Your work so far is correct. At least, those are the eigenvalues I get. You'll notice there are only two eigenvalues. What are the algebraic multiplicities of those two eigenvalues?
A = ( -1 1 -1)
( 0 c 0)
(3+2c-c^2 b 3)
can only be diagonalised for some values of the parameters b and c . Construct a table
showing the values of b and c for which diagonalisation is possible. You must show
For this I have done the following:
P(ƛ) = -(c -ƛ)((-1-ƛ)(3-ƛ)+3+2c-c^2)
For what I finally got
c-ƛ =0 -> ƛ =c
Then ƛ^2 -2ƛ +2C -ƛ^2=0
and then getting the value for ƛ
ƛ = 1+-(1-c)
Is this right?
And what should I do next with this problem?
Thank you for your time.
That is the definition of diagonalizable, I grant you. However, it's not much use in the present circumstance, I'm afraid. Have you seen this one:
A matrix A is diagonalizable if and only if, for every eigenvalue of A, its geometric multiplicity equals its algebraic multiplicity.
You always know that the geometric multiplicity of an eigenvalue is less than or equal to the algebraic multiplicity. Here, we're saying that they have to be equal for all the eigenvalues. If that happens, you can diagonalize the matrix.
So, how can you use this idea to find out when your matrix is diagonalizable?