0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – JMoravitz May 23 '18 at 4:15
  • $\begingroup$ 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. $\endgroup$ – JMoravitz May 23 '18 at 4:18
  • $\begingroup$ Yes I understood little bit. I ll get back and ask if I am stuck again. @JMoravitz Thanks again . $\endgroup$ – Daman May 23 '18 at 4:36
  • $\begingroup$ @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 $(1,0,-1)$. $\endgroup$ – Daman May 23 '18 at 5:59
  • $\begingroup$ Of course the two determinants won’t be equal. You’re multiplying one of the columns by $-1$. Why does that matter? $\endgroup$ – amd May 23 '18 at 8:48
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Could you please answer my question in comments ? . Thanks for ans btw $\endgroup$ – Daman May 23 '18 at 6:00
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.