Find the eigenvector and eigenvalues for the following 3 x 3 Matrix?

Question

$$ \pmatrix{5 & 8 & 16 \\ 4 & 1 & 8 \\ -4 &-4 & -11} $$

I already got the eigenvalues that is $\lambda = 1$ and $-3$. And I managed to solve the eigenvector corresponding to the value $\lambda = 1$ which is $(k,k/2,-k/2)$. But for $\lambda = -3$, I struggle to solve the system of equations . Please help.

Answer 1

For $\lambda=-3$ we consider $A-\lambda I = A-(-3)I=A+3I$

$\begin{bmatrix}5&8&16\\ 4&1&8\\ -4&-4&-11\end{bmatrix} + \begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix} = \begin{bmatrix} 8&8&16\\4&4&8\\-4&-4&-8\end{bmatrix}$

We try to ask: what is the eigenspace for the eigenvalue $-3$?

I.e. we try to ask: what is the nullspace of $A+3I$?

I.e. we ask for the solution set to $(A+3I)v=0$

Using standard techniques, we can consider the augmented matrix:

$\left[\begin{array}{ccc|c} 8&8&16&0\\4&4&8&0\\-4&-4&-8&0\end{array}\right]$ and row reduce.

Noticing that every row is a multiple of one another, this row reduces to:

$\left[\begin{array}{ccc|c}1&1&2&0\\0&0&0&0\\0&0&0&0\end{array}\right]$

Interpreting this as a system of equations, this implies that:

$\begin{cases} v_1+v_2+2v_3=0\\v_2~\text{can be anything}\\ v_3~\text{can be anything}\end{cases}$

Letting $v_2=s$ and $v_3=t$, this changes the system to the form:

$\begin{cases} v_1+s+2t=0\\v_2=s\\v_3=t\end{cases} \xrightarrow{\text{which becomes}}\begin{cases}v_1=-s-2t\\v_2=s\\v_3=t\end{cases}$

this implies that the vector is of the form $\begin{bmatrix} -s-2t\\s\\t\end{bmatrix} = s\begin{bmatrix}-1\\1\\0\end{bmatrix}+t\begin{bmatrix}-2\\0\\1\end{bmatrix}$

These two vectors then form a basis for the eigenspace for the eigenvalue $-3$.

(of course, there are infinitely many correct answers for the problem, so long as the two vectors you choose span the same space as the two vectors I give above)

Answer 2

Unless I'm mistaken, the matrix for the eigenvalue -3 should have rows 8,8,16; 4,4,8; and -4,-4,-8. The second and third rows are multiples of the first. Appropriate row operations eliminate them. You end with the equation x+y+2z=0, which has solutions equal to the span of the vectors (-1,1,0) and (-2,0,1). I hope this helps.