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?
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.
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!
It is a real and symmetric matrix so it is diagonalizable thanks to the spectral theorem.
It depends on the number of eigenvalues you've found. If a matrix has distinct eigenvalues then it is diagonalizable. If it has less than distinct eigenvalues then it may or may not be diagonalizable.