- #1
kehler
- 104
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Homework Statement
a) Consider a real matrix, A =
[a b]
[c d]
Give simple algebraic criteria in terms of a,b,c,d for when the following are true:
i)there exists a real matrix P that diagonalises A,
ii)there exists a complex matrix P that diagonalises A, but no real matrix P that does
b) Let A be a 2 x 2 matrix which cannot be diagonalized by any matrix P (real or not). Prove there is an invertible real 2 x 2 matrix P such that
P-1AP =
[ lambda 1 ]
[ 0 lambda]
The Attempt at a Solution
Here's what I have so far (the 'x's are meant to be lambdas):
a)To find the eigenvalues, let det(A-xI) = 0
So, (a-x)(d-x)-bc = 0
x2 - (a+d)x -bc + ad = 0
Using the quadratic formula,
x = ((a + d) +- sqrt(a2 + d2 -2ad +4bc) ) / 2
i) When there are two real and linearly independent eigenvectors, A is diagonalisable by a real matrix P. This occurs when A has two distinct real eigenvalues.
So a2 + d2 -2ad +4bc > 0
ii) When there are two complex and linearly independent eigenvectors, A is diagonalisable by a complex matrix P. This occurs when A has two distinct complex eigenvalues.
So a2 + d2 -2ad +4bc < 0
When a2 + d2 -2ad +4bc = 0, a has one eigenvalue ((a+d)/2a) of multiplicity 2. Row reducing (A-xI) for this eigenvalue produces at most one free variable, and so there will not be two linearly independent eigenvectors. Thus, A is not diagonalisable when a2 + d2 -2ad +4bc = 0.
Is this correct?? :S I don't quite know what the question means by 'simple algebraic criteria'..
b) I have no clue how to do this part. If A can't be diagonalised, it will not have two linearly independent eigenvectors. What then will the columns of P be? :S
How should I go about proving this?
Any help would be much appreciated :)