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.

 January 25th, 2014, 06:45 AM #5 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: square of invertible matrix Matrix $A$ is invertible if and only if its determinant $det A$ is different from 0. Observe also that $det(A^2)= det A . det A$ so if $det A \neq 0$ then also $det (A^2) \neq 0$. This is a way to verify that $A^2$ is invertible if $A$ is.
January 25th, 2014, 04:43 PM   #6
Senior Member

Joined: Mar 2012

Posts: 294
Thanks: 88

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 A² is also invertible. $\mathbf{A}^2 * \mathbf{A}^{-1}* \mathbf{A}^{-1}= \mathbf{A}^{2} * \mathbf{A}^{-2} = \mathbf{A}^{0} = \mathbf{I}$ $\mathbf{A}^{-1}* \mathbf{A}^{-1} * \mathbf{A}^2= \mathbf{A}^{-2} * \mathbf{A}^{2} = \mathbf{A}^{0} = \mathbf{I}$ I'm pretty sure this is correct, or did I miss something?
This is not bad, but it's better to avoid relying on $A^2 * A^{-2}= A^0$ for a matrix. Instead:

$A^2*(A^{-1}*A^{-1})= A*(A*A^{-1})*A^{-1} = A*I*A^{-1} = A*A^{-1} = I$

This shows that the matrix $A^{-1}*A^{-1}$ 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

$B^{-1}A^{-1}$ is demonstrably an inverse for AB (on both sides).

Also, since $(A^{-1})^{-1}= A$, 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:

$A^0= I,A^{-k} = (A^{-1})^k$. We then have:

$A^kA^m= A^{k+m}$ for ANY integers k and m, as well as:

$(A^k)^m= A^{km}$ (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.

January 25th, 2014, 11:36 PM   #7
Senior Member

Joined: Jun 2013
From: London, England

Posts: 1,316
Thanks: 116

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.

 Tags invertible, matrix, square

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Magnesium Linear Algebra 2 December 11th, 2013 02:09 AM shine123 Linear Algebra 1 September 21st, 2012 08:47 AM problem Linear Algebra 3 August 31st, 2011 05:30 AM 450081592 Linear Algebra 1 June 24th, 2010 07:13 AM Victorious Linear Algebra 1 January 5th, 2009 03:54 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top