Invertible matrices

BrownianMan · #1 $Old$ November 6th, 2009, 08:03 PM

Let A, B, and C denote n x n matrices. Show that if A, C, and ABC are invertible, then B is invertible.

Any help would be appreciated!

Debsta · #2 $Old$ November 6th, 2009, 08:42 PM

A^-1 exists and B^-1 exists. ABC and so (ABC)^-1 exists.
Now (ABC) x (C^-1 x B^-1 x A^-1) = I (by associativity of matrix mutiplication)
So (ABC)^-1 = C^-1 x B^-1 x A^-1.
Therefore, since A^-1 exists and B^-1 exists and (ABC)^-1 exists, the so does B^1.
Therefore B is invertible.

HallsofIvy · #3 $Old$ November 7th, 2009, 05:04 AM

Quote:

Originally Posted by Debsta $View Post$

A^-1 exists and B^-1 exists.

Did you mean "and C^-1 exists"?

Quote:

ABC and so (ABC)^-1 exists.
Now (ABC) x (C^-1 x B^-1 x A^-1) = I (by associativity of matrix mutiplication)

That is only true if B^-1 exists, which is what you are trying to prove.

Quote:

So (ABC)^-1 = C^-1 x B^-1 x A^-1.
Therefore, since A^-1 exists and B^-1 exists and (ABC)^-1 exists, the so does B^1.
Therefore B is invertible.

B is invertible if and only if det(B) is not 0. Since ABC is invertible, det(ABC)= det(A)det(B)det(C) is not 0 so det(B) cannot be 0.
Actually, given that ABC is invertible, it follows that each of A, B, and C is invertible. You don't need to be given that A and C are also invertible.

tonio · #4 $Old$ November 7th, 2009, 05:11 AM

Quote:

Originally Posted by Debsta $View Post$

A^-1 exists and B^-1 exists. ABC and so (ABC)^-1 exists.
Now (ABC) x (C^-1 x B^-1 x A^-1) = I (by associativity of matrix mutiplication)
So (ABC)^-1 = C^-1 x B^-1 x A^-1.
Therefore, since A^-1 exists and B^-1 exists and (ABC)^-1 exists, the so does B^1.
Therefore B is invertible.

This cannot be correct since you're using in the "proof" what you must prove, namely: that $B^{-1}$ exists! If you begin by saying " $A^{-1}\,\,and\,\,B^{-1}$ exist" then all the following is already wrong, unless that was a typo and you actually meant $C^{-1}$ instead of $B^{-1}$ , but then you use $B^{-1}$ in the next line so it is wrong, not to mention that your conclusion is that $B^1=B$ exists, which of course is trivially true.

I think we could argue as follows: since $ABC$ is invertible there exists an invertible matrix $D$ s.t. $ABC\cdot D=I$ , but using associativity of matrix multiplication and the fact that $A^{-1}\,,\,\,C^{-1}$ exist we can get:

$A(BC)D=I\Longrightarrow BCD=A^{-1}\Longrightarrow BC=A^{-1}D^{-1}\Longrightarrow B=A^{-1}D^{-1}C^{-1}$ , and as B is the product of invertible matrices then it itself is invertible. Q.E.D.

Tonio

Debsta · #5 $Old$ November 7th, 2009, 03:19 PM

Yes sorry ...I DID mean "and C^-1 exists" in the first line. And the second last line should be "Therefore since A^-1 and C^-1 exists and (ABC)^-1 exists, then so does B^-1."
Note to self: Proofread before hitting send button.

With those changes I think the proof is valid.

tonio · #6 $Old$ November 7th, 2009, 04:15 PM

Quote:

Originally Posted by Debsta $View Post$

Yes sorry ...I DID mean "and C^-1 exists" in the first line. And the second last line should be "Therefore since A^-1 and C^-1 exists and (ABC)^-1 exists, then so does B^-1."
Note to self: Proofread before hitting send button.

With those changes I think the proof is valid.

