Eigenvector

**Revy** · December 8th 2010, 12:23 PM

I have matrix $A=\begin{pmatrix}0&1&0\\2&-2&-1\\-3&0&0\end{pmatrix}$
and I get that $det(A-LE)=-L^3+2L^2+2L+3$
so
$L^3-2L^2-2L-3=0$
//after this our lecturer takes random numbers and checks if they can be eigenvalues.
For this problem I'm sure that eigenvalue is only $3$

//but what if I get $L^3-7L^2-36=0$ ?
//I have to check all numbers till 6
//so how did he find that for this one $L$ is $6, 3$ and $2$ so fast?
// EH... NEVERMIND xD

further, to find eigenvector I have
$\begin{pmatrix}-3&1&0&|0\\2&-5&-1&|0\\-3&0&-3&|0\end{pmatrix}$
but I can't find a way to substitute so that I would have 2 even rows or columns, so maybe there is another way I don't know?

some update:
I ended up at
$\begin{pmatrix}-3&2&-1&|0\\0&-3&-3&|0\end{pmatrix}=\begin{pmatrix}1&-2/3&1/3&|0\\0&1&1&|0\end{pmatrix}=\begin{pmatrix}1&-0&1&|0\\0&1&1&|0\end{pmatrix}=\begin{pmatrix}1&0 &|-C\\0&1&|-C\end{pmatrix}$

$x_1 = -C$
$x_2 = -C$
$x_3 = C$

So, after getting someone annoying for updating my post so many times xD
Q: HOW DO I STANDARDIZE EIGENVECTOR AND CHECK IF ANSWER IS CORRECT?
someone told me: "you need to find vector length or smth" =/

$|x|=\sqrt{(-C)^2+(-C)^2+C^2}=\sqrt{3}C$
$x_N=\begin{pmatrix}-1/\sqrt{3}\\-1/\sqrt{3}\\1/\sqrt{3}\end{pmatrix}$ ???????

IF CORRECT, HOW DO I CHECK IF IT IS SO?

**Prove It** · December 8th 2010, 04:07 PM

For your first, there are also two complex eigenvalues, $\displaystyle \frac{-1+\sqrt{3}i}{2}$ and $\displaystyle \frac{-1 - \sqrt{3}i}{2}$ . All the eigenvalues are important.

For the second $\displaystyle \lambda^3 - 7\lambda^2 - 36 = 0$ , this does not factorise, in fact, it is not solvable exactly. And a simple check of the graph would see there is only one real root, so his answer can not possibly be correct.

**HallsofIvy** · December 9th 2010, 03:43 AM

Originally Posted by Revy

I have matrix $A=\begin{pmatrix}0&1&0\\2&-2&-1\\-3&0&0\end{pmatrix}$
and I get that $det(A-LE)=-L^3+2L^2+2L+3$
so
$L^3-2L^2-2L-3=0$
//after this our lecturer takes random numbers and checks if they can be eigenvalues.
For this problem I'm sure that eigenvalue is only $3$

I doubt that he takes "random" numbers- he probably knows them ahead of time. Here, however, the "rational root theorem" says that any rational number satisfying this equation must be 1, -1, 3, or -3. 3 is the only one of those that satisfies it. However, as Prove It says, there are two complex roots as well. Divide $L^3- 2L^3- 2L+ 3$ by L- 3 to get the quadratic equation they must solve and use the quadratic formula.

//but what if I get $L^3-7L^2-36=0$ ?
//I have to check all numbers till 6

what do you mean by "all number till 6". Surely you understand that roots are not necessarily integers! By the "rational root theorem" again, the only possible rational solutions must be factors of 36- but there are a bunch of those!

//so how did he find that for this one $L$ is $6, 3$ and $2$ so fast?

I can only say that he must be very clever if he convinced you that 6, 3, and 2 were solutions!
$6^3- 7(6^2)- 36= 216- 252- 36= -36- 36= -72$ , not 0.
$3^3- 7(3^2)- 36= 27- 63- 36= -36- 36= -72$ , not 0.
$2^3- 7(2^2)- 36= 8- 28- 36= -56$ , not 0.
6 and 3 satisfy the equation $L^3- 7L^2+ 36= 0$ but I don't see how to include "2" in there.

// EH... NEVERMIND xD

further, to find eigenvector I have
$\begin{pmatrix}-3&1&0&|0\\2&-5&-1&|0\\-3&0&-3&|0\end{pmatrix}$
but I can't find a way to substitute so that I would have 2 even rows or columns, so maybe there is another way I don't know?

I'm not sure what you mean by this. The definition of "eigenvalue" is a number, L such that Av= Lv for some non-zero vector v. To find v, solve Av= Lv.

Since you are talking, again, about $A= \begin{bmatrix}0 & 1 & 0 \\ 2 & -2 & -1\\ - 3 & 0 & 0\end{bmatrix}$ with eigenvalue 3, you want to solve
$\begin{bmatrix}0 & 1 & 0 \\ 2 & -2 & -1\\ - 3 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}3x \\ 3y \\ 3z\end{bmatrix}$
$\begin{bmatrix}y \\ 2x- 2y- z \\ -3x\end{bmatrix}= \begin{bmatrix}3x \\ 3y \\ 3z\end{bmatrix}$
which means we must have y= 3x, 2x- 2y- z= 3y, and -3x= 3z. We can replace y and z in the second equation with 3x and -x, respectively. Then it becomes 2x- 2(3x)- (-x)= 0 no matter what x is. That is, any vector of the form (x, 3x, -x)= x(1, 3, -1) will satisfy that: any multiple of (1, 3, -1) is an eigenvector.

some update:
I ended up at
$\begin{pmatrix}-3&2&-1&|0\\0&-3&-3&|0\end{pmatrix}=\begin{pmatrix}1&-2/3&1/3&|0\\0&1&1&|0\end{pmatrix}=\begin{pmatrix}1&-0&1&|0\\0&1&1&|0\end{pmatrix}=\begin{pmatrix}1&0 &|-C\\0&1&|-C\end{pmatrix}$

$x_1 = -C$
$x_2 = -C$
$x_3 = C$

So, after getting someone annoying for updating my post so many times xD
Q: HOW DO I STANDARDIZE EIGENVECTOR AND CHECK IF ANSWER IS CORRECT?
someone told me: "you need to find vector length or smth" =/

$|x|=\sqrt{(-C)^2+(-C)^2+C^2}=\sqrt{3}C$
$x_N=\begin{pmatrix}-1/\sqrt{3}\\-1/\sqrt{3}\\1/\sqrt{3}\end{pmatrix}$ ???????

IF CORRECT, HOW DO I CHECK IF IT IS SO?

You don't need to normalize unless the problem specifically asked for a unit eigenvector. Just check to see if, in fact,
Av= Lv. You can keep "C" or factor it out: (-C, -C, C)= C(-1, -1, 1) so (-1, -1, 1) is such a vector.
Try $\begin{bmatrix}0 & 1 & 0 \\ 2 & -2 & -1\\ - 3 & 0 & 0\end{bmatrix}\begin{bmatrix}-1 \\ -1 \\ 1\end{bmatrix}= \begin{bmatrix}0 -1+ 0 \\ -2+ 2- 1 \\ 3+ 0+ 0\end{bmatrix}= \begin{bmatrix}-1 \\ -1 \\ 3\end{bmatrix}$
which is NOT "3" times the vector (-1, -1, 1), nor any number times that vector, so it is NOT an eigenvector.

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