Question

\[A=\left(
\begin{array}{ccc}
2 & -3 & 1 \\
1 & -2 & 1 \\
1 & -3 & 2
\end{array}
\right)\]

Answer 1

\[x=\left(
\begin{array}{c}
1 \\
1 \\
1
\end{array}
\right)\]

Answer 2

\[y=\left(
\begin{array}{c}
3 \\
1 \\
0
\end{array}
\right)\]

Answer 3

\[z=\left(
\begin{array}{c}
-1 \\
0 \\
1
\end{array}
\right)\]

Answer 4

x , y, z are eigenvector of A

Answer 5

Find two matrices X and D such that X*D=A*X

Answer 6

@Zarkon

Answer 7

this is not unique...choose X to be the zero matrix

Answer 8

may be they want something other than trivial case?

Answer 9

D is the diagonal of eugene values; and X is a matrix of columns of (x y z) if those are the eugene vectors

Answer 10

so how do we find eigenvalue from eigenvector

Answer 11

\[D=\begin{pmatrix}e_1&0&0\\0&e_2&0\\0&0&e_3 \end{pmatrix}\]
\[X=\begin{pmatrix}\vec e_1&\vec e_2&\vec e_3 \end{pmatrix}\]

Answer 12

the values are unique; just value out A is my first thought

Answer 13

unique is prolly a bad term

Answer 14

We could use the definition

Ax = Lx , and most times the L is usually a moot point

Answer 15

i still think Evalueing A is my safest idea

Answer 16

I tried X D= A X and it doesn't seem to match

Answer 17

\[det\begin{pmatrix}
2-n & -3 & 1 \\
1 & -2-n & 1 \\
1 & -3 & 2-n
\end{pmatrix}\]

should get us the Evalues

Answer 18

eigen values :]

Answer 19

oh,right by finding determinant

Answer 20

got it , I got

\[\lambda1=0,\lambda2=1,\lambda3=1\]

Answer 21

good,

then we rref A by subtracting lambda from the diag to relate it to th vectors

Answer 22

got it , it matches, thanks so much Amistre

Answer 23

youre welcome