Eigenvectors

[scrum]

I'm a little lost with exactly how these go but here goes

I have a matrix and i'm asked for eigenvalues and eigenvectors
1 0 0
-1 3 0
3 2 -2

From the book I know that the values for this triangular matrix will just be -2 3 and 1

To solve for the eigenvectors for the -2, subtract my eigenvalue from each diagonal
3 0 0
-1 5 0
3 2 0
so now i have this matrix. it seems to me this makes v1 = 0 and then v2 = 0 and then everything equals zero. the example in the book is no help as no column gets filled with zeros.

 

[galactus]

To find the eigenvectors, plug your eigenvalues back in for lambda in the matrix and solve.

\(\displaystyle \begin{bmatrix}{\lambda}-1&0&0\\1&{\lambda}-3&0\\-3&-2&{\lambda}+2\end{bmatrix}\)


Here is a for instance. Then, you try yours above.

Let's take the matrix \(\displaystyle \begin{bmatrix}1&-3\\-4&2\end{bmatrix}\)

By definiton \(\displaystyle x=\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}\) is an eigenvector of A iff x is a nontrivial solution of \(\displaystyle ({\lambda}I-A)x=0\)

That mean if:

\(\displaystyle \begin{bmatrix}{\lambda}-1&-3\\-4&{\lambda}-2\end{bmatrix}\cdot\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\)

Now, if we sub in one of its eigenvalues, -2, we get:

\(\displaystyle \begin{bmatrix}-3&-3\\-4&-4\end{bmatrix}\cdot \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\)

If we solve this system, we get \(\displaystyle \boxed{x_{1}=-t, \;\ x_{2}=t}\).

So, the eigenvectors corresponding to \(\displaystyle {\lambda}=-2\) are the non zero solutions of the form:

\(\displaystyle x=\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}=\begin{bmatrix}-t\\t\end{bmatrix}\)

See?. Now, try yours. Hope that was a nice tutorial.

To check if you have them right, try using \(\displaystyle Ax={\lambda}x\). Where x is the eigenvector and lambda is a corresponding eigenvalue.

 

[scrum]

Thanks for the tutorial and the help. i tried it and i got the one you gave me but i still can't get mine.

Now, if we sub in one of its eigenvalues, -2, we get:

\(\displaystyle \begin{bmatrix}-3&-3\\-4&-4\end{bmatrix}\cdot \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\)

If we solve this system, we get \(\displaystyle \boxed{x_{1}=-t, \;\ x_{2}=t}\).

So, the eigenvectors corresponding to \(\displaystyle {\lambda}=-2\) are the non zero solutions of the form:

\(\displaystyle x=\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}=\begin{bmatrix}-t\\t\end{bmatrix}\)

See?. Now, try yours. Hope that was a nice tutorial.

This is the step I still cannot figure out.

I have

[-3 0 0] [x1] [0]
[1 -5 0] * [x2] = [0]
[-3 -2 0] [x3] [0]

but i don't know what to do with it. it would seem that x1 = 0 and then that makes x2 = 0 and x3 can be anything so it's 0,0,0 but that's not right.

 

[Deleted member 4993]

Thanks for the tutorial and the help. i tried it and i got the one you gave me but i still can't get mine.

Now, if we sub in one of its eigenvalues, -2, we get:

\(\displaystyle \begin{bmatrix}-3&-3\\-4&-4\end{bmatrix}\cdot \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\)

If we solve this system, we get \(\displaystyle \boxed{x_{1}=-t, \;\ x_{2}=t}\).

So, the eigenvectors corresponding to \(\displaystyle {\lambda}=-2\) are the non zero solutions of the form:

\(\displaystyle x=\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}=\begin{bmatrix}-t\\t\end{bmatrix}\)

See?. Now, try yours. Hope that was a nice tutorial.

This is the step I still cannot figure out.

I have

[-3 0 0] [x1] [0]
[1 -5 0] * [x2] = [0]
[-3 -2 0] [x3] [0]

but i don't know what to do with it. it would seem that x1 = 0 and then that makes x2 = 0 and x3 can be anything so it's 0,0,t but that's not right.

Sometimes - the value of 't' is chosen in such a way that the magnitude of the vector is unity - in this case then t = 1.