Jordan Forms Problem: Finding Det(A) and Eigenvalues

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To find the determinant of matrix A for eigenvalue calculations, it's essential to use the Cayley-Hamilton theorem and understand the relationship between algebraic and geometric multiplicities. The discussion emphasizes that if a matrix is diagonalizable, the algebraic multiplicity equals the geometric multiplicity. The equation provided allows for the calculation of eigenvalues and eigenvectors, with the geometric multiplicity determined by the dimension of the kernel of (A-λI). The conversation concludes with a confirmation that the dimension of the kernel indeed represents the geometric multiplicity.
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Homework Statement


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The Attempt at a Solution


I need an efficient way to find the det(A) so I can find the eigenvalues together with the trace or will use Cayley-Hamilton Thm. I can find the algebraic multiplicities but cannot find the geometric ones. If a matrix is diagonalisable; then AM=GM?
 

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You do NOT need to find the determimant of that matrix. You are given one eigenvector and asked to show that it is an eigenector and find the corresponding eigenvalue. Solve the equation
\begin{bmatrix} 7 & -1 & 3 & 1 \\ -2 & 6 & -3 & -1 \\ -4 & 2 & -1 & -2 \\4 & -2 & 6 & 7 \end{bmatrix} \begin{bmatrix}1 \\ 2 \\ 0 \\ 0\end{bmatrix}= \lambda \begin{bmatrix}1 \\ 2 \\ 0 \\ 0\end{bmatrix}
for \lambda.

You should then be able to reduce the matrix to 3 by 3 (or smaller) to find the remaining eigenvalues and eigenvectors. An n by n matrix is diagonalizable if and only if it has n independent eigenvectors.
 
How would I find the geometric multiplicities? I can do the rest of the question
 
Assuming you already found λ for the eigenvector.

Did you write out A-λI?

You should be able to find other eigenvector solutions easily.
That is, solutions of (A-λI)v=0.

This will give you the geometric multiplicity of this λ.

You should also find that the remaining eigenvalue is easy to deduce as well.
 
I like Serena said:
Assuming you already found λ for the eigenvector.

Did you write out A-λI?

You should be able to find other eigenvector solutions easily.
That is, solutions of (A-λI)v=0.

This will give you the geometric multiplicity of this λ.

You should also find that the remaining eigenvalue is easy to deduce as well.

Ok, thanks. Really should have remembered that. So, the dim{Ker(A-λI)}=Geometric Multiplicity?
 
Yes.
 
I like Serena said:
Yes.

Thanks again.
 

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