# How to find the eigenvectors of this matrix?

Given

$$A = \begin{bmatrix} -9 & 4 & 4\\ -8 & 3 & 4 \\ -16 & 8 & 7 \end{bmatrix}$$

I calculated eigenvalues $\lambda= -1,-1,3$. The algebraic multiplicity (AM) and geometric multiplicity (GM) of $\lambda=-1$ are $2$, which tells us that there will be two linearly independent eigenvectors.

I am not sure how to find the eigenvectors. Usually, I take one variable and equate it to $t$ and then solve it for the other two. I am not quite sure how to find eigenvectors when we have two free variables.

Steps:

$$(A-\lambda I)=0$$

$$\begin{bmatrix} -8 & 4 & 4\\ -8 & 4 & 4 \\ -16 & 8 & 8 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$$

$$\begin{bmatrix} 8 & -4 & -4\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$$

$$2x-y-z=0$$

I don't know how to proceed from here.

• So, $2x-y-z=0$. Let us define some parametric variables. Out of convenience, lets use the parameters $s$ and $t$ and let us use $y=s$ and $z=t$. We have then $\begin{cases}x=\frac{1}{2}s+\frac{1}{2}t\\y=1s+0t\\z=0s+1t\end{cases}$. Can you continue now? – JMoravitz May 23 '18 at 4:15
• Note: there are infinitely many correct answers. The answer I lead you to above is just one of many. You could have instead chosen to set $x=s$ and $y=t$ for example which would lead you to a different pair of eigenvectors. The point is that any correct answer will have the span of the two vectors all corresponding to the same eigenspace. – JMoravitz May 23 '18 at 4:18
• Yes I understood little bit. I ll get back and ask if I am stuck again. @JMoravitz Thanks again . – Daman May 23 '18 at 4:36
• @JMoravitz I have a problem. I am stuck. Three eigenvectors came $(1,1,0),(1,0,-1),(1,1,2)$. I have to Diagonalize it. My question is will it change determinant value if I take second vector$(-1,0,1)$ in my second column instead of $(1,0,-1)$. – Daman May 23 '18 at 5:59
• Of course the two determinants won’t be equal. You’re multiplying one of the columns by $-1$. Why does that matter? – amd May 23 '18 at 8:48

so $x=\lbrace\frac{y}{2}+\frac{z}{2}$

let be $y=1$, $z=0$, then

$v1=\begin{bmatrix}\frac {1}{2} \\ {1}\\ {0} \end{bmatrix}$

let be $y=0$, $z=1$, then $v2=\begin{bmatrix}\frac {1}{2} \\ {0}\\ {1} \end{bmatrix}$

later, you only substitute $\lambda (i)$ and $v1$,$v2$ in the equations $\ (A-\lambda I)v1=0$

and $\ (A-\lambda I)v2=0$

• Could you please answer my question in comments ? . Thanks for ans btw – Daman May 23 '18 at 6:00

The null space of a matrix is the orthogonal complement of its row space. The latter is clearly spanned by $(8,-4,-4)^T$, so you’re looking for a linearly independent pair of vectors orthogonal to this one. These can be found by inspection: for any vector $(a,b,c)^T$, the vectors $(0,c,-b)^T$, $(-c,0,a)^T$ and $(b,-a,0)^T$ are all orthogonal to it, and if $(a,b,c)^T\ne0$, then at least two of those orthogonal vectors are also nonzero.