Question

eigen values and vectors

Answer 1

\[A=\left[\begin{matrix}2 & 1 \\ 1 & 2\end{matrix}\right]\]
please show steps to get to the eigenvalues 3 and 1

Answer 2

for some reason my lecture notes are conflicting with my past exam papers on this...should be really easy but my answer was wrong....i thought it was 2 and 3

Answer 3

\[det(A-\lambda I)=0 =>det\left(\begin{matrix}2- \lambda & 1 \\ 1 & 2-\lambda\end{matrix}\right)=(2-\lambda)(2-\lambda)-1=\lambda^2-4\lambda+3=(\lambda-3)(\lambda-1)\]
So the eigenvalues are 3 and 1.
Do you follow the working now?

Answer 4

Sorry, the final factorisation was cut off there:
\[(\lambda-3)(\lambda-1)\]

Answer 5

ohhhh damn! i forgot to minus the 1

Answer 6

Ah, easy mistake :)

Answer 7

just looking at the stuff in brackets....man i am popular

Answer 8

nice profile pic traxter

Answer 9

Thanks @Algebraic! :D He is my hero!

Answer 10

@traxter
Could hardly have chosen better... :)

Answer 11

His quote:
"I learned very early the difference between knowing the name of something and knowing something"
is what I live my studies by :)

Answer 12

\[\textbf Ax=\lambda x\]\[(\textbf A-\lambda \textbf I)x=0\]\[\left|\textbf A-\lambda \textbf I\right|=0\]\[\left|\left[\begin{matrix}2 & 1 \\ 1 & 2\end{matrix}\right]-\lambda\left[ \begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right]\right|=0\]\[\left|\left[\begin{matrix}2 & 1 \\ 1 & 2\end{matrix}\right]-\left[ \begin{matrix} \lambda & 0 \\ 0 & \lambda \end{matrix}\right]\right|=0\]\[\left|\left[\begin{matrix}2-\lambda & 1 \\ 1 & 2-\lambda\end{matrix}\right]\right|=0\]\[(2-\lambda)^2-1=0\]
\[4-4\lambda+\lambda^2-1=0\]\[3-4\lambda+\lambda^2=0\]\[(\lambda-1)(\lambda-3)=0\]\[\lambda_1=1\qquad\lambda_2=3\]

Answer 13

overkill! But i understand now...silly mistake...guess its getting late

Answer 14

now the vectors?

Answer 15

the vectors are 1,1 and -1,1. thanks!

Answer 16

for \(\lambda_1=1\)
\[\textbf Ax=\lambda_1 x\]
\[\left[\begin{matrix}2 & 1 \\ 1 & 2\end{matrix}\right]\left[\begin{matrix}x_1 \\ x_2\end{matrix}\right]=\left[\begin{matrix}x_1 \\ x_2\end{matrix}\right]\]
\[2x_1+x_2=x_1\]\[x_1+x_2=0\]\[x_2=-x_1\]
let\[t=x_1=-x_2\]
\[\left[\begin{matrix}x_1 \\ x_2\end{matrix}\right]=t\left[\begin{matrix}1 \\ -1\end{matrix}\right]\]
________

for \(\lambda_1=3\)
\[\textbf Ax=\lambda_1 x\]
\[\left[\begin{matrix}2 & 1 \\ 1 & 2\end{matrix}\right]\left[\begin{matrix}x_1 \\ x_2\end{matrix}\right]=3\left[\begin{matrix}x_1 \\ x_2\end{matrix}\right]\]
\[2x_1+x_2=3x_1\]\[-x_1+x_2=0\]\[x_1=x_2\]
let\[u=x_1=x_2\]
\[\left[\begin{matrix}x_1 \\ x_2\end{matrix}\right]=u\left[\begin{matrix}1 \\ 1\end{matrix}\right]\]