Proving Invertibility of Matrix Products: Induction and Artin's Algebra

[IKonquer]

I have very little experience with proofs, and I am trying to learn algebra on my own. The following problem is found within the text of Michael Artin's Algebra.

The first problem encountered was the following: Let A, B be n x n matrices. If both are invertible, then so is their product AB, and $$(AB)^{-1} = B^{-1}A^{-1}$$

To show this is true for the right inverse I said,

Let X be an invertible matrix.

$$\begin{flalign*}<br /> XX^{-1} = I\\<br /> (AB)(AB)^{-1} = I\\<br /> A(BB^{-1})A^{-1} = I \\<br /> AIA^{-1} = I\\<br /> AA^{-1} = I\\<br /> I = I\\<br /> \end{flalign*}$$

Also the same can be said for the left inverse, which makes sense to me.

I'm having a lot of trouble understanding the following: If $$A_{1}, ... , A_{m}$$ are invertible, then so is the product $$A_{1} ... A_{m}$$ and its inverse is $$A_{m}^{-1} ... A_{1}^{-1}$$

Let m = 1

$$A_{1}A^{-1} = I$$

This shows that $$A_{1}$$ is invertible, and its product, which is itself, is invertible.

For m + 1

Let $$A_{1}, ... , A_{m}, A_{m+1}$$ be invertible matrices and let P be the product of $$A_{1}, ... , A_{m}$$.

"By the induction hypothesis, P is invertible." - I don't quite understand why you can say that. Could someone explain why this is true?

 

[IKonquer]

Oh wait, I think I see why P is invertible.

Similarly to this:
$$A_{1}A^{-1} = I$$

Since P is just a product of matrices, it is just another matrix.

$$\begin{flalign*}<br /> P_{1}P^{-1} = I\\<br /> (A_{1} ... A_{m})(A_{m}^{-1} ... A_{1}^{-1}) = I\\<br /> A_{1} ... ( A_{m}A_{m}^{-1}) ... A_{1}^{-1} = I\\<br /> I = I \\<br /> \end{flalign*}$$

Even though this may be painfully obvious to all of you, am I doing this right?

Also is there a way to delete a topic I started?

 

[HallsofIvy]

It is better not to delete threads after you don't want them any more. Other people can learn from them.

You have done proofs by induction before? To prove "A_n is true for all n", n a positive integer, you prove two things: "A_1 is true" and "if A_k is true then A_{k+1} is true". The "induction hypothesis" is "if A_k is true".

To prove "if A_1, A_2, ..., A_n are all invertible, then (A_1A_2...A_n) is invertible and its inverse is (An^{-1}...A_2^{-1}A_1^{-1})", the "base case" is "If A_1 is invertible, then A_1 is invertible and its inverse is A_1^{-1}" which is trivial.

The "induction hypothesis" is "if A_1, A_2, ..., A_k are invertible then (A_1A_2...A_k) is invertible and its inverse is (A_k^{-1}...A_2^{-1}A_1^{-1})" and you then use that to prove "if A_1, A_2, ..., A_k, A_{k+1} are invertible then (A_2A_2...A_kA_{k+1}) is invertible and its inverse is (A_{k+1}^{-1}A_k^{-1}...A_2^{-1}A_1^{-1})"

Since they have defined P to be (A_1A_2...A_k), it is invertible by hyopthesis- the whole point of induction is to avoid doing the kind of "..." argument you have in your second post.

 

[IKonquer]

I understand proof by induction when you have numbers. But the use of matrices really complicates it for me. Could someone please write out a way to prove:

If A_{1}, ... , A_{m} are invertible, then so is the product: A_{1} ... A_{m} and its inverse is A_{m}^{-1} ... A_{1}^{-1}

Much appreciated.

 

[Dick]

I understand proof by induction when you have numbers. But the use of matrices really complicates it for me. Could someone please write out a way to prove:

If A_{1}, ... , A_{m} are invertible, then so is the product: A_{1} ... A_{m} and its inverse is A_{m}^{-1} ... A_{1}^{-1}

Much appreciated.

I think induction is overkill in a simple proof like this and the '...' is perfectly understandable. Unless your course expects you to be super formal about proofs, and if you are little fuzzy about induction, I doubt it is.