1. ## Linear Algebra

Hi Everyone,

Just needed to check these since I dont particularly enjoy linear algebra, and im a little rusty.

(1) If I have an n*n matrix that has n different eigenvalues, that automatically makes the matrix diagonalizable, does it not?

(2) If an n*n matrix has n eigenvalues, some of which share the same value, can it still be diagonalizable?

2. Originally Posted by Maccaman
(1) If I have an n*n matrix that has n different eigenvalues, that automatically makes the matrix diagonalizable, does it not?
Yes. Because distinct eigenvectors to distinct eigenvalues are linearly independent. Therefore, there is an eigenbasis. This makes the matrix diagnolizable.

(2) If an n*n matrix has n eigenvalues, some of which share the same value, can it still be diagonalizable?
Yes

3. Okay...........

But how about the following 9*9 matrix?

$\begin{pmatrix}0&0&0&1&0&0&0&0&20\\0&0&1&0&0&0&0&1 9&0\\0&1&0&3&0&1&7&0&0\\1&0&3&0&0&0&0&0&0\\0&0&0&0 &0&0&0&4&0\\0&0&1&0&0&0&0&12&0\\0&0&7&0&0&0&8&200& 0\\0&19&0&0&4&12&200&4&1\\20&0&0&0&0&0&0&1&10\end{ pmatrix}$

I dont think it is diagnolizable, but im not sure if the reason i think its not diagnolizable is actually correct.

Matlab gives eigenvalues
-195.49185235650116965
-15.656116624466667631
-3.0548869783011998180
-0.20599841277256770387e-16
0.42029977764762430949e-3
0.10046089267634544795
3.0880586940309537880
25.630116654326307923
207.38379941845772692
Remark the fourth one is very small. It may be 0. That would mean the determinant is 0 and the matrix would not be invertible.

5. Hello,

Note the second and 6th rows that have zeros in the same place.
So they're good candidates to find a linear combination.

There's already 1 at the same place. Remain 19 and 12 in the 8th column.
Find a row that will eliminate the difference between 19 and 12, namely row 5.

row2 = row6 + (7/4) row5

hence the rows are not linearly independent and hence the matrix is not invertible.

6. Originally Posted by Moo
Hello,

Note the second and 6th rows that have zeros in the same place.
So they're good candidates to find a linear combination.

There's already 1 at the same place. Remain 19 and 12 in the 8th column.
Find a row that will eliminate the difference between 19 and 12, namely row 5.

row2 = row6 + (7/4) row5

hence the rows are not linearly independent and hence the matrix is not invertible.
It is interesting to see that a computer program made such a big mistake in computing eigenvalues for large matrices. It seems to be a numerical problem. But even if the matrix is not invertible does not imply it is non-diagnolizable.

EDIT: I did not realize the symmetry, but flyingsquirrel realized it!

That's not what I meant.
If you can compute eigenvalues then the matrix is diagonalizable isn't it?

8. Originally Posted by Maccaman
But how about the following 9*9 matrix?

$\begin{pmatrix}0&0&0&1&0&0&0&0&20\\0&0&1&0&0&0&0&1 9&0\\0&1&0&3&0&1&7&0&0\\1&0&3&0&0&0&0&0&0\\0&0&0&0 &0&0&0&4&0\\0&0&1&0&0&0&0&12&0\\0&0&7&0&0&0&8&200& 0\\0&19&0&0&4&12&200&4&1\\20&0&0&0&0&0&0&1&10\end{ pmatrix}$

I dont think it is diagnolizable, but im not sure if the reason i think its not diagnolizable is actually correct.
It is a real and symmetric matrix so it is diagonalizable thanks to the spectral theorem.
Originally Posted by vincisonfire
If you can compute eigenvalues then the matrix is diagonalizable isn't it?
It depends on the number of eigenvalues you've found. If a $n\times n$ matrix has $n$ distinct eigenvalues then it is diagonalizable. If it has less than $n$ distinct eigenvalues then it may or may not be diagonalizable.