Solving Repeated Eigenvalues - [2,2,1;1,3,1;1,2,2]

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In summary: However, because this equation is nonlinear, there are an infinite number of such vectors, all of which are valid solutions to the equation. For your eigenvalue two, you will not find a corresponding eigenvector in this space. Instead, you will find that the equation leads to a single vector in the space, which is the so called eigenvector of negation, or v_2=-x.
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Homework Statement



Obtain the eigenvalues and corrosponding eigenvectors for the matrix: [2,2,1;1,3,1;1,2,2]

Homework Equations





The Attempt at a Solution



I can solve for the eigenvalues, 5, 1, and 1

I can solve the eigenvalue 5 for the eigenvector B[1;1;1]

Yet somehow, the 1 eigenvalues are supposed to give me two different eigenvectors. Unfortunately, this makes no sense to me, which is why I'm here...

Can someone please explain to me how to finish solving this question?
 
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  • #2
Since 1 is an eigenvaue, the equation
[tex]\left[\begin{array}{ccc}2 & 2 & 1\\1 & 3 & 1\\ 1 & 2 & 2\end{array}\right]\left[\begin{array}{c} x \\ y \\ z\end{array}\right]= \left[\begin{array}{c}x \\ y \\ z\end{array}\right][/tex]
must have a non-trivial solution. Multiplying that out and looking at individual rows, we must have 2x+ 2y+ z= x, x+ 3y+ z= y, x+ 2y+ 2z= z. Because 1 is an eigenvalue, those are not independent and must have an infinite number of solutions. In fact, in this particular problem, the three equations those reduce to, x+ 2y+ z= 0, x+ 2y+ z= 0, and x+ 2y+ z= 0, are all the same equation. You can solve for z as a function of x and y: z= -x- 2y. That means you can choose any numbers you like for x and y, then solve for z to get an eigenvalue. Can you choose them so that you get two independent vectors? (I always like to use simple numbers like 0 and 1.)
 
  • #3
I call your matrix B. Then I presume you solved [itex]\det(B-\lambda\mathbb{I}_{3\times 3})=0[/itex] to find the eigenvalues [itex]\lambda[/itex]. You found [itex]\lambda_1=5 \text{ and } \lambda_2=1[/itex].

Once you have done that I further assume you tried to find the eigenvector [itex]v_i[/itex] associated to the eigenvector [itex]\lambda_i[/itex] by solving [itex](B-\lambda_i\mathbb{I}_{3\times 3})v_i=0[/itex]. For your eigenvalue one you will find that this equation leads to a two-dimensional solution space for [itex]v_2[/itex], the so called eigenspace. ANY vector in this two-dimensional space will be an eigenvector with eigenvalue one.
 
Last edited:

1. What are repeated eigenvalues?

Repeated eigenvalues are eigenvalues that appear more than once in the characteristic polynomial of a matrix. This means that the matrix has fewer distinct eigenvalues than its dimension.

2. Why is it important to solve for repeated eigenvalues?

Solving for repeated eigenvalues is important in understanding the behavior and properties of a matrix. It can help identify symmetry and other patterns in the matrix, as well as determine its diagonalizability.

3. How do you solve for repeated eigenvalues?

To solve for repeated eigenvalues, we first find the eigenvalues of the matrix using the characteristic polynomial. If there are repeated eigenvalues, we then use the eigenvectors corresponding to those eigenvalues to find the generalized eigenvectors. The generalized eigenvectors can then be used to find the Jordan canonical form of the matrix.

4. Can a matrix have more than one repeated eigenvalue?

Yes, a matrix can have multiple repeated eigenvalues. This means that there are several eigenvalues that have the same algebraic multiplicity, or number of times they appear in the characteristic polynomial.

5. How do you know if a matrix has repeated eigenvalues?

We can determine if a matrix has repeated eigenvalues by finding the eigenvalues of the matrix and checking if any appear more than once in the characteristic polynomial. Alternatively, we can also check if the geometric multiplicity, or dimension of the eigenspace, is less than the algebraic multiplicity of an eigenvalue.

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