# eigenvalues

• June 11th 2009, 04:56 AM
gconfused
eigenvalues

The matrix is:
$
\left[\begin{array}{rrr}-4&0&3\\1&1&-1\\-6&0&5\end{array}\right]
$

but then I keep getting a weird answer sighs (what i did was subtract lamda from the diagonal: a11, a22, a33)

the eigenvalues are supposed to be 2, -1 and 1.

• June 11th 2009, 05:20 AM
scorpion007
Finding the det of the following:

$\begin{vmatrix} -4-\lambda&0&3\\1&1-\lambda&-1\\-6&0&5-\lambda\end{vmatrix} = (-4-\lambda )(1-\lambda )(5-\lambda ) + 3(-1+\lambda )\cdot (-6) = 0$

$\implies - (\lambda - 2) (\lambda - 1) (\lambda + 1) = 0$

Hence $\lambda_1 = 2, \lambda_2 = 1, \lambda_3 = -1$

QED.
• June 11th 2009, 05:20 AM
TheAbstractionist
$\begin{vmatrix}\lambda+4&0&-3\\[3mm]
-1&\lambda-1&1\\[3mm]
6&0&\lambda-5\end{vmatrix}$

$=\ (\lambda+4)\begin{vmatrix}\lambda-1&1\\[1mm]
0&\lambda-5\end{vmatrix}-3\begin{vmatrix}-1&\lambda-1\\[1mm]
6&0\end{vmatrix}$

$=\ (\lambda+4)(\lambda-1)(\lambda-5)+18(\lambda-1)$

$=\ (\lambda-1)\left[(\lambda+4)(\lambda-5)+18\right]$

$=\ (\lambda-1)(\lambda^2-\lambda-20+18)$

$=\ (\lambda-1)(\lambda^2-\lambda-2)$

$=\ (\lambda-1)(\lambda+1)(\lambda-2)$

Looks pretty straightforward to me. (Wink)
• June 11th 2009, 05:22 AM
gconfused
i dunno, i think that i must have messed up my algebra along the way.. thanks guys