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Amanda's question at Yahoo! Answers (Eigenvalues and eigenvectors)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Here is the question:

So, I'm attempting to work on this question:

Given that matrix A =
0 1 1
1 0 1
1 1 0

and has λ = -1 as one of its eigenvalues, find the corresponding eigenvectors.

The answers are
-1
1
0

and

-1
0
1

BUT.. I don't understand how they got that because can't you bring it down to the reduced form of
1 0 0
0 1 0
0 0 1

So how do you get the corresponding eigenvalues from that? I'm so confused! Help please :)
Here is a link to the question:

How to I find corresponding eigenvectors? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Amanda,

I don't understand either how they got those eigenvalues, surely is a typo. Using the transformations $R_2\to R_2-R_1$, $R_3\to R_3-R_1$ and $C_1\to C_1+C_2+C_3$ we get: $$\begin{aligned}
\begin{vmatrix}{-\lambda}&{\;\;1}&{\;\;1}\\{\;\;1}&{-\lambda}&{\;\;1}\\{\;\;1}&{\;\;1}&{-\lambda}\end{vmatrix}&=\begin{vmatrix}{-\lambda}&{\;\;1}&{\;\;1}\\{\;\;1+\lambda}&{-\lambda}-1&{\;\;0}\\{\;\;1+\lambda}&{\;\;0}&{-\lambda-1}\end{vmatrix}\\&=\begin{vmatrix}{-\lambda+2}&{\;\;1}&{\;\;1}\\{\;\;0}&{-\lambda}-1&{\;\;0}\\{\;\;0}&{\;\;0}&{-\lambda-1}\end{vmatrix}\\&=(-\lambda+2)(\lambda+1)^2=0\\&\Leftrightarrow \lambda=2\mbox{ (simple) }\vee \;\lambda=-1\mbox{ (double) }
\end{aligned}$$ The eigenvectors are: $$\ker (A-2I)\equiv \left \{ \begin{matrix}-2x_1+x_2+x_3=0\\x_1-2x_2+x_3=0\\x_1+x_2-2x_3=0\end{matrix}\right. $$ As $\lambda=2$ is simple, $\dim(\ker(A-2I))=1$ and easily we find a basis of this eigenspace: $B_2=\{(1,1,1)\}$. On the other hand: $$\ker (A+I)\equiv \left \{ \begin{matrix}x_1+x_2+x_3=0\\x_1+x_2+x_3=0\\x_1+x_2+x_3=0\end{matrix}\right.$$ Now, $\dim(\ker(A+I))=3-\mbox{rank }(A+I)=3-1=2$ and easily we find a basis of this eigenspace: $B_{-1}=\{(-1,1,0),(-1,0,1)\}$.