# Eigenvalues of a diagonalizable matrix

1. Oct 1, 2012

### mathmathRW

1. The problem statement, all variables and given/known data

I need to prove or disprove this statement:
"If all eigenvalues of a diagonalizable matrix A are equal and have the same value c, then A=cI."

2. Relevant equations

3. The attempt at a solution
I have tried coming up with a diagonalizable matrix that has eigenvalues with all the same value, but so far the only ones I can come up with are ones that fit A=cI. In other words, I can't disprove it. However....I have no idea how to prove it is true either!

2. Oct 1, 2012

### micromass

Staff Emeritus
Take an arbitrary diagonal matrix. Can you prove that the diagonal entries are eigenvalues? Can you prove that these entries are exactly the eigenvalues?

3. Oct 2, 2012

### HallsofIvy

Staff Emeritus
What micromass is suggesting is fairly simple but not enough, by itself, to answer this this question. It is not enough to prove that a diagonal matrix having a single eigenvalue has that eigenvalue on its diagonal. You must prove that any diagonalizable matrix having a single eigenvalue is diagonal.

Since a matrix, M, is said to be "diagonalizable" if and only if there exist an invertible matrix, U, such that UMU-1= D, with D diagonalizable, and that is equivalent to M= U-1DU it would be sufficient to prove: "If D is a diagonal matris and U is any invertible matrix, then U-1DU is also diagonal."

4. Oct 11, 2012

### gdr384

eigenvalues compress information in a matrix into a compact form. This information is retrieved and found through a few handy rules and equations.

(matrix A, identity matrix I, e eigenvalues, v eigenvectors, 0 zero vector)

first off the rules that are in your tool box:

inverse(A)*A=I
A=eI
*****your answer is that simple, if every element in the vector of eigenvalues is the same the it is the same as a constant(not a vector, multiplyed into that matrix)

if proof is needed:
for an nXn matrix their will be n eigen values and vectors associated with them

to find eigen values:

set Av=eIv
then (A-eI)v=0
set B=(A-eI) and take the determinate of both sides(note:the determinate of B is 0)
by taking the determinate you obtain a polynomial
this polynomial is the characteristic polynomial of the matrix, its roots are the eigenvalues of the matrix
(roots are found by setting the characteristic polynomial equal to zero)

the eigenvectors are found by returning to the rule (A-eI)v=0
(note: in solving an nXn matrix and thus nX1 e and v, the form of the solution will take n iterations in the fact you must find each vector with its eigen value as (A-e(1)*I)*v(1)=0 all the way to (A-e(n)*I)*v(n)=0

because this is a fact: Av=eIv
A=eI
and if e is a nX1 vector of the same value E
then A=EI