Algebraic Multiplicity and Eigenspace

In summary, the eigenspace for lambda=5 in the matrix A is two-dimensional if and only if there exists a non-zero value for h such that the rank of A-lambda*I is 2. This can be achieved by setting h=6, resulting in a rank of 2 and a two-dimensional eigenspace.
  • #1
Hashmeer
16
0

Homework Statement


Find h in the matrix A such that the eigenspace for lambda=5 is two-dimensional.

A= [5,-2,6,-1] [0,3,h,0] [0,0,5,4] [0,0,0,1]
A-lambda*I(n) = [0,-2,6,-1] [0,-2,h,0] [0,0,0,4] [0,0,0,-4]


Homework Equations





The Attempt at a Solution


I'm not really sure how to do this. Is there some relation between algebraic multiplicity and the eigenspace of a matrix that would help?

I tried solving this by row operations to solve for the eigenvectors in the hope that I would be able to eliminate values of h that would have resulted in more or less than a 2 dimensional eigenspace, but it didn't work out. This is the matrix I got (keep in mind this is after applying the lambda value to the diagonals) A= [0,-2,6,-1] [0,0, h-6,1] [0,0,0,4] [0,0,0,0].

I'm pretty confused about this any help would be greatly appreciated. Thanks.
 
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  • #2
Hashmeer said:

Homework Statement


Find h in the matrix A such that the eigenspace for lambda=5 is two-dimensional.

A= [5,-2,6,-1] [0,3,h,0] [0,0,5,4] [0,0,0,1]
A-lambda*I(n) = [0,-2,6,-1] [0,-2,h,0] [0,0,0,4] [0,0,0,-4]


Homework Equations





The Attempt at a Solution


I'm not really sure how to do this. Is there some relation between algebraic multiplicity and the eigenspace of a matrix that would help?

I tried solving this by row operations to solve for the eigenvectors in the hope that I would be able to eliminate values of h that would have resulted in more or less than a 2 dimensional eigenspace, but it didn't work out. This is the matrix I got (keep in mind this is after applying the lambda value to the diagonals) A= [0,-2,6,-1] [0,0, h-6,1] [0,0,0,4] [0,0,0,0].

I'm pretty confused about this any help would be greatly appreciated. Thanks.


Hey there! the eigenspace of having lambda=5 is exactly the nullspace of A-lambda I . and since it is 2 dimensional , it suggest that the rank of the matrix is ?
 
  • #3
So the rank = 2 since rank = # columns (4 in this case) - dimNul A (in this case 2). So if the rank is to equal 2 then I will need another free variable, or I need to remove a pivot position. So since h-6 is in a pivot position I can easily make it a nonpivot column by setting h=6. This would ensure that the dimension of the null space is 2.

Am I going the right way with this? It makes sense to me if this is right. Thanks for the help!
 
  • #4
Hashmeer said:

Homework Statement


Find h in the matrix A such that the eigenspace for lambda=5 is two-dimensional.

A= [5,-2,6,-1] [0,3,h,0] [0,0,5,4] [0,0,0,1]
A-lambda*I(n) = [0,-2,6,-1] [0,-2,h,0] [0,0,0,4] [0,0,0,-4]
So any eigenvector of A corresponding to eigenvalue 5 must satisfy
[tex]\begin{bmatrix}0 & -2 & 6 & 1 \\ 0 & -2 & h & 0 \\ 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & -4\end{bmatrix}\begin{bmatrix}w \\ x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0\end{bmatrix}[/tex].

That gives the equations -2x+ 6y+ z= 0, -2x+ hy= 0, 4z= 0, and -4z= 0. The last two obviously give z= 0 so the first two equations become -2x+ 6y= 0 and -2x+ hy= 0. One obvious eigenvector is (u, 0, 0, 0). There will be another, independent, eigenvector, and so the eigenspace will be two dimensional if and only if there exist non-zero x and y satifying both -2x+ 6y= 0 and -2x+ hy= 0.


Homework Equations





The Attempt at a Solution


I'm not really sure how to do this. Is there some relation between algebraic multiplicity and the eigenspace of a matrix that would help?

I tried solving this by row operations to solve for the eigenvectors in the hope that I would be able to eliminate values of h that would have resulted in more or less than a 2 dimensional eigenspace, but it didn't work out. This is the matrix I got (keep in mind this is after applying the lambda value to the diagonals) A= [0,-2,6,-1] [0,0, h-6,1] [0,0,0,4] [0,0,0,0].

I'm pretty confused about this any help would be greatly appreciated. Thanks.
 

1. What is the difference between algebraic multiplicity and geometric multiplicity?

The algebraic multiplicity of an eigenvalue is the number of times the eigenvalue appears as a root of the characteristic polynomial. The geometric multiplicity, on the other hand, is the dimension of the corresponding eigenspace. In other words, the algebraic multiplicity counts the number of times an eigenvalue is repeated as a root, while the geometric multiplicity measures the number of linearly independent eigenvectors associated with that eigenvalue.

2. How do you calculate the algebraic and geometric multiplicities of an eigenvalue?

The algebraic multiplicity can be found by looking at the degree of the characteristic polynomial and counting the number of times the eigenvalue appears as a root. The geometric multiplicity can be determined by finding the nullity of the matrix (A - λI), where A is the matrix corresponding to the linear transformation and λ is the eigenvalue in question.

3. What is the significance of the algebraic and geometric multiplicities?

The algebraic multiplicity represents the total number of times an eigenvalue is repeated as a root, while the geometric multiplicity represents the dimension of the eigenspace. The geometric multiplicity is always less than or equal to the algebraic multiplicity, and when they are equal, it means that the eigenspace is spanned by a single eigenvector. These multiplicities are important in understanding the behavior of a linear transformation and in diagonalizing matrices.

4. Can an eigenvalue have different algebraic and geometric multiplicities?

Yes, it is possible for an eigenvalue to have different algebraic and geometric multiplicities. This occurs when the eigenspace is not spanned by a single eigenvector, but rather by a combination of linearly independent eigenvectors. In this case, the geometric multiplicity will be less than the algebraic multiplicity.

5. How can you use the algebraic and geometric multiplicities to determine the diagonalizability of a matrix?

A square matrix is diagonalizable if and only if the sum of the geometric multiplicities of all distinct eigenvalues is equal to the dimension of the matrix. In other words, if all the eigenvalues have geometric multiplicity equal to their algebraic multiplicity, then the matrix is diagonalizable. If any eigenvalue has a geometric multiplicity less than its algebraic multiplicity, then the matrix is not diagonalizable.

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