Question

Find eigenvalues for A=[[-4,2,2],[2,-4,2,],[2,2,-4]]

Answer 1

what have you tried so far ?

Answer 2

I've tried finding the determinant of\[A- \lambda I\] but it keeps coming up with complex or decimal values, when it should give -6,-6,0.

Answer 3

i.e. \[\det \left[\begin{matrix}-4- \lambda & 2 & 2 \\ 2 & -4- \lambda & 2 \\ 2 & 2 & -4- \lambda \end{matrix}\right]\]

Answer 4

\[\left|\begin{matrix}-4- \lambda & 2 & 2 \\ 2 & -4- \lambda & 2 \\ 2 & 2 & -4- \lambda \end{matrix}\right|
\\ ~\\~ = (-4-\lambda)[(4-\lambda)^2 - 4] - 2[2(-4-\lambda) - 4) + 2[4-2(-4-\lambda)] \\~\\~

\]

Answer 5

Should the second 4 be negative?

Answer 6

\[
= (-4-\lambda)[(4\color{Red}{+}\lambda)^2 - 4] - 2[2(-4-\lambda) - 4) + 2[4-2(-4-\lambda)]
\]

like this ?

Answer 7

wolfram does give \(\lambda = -6, 0\) :
http://www.wolframalpha.com/input/?i=solve++%28-4-%5Clambda%29%28%284%2B%5Clambda%29%5E2+-+4%29-+2%282%28-4-%5Clambda%29+-+4%29+%2B+2%284-2%28-4-%5Clambda%29%29+%3D0

Answer 8

so it seems our characteristic equation is okay, we just need to simplify and solve it...

Answer 9

Why are both the 4 and the lambda positive in that equation (in the brackets where you changed the - to +) when they're negative in the matrix?

Answer 10

because \((-4-\lambda)(-4-\lambda) = (-(4+\lambda))(-(4+\lambda)) = (4+\lambda)(4+\lambda ) = (4+\lambda)^2\)

Answer 11

Okay, that makes sense.

Answer 12

How do we simplify/solve it, because I keep making more complicated and/or incorrect equations.

Answer 13

\[\begin{align} \\ &= (-4-\lambda)[(4\color{Red}{+}\lambda)^2 - 4] - 2[2(-4-\lambda) - 4) + 2[4-2(-4-\lambda)] \\~\\
&= (-4-\lambda)[\lambda^2 + 8\lambda + 12] + 2[2\lambda+8] + 2[12+2\lambda]
\end{align}\]

Answer 14

see if that looks okay ^^

Answer 15

Yep, except the third term is missing the minus 4. It should be the same as the last term.

Answer 16

Oh yes i do more mistakes than you haha !

Answer 17

\[\begin{align} \\ &= (-4-\lambda)[(4\color{Red}{+}\lambda)^2 - 4] - 2[2(-4-\lambda) - 4) + 2[4-2(-4-\lambda)] \\~\\
&= (-4-\lambda)[\lambda^2 + 8\lambda + 12] + 2[2\lambda+\color{red}{12}] + 2[12+2\lambda]
\end{align}\]


looks okay now ?

Answer 18

Haha, I've been through this matrix 5 or 6 times and come up with wrong and different things everytime, you're doing fine :P
Looks good.

Answer 19

next may be factor the quadratic in first term

Answer 20

\[\begin{align} \\ &= (-4-\lambda)[(4\color{Red}{+}\lambda)^2 - 4] - 2[2(-4-\lambda) - 4) + 2[4-2(-4-\lambda)] \\~\\
&= (-4-\lambda)[\lambda^2 + 8\lambda + 12] + 2[2\lambda+\color{red}{12}] + 2[12+2\lambda] \\~\\

&= (-4-\lambda)[(\lambda +6)(\lambda+2)] + 4[\lambda+\color{red}{6}] + 4[6+\lambda] \\~\\
\end{align}\]

Answer 21

I've expanded it out to get\[-48-32\lambda -4\lambda ^{2}-12\lambda +8\lambda ^{2}-\lambda ^{3}+48-8\lambda\] Does that look right?

Answer 22

It simplifies to \[\lambda ^{3}+12\lambda ^{2} +36\lambda\]
\[=\lambda(\lambda ^{2}+12 \lambda +36)\]

Answer 23

\[=\lambda (\lambda +6)(\lambda +6)\]

Answer 24

So lambda = -6, -6, 0
Which is the answer!

Answer 25

looks perfect !

Answer 26

Thank you so much, I was so stuck!

Answer 27

np :) also there is an interesting property for eigen values :
they add up to the sum of elements in diagonal

Answer 28

\[
\left[\begin{matrix} \color{Red}{-4} & 2 & 2 \\ 2 & \color{Red}{-4} & 2 \\ 2 & 2 & \color{Red}{-4} \end{matrix}\right]

\]

Answer 29

\(\large \color{Red}{-4-4-4} = -6-6+0 = -12\)

Answer 30

Thanks, that'll help with checking my answers!

Answer 31

this is in general true for all matrices

Answer 32

if you notice this is really a special matrix :
symmetrical matrix with constant diagonal

Answer 33

also this is singular - thats the reason we have got 0 as one of the eigen values