# Homework Help: Orthogonal vector proof

1. Dec 12, 2012

### Clandry

Let A be an n × n invertible matrix. Show that if
i ≠ j, then row vector i of A and column vector
j of A-1 are orthogonal.

I'm lost in regards to where to lost.

I want to show that a vector from row vector i from A is orthogonal to a column vector j from A.
Orthogonal means the dot product is 0 or the angle between the 2 vectors is 90 degrees.

After stating the obvious I'm stuck. I think I need to start with figuring out what the relationship between a row vector from A and a column vector from A^-1 is, but how do I do that?

2. Dec 12, 2012

### Michael Redei

How would you show that the dot product of (row vector i of A) and (column vector j of A-1) is equal to 0? And in what kind of operation would you perform several such calculations between row vectors of one matrix and column vectors of another?

3. Dec 12, 2012

### Clandry

If v represents the row vector and w represents the column vector.
I must show that v*wt=0. * means dot product.

The dot product would be then computed by multiplying each element in v by each corresponding element in w and taking the sum of all those products. But I'm not sure how this will show what the problem wants.

4. Dec 12, 2012

### Michael Redei

Apart from calculating dot products, when does one multiply the elements of a row with the elements of a column and take the sum of the results?

If that's no help, try this: how would you prove that A-1 really is an inverse of A? What kind of calculation could you perform to show this?

5. Dec 12, 2012

### Clandry

a-1*a=i?

6. Dec 12, 2012

### Michael Redei

How does one multiply one matrix by another? Say you wanted to multiply A by A-1. Can you explain how you'd obtain the element in row i and column j of the resulting matrix?

7. Dec 12, 2012

### Clandry

OH! I see what you're saying. You'd multiply the row by the column and sum up the products.
This would explain how to multiply the 2 vectors together.

8. Dec 12, 2012

### Michael Redei

All you need now is some criterion to decide which elements in AA-1 can be zero and which can't.

9. Dec 12, 2012

### Clandry

Okay Thanks for the help.

I am still having trouble. Or maybe I understand it but just not realize it.

In AA-1=I, everything off the diagonal is 0. Everything on the diagonal is nonzero, but that's when i=j. Everything off the diagonal is 0 b/c i≠j.

Is that the idea?

10. Dec 13, 2012

### Michael Redei

That's exactly the right idea. You can safely ignore the diagonal, since your initial question was only about i≠j.