Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm currently working on the question:

Suppose that $T:V\rightarrow V $ is a linear map on a 4 dimensional vector space $V$, and that $T$ has 3 different eigenvalues, one of which has a 2-dimensional eigenspace.

i) Must $T$ be diagonalizable?

ii) Can $T$ be diagonalizable?

My thoughts are yes, to both i) and ii). My justification is that the sum of the dimensions of the eigenspaces ($1+1+2$) is equal to the dimension of $V$ which is 4. Here I am assuming that the other two eigenspaces are of dimension 1.

Is my reasoning correct?

share|improve this question

1 Answer 1

up vote 0 down vote accepted

(It's not hard to prove the other eigenspaces have dimension 1, if you want)

The main thing is whether or not you've considered whether the different eigenspaces are linearly dependent or not.

For example, if you take the three one-dimensional subspaces of $\mathbb{R}^3$, the first having basis vector $[1,0,0]$, the second $[1,1,0]$, and the third $[0,1,0]$, then the sum of their dimensions is 1+1+1, but collectively they don't add up to a three-dimensional vector space!

If you've thought about this issue, and think in your problem that it's sufficiently obvious that it doesn't need explanation, then you're probably fine. But it's possible your professor may be skeptical and you should write on it.

If you haven't thought about this issue, then your proof isn't fine and you need to do so! Although you have a chance to get lucky and have your professor think you've thought of it.

Also, I assume you're invoking a theorem that having a complete set of eigenspaces means the matrix is diagonalizable? I'll leave it to your judgement whether the fact you are doing that needs to be said explicitly.

share|improve this answer
    
Thanks for your answer. I thought that distinct eigenvectors (as in, having different eigenvalues) are always linearly independent though? As for the two-dimensional eigenspace mentioned, wouldn't both eigenvectors be linearly independent of eachother, otherwise it wouldnt be two-dimensional? –  Mel May 3 '13 at 11:40
    
@Mel: Yes. The issue is obvious if you know your stuff. However, the question is also "obvious" if you don't know your stuff and are overlooking an important concept. Part of the point of an assignment is to let your teacher tell the difference between the two. :) –  Hurkyl May 3 '13 at 11:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.