How to determine if a 3x3 matrix is diagonalisable or not?

[neelakash]

Can anyone tell me how can we determine if a 3x3 matrix is diagonalisable or not?It is not a homework problem...But I need to know this.Say I am given a 3x3 real matrix...And I want to see if it is diagonalizable or not without brute evaluation...Then how can I dio this?

 

[mathwonk]

there are some special matrices which are automatically diagonalizable with no calculation, namely symmetric ones, and i guess over C, ones which commute with their adjoints.

in general, one needs to do some computation, find the chracteristic polynomial, the eigenvalues, and then see whether the number of independent eigenvectors for each of the eigenvalues equals the multiplicity of the eigenvalue as a root of the characteristic polynomial.

i hope this is right, i forget quickly, and it has been 6 months since i taught this course.

 

[Office_Shredder]

Pretty close mathwonk. You need a number of eigenvectors equal to the dimension of the space the matrix is mapping on/from. For example, on a 4x4 matrix, if 2 is the only eigenvalue, as a double root of the characteristic polynomial, even if you have two linearly independent eigenvectors for 2, you still don't have enough as you need enough to match the dimension of the space (4 in this case). Basically, watch out for the fact that not all the roots of the characteristic polynomial are real

 

[fopc]

Can anyone tell me how can we determine if a 3x3 matrix is diagonalisable or not?It is not a homework problem...But I need to know this.Say I am given a 3x3 real matrix...And I want to see if it is diagonalizable or not without brute evaluation...Then how can I dio this?

The simplest statement I can think of about the diagonizability of a matrix is
that the algebraic multiplicity must equal the geometric multiplicy
for each eigenvalue. It's just another way of saying that there are
as many eigenvectors as there are eigenvalues.
Of course this is just a statement.

The only test I'm aware of is to compute all the eigenvectors, and look to see whether there
are enough.

 

[radou]

Another useful fact is that the geometric multiplicity is less or equal to the algebraic multiplicity, for every eigenvalue. Further on, the geometric multiplicity is greater or equal to one, and hence, in some cases, one can, knowing only the algeraic multiplicities, directly see what the geometric miltiplicities are, and conclude about the possibility of diagonalization. For example, if your characteristic polynomial is of the form p(x) = x(1 - x)(2 - x) (doesn't really matter now), you see that the spectre of the matrix is {0, 1, 2}, and a(0) = a(1) = a(2) = 1, and hence g(0) = g(1) = g(2), so the matrix can be diagonalized.

 

[mathwonk]

here is an example. suppose the charcteristic polynomial of T is (X-a)^n.

that means that (T-a)^n = 0. Diagonalizability emans that actually T-a = 0 already. so you have to compoute the kernel;s of the various powers of T-a to see how far T is from diagonalizability.more generally, if the characteristic polynomial is ∏ (X-ai)^ni, that means that ∏ (T-ai)^ni = 0 and diagonalizability emans that already ∏(T-ai) = 0.

so again, for eqch ai you have to compute the kernel of the various powers of T-ai.

shredder, please read my post again and see if it isn't the same condition as yours.

 

[daniel_i_l]

In addition to what has been said, all symmetric matrices over R are diagonalizable.
To check for a nxn matrix over F, you have to find a basis for F^n where all the vectors in the basis are e-vectors.

 

[matt grime]

Again, daniel, all of what you wrote has been said (in the very first reply by mathwonk).