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

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The discussion presents a question about finding a 3x3 matrix with the characteristic equation (1-lambda)^3=0, focusing on the dimensions of the eigenspace corresponding to lambda=1. Three matrices are provided as examples: the first matrix has an eigenspace dimension of one, the second has a dimension of two, and the third has a dimension of three. The calculations for the dimensions are based on the rank of the matrices subtracted from three. This illustrates how different matrix structures can yield varying eigenspace dimensions while maintaining the same characteristic polynomial. The examples effectively demonstrate the relationship between matrix rank and eigenspace dimension.
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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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) $\dim V_1=3-\mbox{rank }(A-I)=3-\mbox{rank }\begin{bmatrix}{0}&{0}&{0}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{bmatrix}=3-0=3$
 
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