Jordan Forms Problem: Finding Det(A) and Eigenvalues

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Homework Help Overview

The discussion revolves around a problem involving the determination of the determinant of a matrix, finding eigenvalues, and understanding the relationship between algebraic and geometric multiplicities. The context includes the use of the Cayley-Hamilton theorem and the diagonalizability of matrices.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss methods for finding the determinant and eigenvalues, with one participant questioning the necessity of calculating the determinant. There are inquiries about how to find geometric multiplicities and the relationship between algebraic and geometric multiplicities in the context of diagonalizability.

Discussion Status

Some participants have provided guidance on how to approach the problem, suggesting methods for finding eigenvalues and eigenvectors. There is an ongoing exploration of the relationship between the dimensions of the kernel of the matrix and geometric multiplicity, with participants confirming each other's understanding.

Contextual Notes

There is a mention of a specific matrix and an eigenvector provided in the problem statement, which may limit the generalizability of the discussion. Participants are also navigating assumptions about the properties of the matrix in question.

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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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