# Sum of dimensions of eigenspaces?

• Sep 8th 2011, 12:17 PM
virussss123
Sum of dimensions of eigenspaces?
How do I calculate the sum of dimensions of eigenspaces to evaluate whether or not a matrix is diagonalisable?

E.g. dim(span{(-2 0 1)}) + dim(span{(1 0 0)}) ?
They're supposed to be written as vectors but I can't work out how to do that so I apologise for that. Thank you!
• Sep 8th 2011, 01:49 PM
Drexel28
Re: Sum of dimensions of eigenspaces?
Quote:

Originally Posted by virussss123
How do I calculate the sum of dimensions of eigenspaces to evaluate whether or not a matrix is diagonalisable?

E.g. dim(span{(-2 0 1)}) + dim(span{(1 0 0)}) ?
They're supposed to be written as vectors but I can't work out how to do that so I apologise for that. Thank you!

I'm confused. Can you restate the question.
• Sep 8th 2011, 01:51 PM
alexmahone
Re: Sum of dimensions of eigenspaces?
Quote:

Originally Posted by virussss123
How do I calculate the sum of dimensions of eigenspaces to evaluate whether or not a matrix is diagonalisable?

E.g. dim(span{(-2 0 1)}) + dim(span{(1 0 0)}) ?
They're supposed to be written as vectors but I can't work out how to do that so I apologise for that. Thank you!

dim(span{(-2 0 1)}) + dim(span{(1 0 0)}) = 1 + 1 = 2
• Sep 8th 2011, 01:53 PM
virussss123
Re: Sum of dimensions of eigenspaces?
I'm trying to find out if a matrix A is diagonalisable. In order to do this, I need to work out the sum of dimensions of eigenspaces.
Does that make it clearer?
• Sep 8th 2011, 01:54 PM
virussss123
Re: Sum of dimensions of eigenspaces?
Quote:

Originally Posted by alexmahone
dim(span{(-2 0 1)}) + dim(span{(1 0 0)}) = 1 + 1 = 2

Why does it end up as 1+1?
• Sep 8th 2011, 01:57 PM
alexmahone
Re: Sum of dimensions of eigenspaces?
Quote:

Originally Posted by virussss123
Why does it end up as 1+1?

For any non-zero vector v, dim(span(v)) = 1 because a basis of span(v) is {v}, which contains only one vector.
• Sep 8th 2011, 02:00 PM
virussss123
Re: Sum of dimensions of eigenspaces?
Quote:

Originally Posted by alexmahone
For any vector v, dim(span(v)) = 1 because a basis of span(v) is {v}, which contains only one vector.

That makes sense, thanks. So if there are 3 vectors, would it make the sum 3? i.e if there are 3 eigenvalues resulting in 3 different eigenspaces, would the sum of dimensions of eigenspaces be 3? Thank you
• Sep 8th 2011, 02:02 PM
alexmahone
Re: Sum of dimensions of eigenspaces?
Quote:

Originally Posted by virussss123
So if there are 3 vectors, would it make the sum 3?

I guess so, as long as none of the vectors is the zero vector, in which case dim(span(v)) = 0.
• Sep 8th 2011, 02:04 PM
virussss123
Re: Sum of dimensions of eigenspaces?
Quote:

Originally Posted by alexmahone
I guess so, as long as none of the vectors is the zero vector, in which case dim(span(v)) = 0.

That answers everything. Thank you alexmahone!