# Basic invertible matrix proof

1. Apr 30, 2005

### randommacuser

How do I prove that if an nxn matrix A is diagonalizable (is invertible, has rank n, etc.), its columns span Rn?

2. Apr 30, 2005

### EvLer

To prove that columns span Rn, you just need to find the basis of the column-space: reduce the matrix and see if the rank = n, i.e. there has to be n linearly indep. columns, which is very easy to see if the matrix is reduced. Then the basis is the columns of the original matrix, not the columns of the reduced form.
Also, for invertible matrix A, det(A) != 0 (not equal to zero).
Don't know how much that helps.

3. Apr 30, 2005

### Galileo

If A is a matrix with rank n, its column vectors are independent and form a basis, so every vector can be written as a lin. comb. of them. In other words, these vectors span the entire space.

Equivalently, recall that Ax (x some vector in Rn) is a linear combination of the column vectors of A, so Ax=b has a solution if and only if b is in the span of the the column vectors of A.
If A is invertible, the equation $Ax=b$ has a solution for every vector b (namely $x=A^{-1}b$) so the column vectors span Rn.

4. Apr 30, 2005

### Hurkyl

Staff Emeritus
A matrix doesn't have to be invertible to be diagonalizable...

5. Apr 30, 2005

### EvLer

Ohh, yeah
Thanks for pointing that out! At first I was confused by your remark. I had to prove it to myself.
But here's the thing that now bothers me:
given:
A is a matrix, Q - matrix of its eigenvectors, Q' - inverse of Q, D - matrix containing eigenvalues of A by diagonal.
So A is diagonilazable iff A = QDQ'.

Then let's say I take determinant of both sides like so:
det(A) = det(QDQ')
Then det(A) = det(D), since det(QQ') = 1.
But if, as you pointed out, A is not invertible necessarily, then det(A) = 0 but det(D) cannot be = 0!
What am I missing here? It is not supposed to contradict each other

6. Apr 30, 2005

### matt grime

Why can't the determinant of a Diagonal matrix be zero?

Last edited: Apr 30, 2005
7. Apr 30, 2005

### EvLer

this
although I see your point, because eigenvalue can be 0, while eigenvector cannot. I am still confused though, because it is defined exactly the same way in my text-book as well, i.e. Q has to be invertible.

8. Apr 30, 2005

### Data

Is the 0 matrix diagonalizable? Invertible? Once you've answered these, can you guess at the link (and even better, you can try to prove your conjecture!) between diagonalizability and invertibility?

9. Apr 30, 2005

### matt grime

So what, why is this confusing, what's that link got to do with what I wrote? At no point does it state either A or D are invertible, indeed A is invertible if and only if D is.

10. Apr 30, 2005

### EvLer

Was working on conjecture as Data suggested. Came to same conclusion (yooo-hooo, it worked!)
Thanks for confirming it

P.S. Oh, anwering your question: I think I've read it wrong, I thought you said that Q does not have to be invertible. Anyway, sorry for confusion and thanks much again.

Last edited: Apr 30, 2005
11. Apr 30, 2005

### Data

or, equivalently you can say that a diagonalizable matrix is invertible iff all its eigenvalues are nonzero. This can, in fact, be generalized too.

Last edited: Apr 30, 2005