Linear algebra:Diagonalization?

Question

Question: Let A = [2 0 -1]
................... ........... [4 1 -4]
................... ........... [2 0 -1]
is A diagonalizable ?
If yes then find P and D such that
D = (p^-1)A(P)
where P^-1 means inverse of P

Answer 1

At this hour of the night, that's -way- too much 'number crunching' for me ☺

But if you type 'linear algebra calculator' or 'matrix calculaotr' you'll het hundreds of sites that have calculators available to do these for you.

Doug

Answer 2

If I am not mistaken, you have to find lambda so that det (A-lambda I) = 0, these are the eigen values of A. If you can find 3 (because it's a 3x3 matrix), then A is diagonalizable.

D= (a_ij), where a_ij = 0 if i is different from j and a_ii are the Eigenvalues. But sorry, I don't recall how P was found. I took Linear Algebra 8 years ago and I only recall the basics. But there is a good article in Wikipedia, you probably have a book too and some solved problems.

http://en.wikipedia.org/wiki/Eigenvalues

Sorry, I have to go now, but I will try to be back to you with this later.

Ilusion

Answer 3

1. determine the eigenvalues of A...

if A has 3 eigenvalues then it is diagonalizable...

at any rate, even if it has fewer than 3 distinct eigenvalues...
determine the eigenvectors of those eigenvalues...

if you have 3 distinct eigenvectors... then you can get P... it's columns are simply the eigenvectors of A.

D is the scalar matrix whose corresponding component is the eigenvalue of the eigenvector on that column.


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