I don't think so since in "the proof" still appears $B^{-1}$ twice more: here "(ABC) x (C^-1 x B^-1 x A^-1) = I" and here "(ABC)^-1 = C^-1 x B^-1 x A^-1} , so in fact you use $B^{-1}$ ALL along your "proof" and this, as already noted, is incorrect.
Anyway, you already have two different approaches to prove what you want.

Tonio

BrownianMan · #7 $Old$ November 9th, 2009, 05:52 PM

We are restricted in what we can use to write the proof for this question. Basically, all we can use is the Theorem that states:

If A is an invertible matrix, then A^-1 is invertible and (A^-1)^-1 = A

If A and B are n x n invertible matrices, then so is AB and the inverse of AB is the product of the inverses of A and B in the reverse order. (AB)^-1 = B^-1 A^-1

If A is an invertible matrix, then so is A^T , and the inverse of A^T is the transpose of A^-1 . That is (A^T)^-1 = (A^1)^T

The product of n x n invertible matrices is invertible, and the inverse is the product of their inverses in the reverse order.

We cannot use (ABC)D = I implies D(ABC) = I.

Also, another part to this question that I find confusing:

Let A, B, and C denote n x n matrices. Show that if A and AB are invertible, B is invertible. Same rules apply - we can only use the Theorem I posted.

Jameson · #8 $Old$ November 9th, 2009, 07:32 PM

Quote:

Originally Posted by BrownianMan $View Post$

We are restricted in what we can use to write the proof for this question. Basically, all we can use is the Theorem that states:

If A is an invertible matrix, then A^-1 is invertible and (A^-1)^-1 = A

If A and B are n x n invertible matrices, then so is AB and the inverse of AB is the product of the inverses of A and B in the reverse order. (AB)^-1 = B^-1 A^-1

If A is an invertible matrix, then so is A^T , and the inverse of A^T is the transpose of A^-1 . That is (A^T)^-1 = (A^1)^T

The product of n x n invertible matrices is invertible, and the inverse is the product of their inverses in the reverse order.

We cannot use (ABC)D = I implies D(ABC) = I.

Also, another part to this question that I find confusing:

Let A, B, and C denote n x n matrices. Show that if A and AB are invertible, B is invertible. Same rules apply - we can only use the Theorem I posted.

This is copied from tonio's post.

$A(BC)D=I\Longrightarrow BCD=A^{-1}\Longrightarrow BC=A^{-1}D^{-1}\Longrightarrow B=A^{-1}D^{-1}C^{-1}$

ABC is invertible and from the 1st rule in your list it's inverse is (ABC)^-1. You can call this D if you wish. It's just shorthand and doesn't change any property of it.

By 1 again, ABCD=I

Now, matrix multiplication is associative. That's a basic property that you should have proved well before this proof.

So ABCD=I implies A(BCD)=I by associative property. By definition of inverse matrices, A and BCD are inverses. So BCD=A^-1 by rule 1. D=(ABC)^-1 is invertible by rule 1 so starting with BCD=A^-1 we get that BCDD^-1=A^-1D^-1 which means BC=A^-1D^-1. Same step now using the fact that C^-1 exists and CC^-1=I. This leaves you with B=A^-1D^-1C^-1 thus B exists.

All I did was explain tonio's post in more detail and reference your rules. The line in Latex sufficiently shows this proof.

BrownianMan · #9 $Old$ November 9th, 2009, 07:48 PM

Thanks.

What about the other part?

Let A, B, and C denote n x n matrices. Show that if A and AB are invertible, B is invertible. Same rules apply - we can only use the Theorem I posted.

Jameson · #10 $Old$ November 9th, 2009, 07:57 PM

Quote:

Originally Posted by BrownianMan $View Post$

Thanks.

What about the other part?

Let A, B, and C denote n x n matrices. Show that if A and AB are invertible, B is invertible. Same rules apply - we can only use the Theorem I posted.

This is the exact same idea! It's actually easier. Look at my previous post. You know A,C and ABC are invertible. You showed B was too. Now you have A and AB being invertible and want to show B is. Same process!

Use the definition of the inverse to show that some D exists where D=(AB)^-1. Then start with ABD=I and from there use the other inverses and grouping to solve for B, then use the last rule on your list to show that B is a product of invertible matrices thus is invertible.