Eigen Vector

**AshleyT** · August 18th 2011, 01:08 AM

EDIT: just writing out my working. 2 mins

.

http://www.wolframalpha.com/input/?i=eigenvalues+{{1%2C+1%2C+1}%2C{2%2C2%2C+2}%2C+{3 %2C+3%2C+3}}

How is {1, 2, 3} and answer(Corresponding to EV = 6)? I get {2, 1, 0) and just can't seem to work out how they get (1, 2, 3)!

Eigen Value 6:

$\begin{matrix}5 & -1 & -1 \\ -2 & 4 & -2 \\ -3 & 3 & -3 \end{matrix}$

So, for
$-3x -3y + 3z = 0$
$x + y - z = 0$
I obtain:
$y\vec{-1, 1, 0) + z\vec{1, 0 , 1)$

$-2x -4y + -2z = 0$
$x -2y + z = 0$
I obtain:
$y\vec{2, 1, 0) + z\vec{-1, 0 , 1)$

...And actually just typing this out i've mega confused myself as there are now 4 different basis and there's meant to be a max of 3 :S.

**FernandoRevilla** · August 18th 2011, 01:18 AM

The matrix is $A=\begin{bmatrix}{1}&{1}&{1}\\{2}&{2}&{2}\\{3}&{3} &{3}\end{bmatrix}$ so, $\ker (A-6I) \equiv\begin{Bmatrix}-5x_1+x_2+x_3=0\\2x_1-4x_2+2x_3=0\\3x_1+3x_2-3x_3=0\end{matrix}$ . Verify that $(1,2,3)$ is a solution and $(-2,1,0)$ it is not.

**AshleyT** · August 18th 2011, 01:41 AM

EDIT: Just realised i massive stupid mistake in calculating all of these! Nvm.

Thankyou!

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