Diagonalisability of a Matrix with Zero Vector Elements

[sassie]

Homework Statement

Is the following matrix diagonalisable?

[0 0 0 0 0 ...0
0 1 0 0 0 ...0
0 0 2 0 0 ... 0
0 0 0 3 0 ...0
. . .
. . .
. . .
. .0
0 0 0 0 0 0...n]

(having trouble showing the mtarix, basically 0,1,2,3...n down the diagonal and zeroes off-diagonal)

Homework Equations

Given.

The Attempt at a Solution

I thought that because it contains the zero vector, the matrix doesn't have n linearly independent columns, thus not diagonalisable? I'm not entirely sure. Plus, is it possible to find the eigenvalues and eigenvectors for the matrix?

 

[gabbagabbahey]

If you mean

\begin{pmatrix}0&0&0&0&\ldots&0\\0&1&0&0&\ldots&0\\0&0&2&0&\ldots&0\\0&0&0&3&\ldots&0\\\vdots&\vdots&\vdots&\vdots&\ldots&\vdots\\0&0&0&0&\ldots&n\end{pmatrix}

then isn't the matrix already diagonalized?

 

[sassie]

Yes, that's the matrix that I mean. :)

So, if the matrix has a zero column, it can still be diagonalisable?
I thought that if it didn't have n linearly independent vectors, it cannot diagonalisable...

 

[gabbagabbahey]

http://en.wikipedia.org/wiki/Diagonalizable_matrix]In[/PLAIN] linear algebra, a square matrix A is called diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix P such that P^{-1}AP is a diagonal matrix.

Obviously, since your matrix is diagonal, choosing P=I_n (the n\times n identity matrix) will show that it is diagonalizable.

I think you are mistaking a matrix's columns for its eigenvectors; if an n\times n matrix doesn't have n linearly independent eigenvectors it isn't diagonalizable.

 

[sassie]

Okay, thank you! I should obviously read my notes better next time! :)

You've been a great help!