# Basic linear algebra

1. Feb 6, 2006

### ak416

Im just working through some problems trying to learn linear algebra on my own. Here's one im having trouble with: Let A be invertible. Prove that A_t is invertible and (A_t)-1 = (A-1)_t ,where _t means transpose. -1 means inverse. I started trying to look at all the entries of general n by m matrices but it looks very tedious. Any trick here?

2. Feb 7, 2006

### Hurkyl

Staff Emeritus
It's an easy problem when you try it with linear algebra, instead of trying to do it as a computational matrix problem! What is the definition of inverse? What are the algebraic properties of transpose?

3. Feb 7, 2006

### ak416

Ya actually its pretty easy using the fact that (AB)_t = B_t A_t because AA^-1 = A^-1A = I_n so taking the transpose of these gives the answer.

Heres another one: Prove that if A is invertible and AB = 0 then B=0. I looked at it through the definition of matrix multiplication and its clear that A cannot be 0, but I still dont see why B has to be zero.

4. Feb 7, 2006

### Hurkyl

Staff Emeritus
Have you tried using the fact A is invertible?

5. Feb 7, 2006

### ak416

Ya of course, but all i know is that there exists an A^-1 st AA^-1 = A^-1A = In so the sum in each entry of AA^-1 = 1 (when row equals column). So A is clearly not zero but when you look at the entries of AB you know that the sum of each entry has to equal zero but that can happen in many ways (with negatives of the sum cancelling out with positives) so im not sure.

6. Feb 7, 2006

### Hurkyl

Staff Emeritus
Well, you know that "invertible" means is that $A A^{-1} = A^{-1} A = I$, right? That would suggest to me that you ought to do something to your equation so that $A A^{-1}$, $A^{-1} A$, or $I$ appear somewhere in it!

7. Feb 7, 2006

### ak416

oo ok that was easy. AB = 0, A^-1AB = 0, B = 0 ! Am i right?

8. Feb 7, 2006

### Hurkyl

Staff Emeritus
Yep!

Of course, the full proof would be:

$$AB = 0$$
$$A^{-1} AB = A^{-1} 0$$
$$A^{-1} AB = 0$$
$$I B = 0$$
$$B = 0$$

where I've made sure to do one step at a time.

I expect you already know these little details, and are aware you're skipping them for brevity, but I just want to make extra sure!

9. Feb 7, 2006

### ak416

Ya thanks for that. I have one more question thats kinda bugging me. Let A and B be nxn matrices st AB is invertible. Prove that A and B are invertible. Give an example to show that arbitrary matrices A and B need not be invertible if AB is invertible.

The first part im not sure I know that AB(AB)^-1 = (AB)^-1(AB) = I . The only thing I can think of is switching (AB)^-1 to B^-1A^-1 but that can only be done once you know there exists A^-1, B^-1. So Whats the best approach? To try to prove that an A^-1 exists, Or to guess and try out something that may be an A^-1 and show that it works.
Also, the second part I dont know what they mean by arbitrary. I thought its a fact that only n by n matrices are invertible.

10. Feb 7, 2006

### Hurkyl

Staff Emeritus
You won't know until you try. (I know how to turn either one of these ideas into a proof, but I don't know how much you've learned yet)

Yep. That's true...

11. Feb 7, 2006

### ak416

well i managed to get (AB)^-1 = I and so AB = I. If i can somehow get BA = I ... (or am i off track?)

edit:never mind kinda messed up my algebra.

Last edited: Feb 7, 2006
12. Feb 7, 2006

### ak416

Ok i managed to do it but I had to use the fact that for matrices, if AB = CB then A = C. Is this true?

13. Feb 7, 2006

### Hurkyl

Staff Emeritus
Unfortunately, no. What was your work? Maybe it can still be salvaged?

14. Feb 7, 2006

### ak416

Well actually its more of a special case. if AB = IB then A = I. Im sure thats true right?

15. Feb 7, 2006

### Hurkyl

Staff Emeritus
Nope. What if, for example, B = 0? (Or is diagonal with some entries zero)

16. Feb 7, 2006

### ak416

Well what i did was: A(B(AB)^-1) = I (im guessing B(AB)^-1 is the inverse of A). Now, B(AB)^-1(AB) = BI = IB and what i thought i could do was just cancel out the B on the right and get B(AB)^-1 A = I (Similarily for the inverse of B). So, any tips?

17. Feb 8, 2006

### Hurkyl

Staff Emeritus
Oh bleh! Of course this one won't be so simple! Sorry, I didn't think far enough ahead

Did you notice that you haven't used the fact A and B are square matrices? That is a crucial assumption, so you will not be able to do the problem unless you use it in some way.

Incidentally, the work you did is still valuable: when you know that AB = I, we say that B is a right inverse of A, and that A is a left inverse of B. (So, the inverse of a matrix A is simply something that is both a right and a left inverse of A)

Left and right inverses can be important when working with nonsquare matrices. For example, you might want to solve the equation Ax=y, but A is not a squrae matrix! Fortunately, if A is left invertible, you could just multiply by one of its left inverses (say, B) and get x=By.

(An example is when solving a system of linear equations when you have more unknowns than equations. Typically, the coefficient matrix is left-invertible. You will probably not be taught to think of it this way, though!)

Bleh, then all the ways I know to do this problem involve something that isn't completely trivial. The most straightforward is with determinants. (If you've learned them)

While you don't have a theorem that AB = CB --> A = C, you do have the following:

If Av = Bv for every vector v, then A = B. (Similarly, if AC = BC for every matrix C, then A = B)

Also, if A is a non-invertible square matrix, then Av=0 has a nontrivial solution. You could probably use that to do this problem too.