0
$\begingroup$

I have a matrix called Ω. This is the matix:

$$\frac 1 2\begin{bmatrix}2 & 0 &0\\0 & 3 &-1 \\0 & -1 & 3\end{bmatrix}$$

It's eigenvalues are known (I have calculated them earlier). They are $l_1 = l_2 = 1$ and $l_3 = 2$

I want to find the eigenvectors from this data.

In the case where the eigenvalue is $2$, this is used:

$Ω ( a_{13} a_{23} a_{33} ) = 2( a_{13} a_{23} a_{33} ) ⇔ … ⇔ a_{13} = 0, a_{23} = −a_{33}$

As you can see, we arrive to some conclusion of $a_{13}, a_{23}$ and $a_{33}$. But how this is calculated?

$\endgroup$
0
2
$\begingroup$

Your conclusion, $$ a_{13} = 0, a_{23} = −a_{33},$$ is correct. Any vector that satisfies that condition, i.e. any vector of the form $$ v=a_{23}(0,1,-1) $$ will be an eigenvector of $\Omega$ with eigenvalue 2.

You'll notice that this doesn't fully determine the eigenvector, because we haven't "found" $a_{23}$ yet. This is an intrinsic feature of eigenvectors: if $x$ is an eigenvector of $A$ with eigenvalue $\lambda$, i.e. if $$ Ax=\lambda x, $$ then for any real number $r$, the vector $y=rx$ will satisfy $$ Ay = A\ rx = r \ Ax=r\ \lambda x = \lambda \ rx = \lambda y, $$ i.e. $y$ will also be an eigenvector of $A$ with eigenvalue $\lambda$. That therefore means that when we're finding eigenvectors, there will always be (at least) one residual normalization constant left over, and that can either be left free, determined arbitrarily, or constrained using some additional normalization requirement (such as e.g. requiring that $||x||=1$).

$\endgroup$
1
$\begingroup$

To find the eigenvectors for an eigenvalue $\lambda$, you want to solve the linear system $(A-\lambda I)x=0$.

In this case, when $\lambda=1$, we want to solve
$$ \begin{bmatrix} 0&0&0\\ 0&1/2&-1/2\\0&-1/2&1/2\end{bmatrix}\begin{bmatrix} x\\ y\\ z\end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}. $$ That is, $$ \begin{bmatrix} 0\\ (y-z)/2\\ (y-z)/2\end{bmatrix}=\begin{bmatrix} 0\\0\\0\end{bmatrix}. $$ So the eigenvectors will be those with coordinates $(x,y,z)$ satisfying $y-z=0$. That is, vectors of the form $(x,y,y)=x(1,0,0)+y(0,1,1)$. In this case the eigenspace is 2-dimensional.

When $\lambda=2$, the system you want to solve, $(A-2I)v=0$, is $$A-2I=\begin{bmatrix} -1&0&0\\ 0&-1/2&-1/2\\0&-1/2&-1/2\end{bmatrix}\begin{bmatrix} x\\ y\\ z\end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}.$$ This corresponds to $x=0$, $y+z=0$, so vectors of the form $(0,y,-y)=y(0,1,-1)$. In this case the eigenspace is one-dimensional.

$\endgroup$

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.