Cutiee pie's question at Yahoo Answers (dimension of eigenspaces).

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In summary, an easy example of a 3x3 matrix with characteristic equation (1-lambda)^3=0 is:(i) $A=\begin{bmatrix}{1}&{1}&{0}\\{0}&{1}&{1}\\{0}&{0}&{1}\end{bmatrix}$ with the eigenspace corresponding to lambda=1 having dimension one.(ii) $A=\begin{bmatrix}{1}&{0}&{1}\\{0}&{1}&{0}\\{0}&{0}&{1}\end{bmatrix}$ with the eigenspace corresponding to lambda=1 having dimension two.(iii) $A=\begin{bmatrix}{1}&
  • #1
Fernando Revilla
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Here is the question:

Give an easy example of a 3x3 matrix with characteristic equation (1-lambda)^3=0 such that:?
a) the eigenspace corresponding to lambda=1 has dimension one
b) the eigenspace corresponding to lambda=1 has dimension two
c) the eigenspace corresponding to lambda=1 has dimension three

Here is a link to the question:

Give an easy example of a 3x3 matrix with characteristic equation (1-lambda)^3=0 such that:? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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  • #2
Hello cutiee pie,

Choose:

(i) $A=\begin{bmatrix}{1}&{1}&{0}\\{0}&{1}&{1}\\{0}&{0}&{1}\end{bmatrix}$ (ii) $A=\begin{bmatrix}{1}&{0}&{1}\\{0}&{1}&{0}\\{0}&{0}&{1}\end{bmatrix}$ (iii) $A=\begin{bmatrix}{1}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0}&{1}\end{bmatrix}$

Clearly, in these cases the characteristic polynomial is $\chi(\lambda)=(1-\lambda)^3$. Besides,

(i) $\dim V_1=3-\mbox{rank }(A-I)=3-\mbox{rank }\begin{bmatrix}{0}&{1}&{0}\\{0}&{0}&{1}\\{0}&{0}&{0}\end{bmatrix}=3-2=1$(ii) $\dim V_1=3-\mbox{rank }(A-I)=3-\mbox{rank }\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{bmatrix}=3-1=2$

(iii) $\dim V_1=3-\mbox{rank }(A-I)=3-\mbox{rank }\begin{bmatrix}{0}&{0}&{0}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{bmatrix}=3-0=3$
 

FAQ: Cutiee pie's question at Yahoo Answers (dimension of eigenspaces).

What is the dimension of an eigenspace?

The dimension of an eigenspace is the number of linearly independent eigenvectors associated with a particular eigenvalue.

Why is the dimension of an eigenspace important?

The dimension of an eigenspace is important because it tells us how many linearly independent solutions there are for a particular eigenvalue. This information is crucial in solving systems of linear equations and understanding the behavior of linear transformations.

How do you find the dimension of an eigenspace?

To find the dimension of an eigenspace, we first need to find the eigenvalues of the matrix or linear transformation in question. Then, for each eigenvalue, we find the corresponding eigenvectors. The number of linearly independent eigenvectors for a particular eigenvalue is the dimension of the eigenspace.

Can the dimension of an eigenspace be greater than the dimension of the original vector space?

Yes, the dimension of an eigenspace can be greater than the dimension of the original vector space. This can happen if there are repeated eigenvalues or if the original vector space is not finite-dimensional.

How does the dimension of an eigenspace relate to the diagonalizability of a matrix?

A matrix is diagonalizable if and only if the dimension of each eigenspace is equal to the multiplicity of its corresponding eigenvalue. In other words, if the dimension of each eigenspace matches the number of times the eigenvalue appears as a root of the characteristic polynomial, then the matrix is diagonalizable.

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