Eigen vectors for matrix with unknown constants?

Question

I have the following matrix: $$\begin{bmatrix}\alpha&0&0\\\beta-\alpha&\beta&0\\1-\beta&1-\beta&1\end{bmatrix}$$

So far I have worked out the polynomial to be: $$(\alpha-\lambda)(\beta-\lambda)(1-\lambda)$$

And have the eigen values: $$\lambda_1=\alpha$$ $$\lambda_2=\beta$$ $$\lambda_3=1$$

But the problem I am having is I am not sure how to get the eigen vectors when $\alpha$ and $\beta$ are involved.

@GitGud I already know the eigenvectors through using online calculators what I don't understand is the method how to work them out. How would i know the last column is clearly an eigenvector? — John Sumner, May 14 '15 at 22:06
I would expect the $3,1$ position to be $1-\alpha$ to fit the pattern — Will Jagy, May 14 '15 at 22:17
Anyway, given an eigenvalue, subtract it from all three entries on the main diagonal (so that one of those entries is now $0$) and find a vector in the null-space of that revised matrix. As long as $1, \alpha, \beta$ are all different, this is all that is necessary — Will Jagy, May 14 '15 at 22:20
see http://math.stackexchange.com/questions/996654/how-to-solve-this-to-find-the-null-space for example, yours is easier — Will Jagy, May 14 '15 at 22:22
http://math.stackexchange.com/questions/88301/finding-the-basis-of-a-null-space — Will Jagy, May 14 '15 at 22:24
@WillJagy The problem I'm having is solving $(\beta-\alpha)x +(\beta-\alpha)y = 0$ and $(1-\beta)x+(1-\beta)y +(1-\alpha)z = 0$ to get the eigen vector — John Sumner, May 14 '15 at 22:29
meanwhile, $\alpha = \beta$ is special, so is $\alpha = 1,$ so is $\beta = 1,$ and so is $\alpha = \beta = 1$ although now easy. — Will Jagy, May 14 '15 at 22:29
For the first case, assume $\alpha, \beta, 1$ are all different, so your pairwise differences are all nonzero. Bye. — Will Jagy, May 14 '15 at 22:31

score 1 · Accepted Answer · answered May 15 '15 at 00:06

You get the eigenvectors just the same as you would normally, except here we need to split into cases where some of the eigenvalues may be equal.

Suppose first that none of the eigenvalues are equal. We need to find vectors $v=(a,b,c)$ such that $Av=\lambda v$. For $\lambda_1=\alpha$ for example, we have the system of equations $$a\alpha=a\alpha$$ $$a(\beta-\alpha)+b\beta= b\alpha $$ $$a(1-\beta)+b(1-\beta)+c= c\alpha$$

The first equation is trivially true, and simplifying the second equation gives $a=\dfrac{\alpha-\beta}{\beta-\alpha}b\hspace{3mm}\implies\hspace{3mm}a=-b$, which we can include with the third equation to get $c=c\alpha\hspace{3mm}\implies\hspace{3mm}c=0$. Hence setting $a=t$, we have the eigenvectors for $\lambda_1=\alpha$ are $\left[\begin{matrix}t\\-t\\0\end{matrix}\right]$ for $t\in\mathbb{R}$.

You can apply the same process to find the eigenvectors for the other eigenvalues.

Now we have to deal with other cases: where $\alpha=\beta\ne1$, where $\alpha=1\ne\beta$, where $\beta=1\ne\alpha$, and finally where $\alpha=\beta=1$.

I will demonstrate the case $\alpha=\beta\ne1$ and leave the other cases to you.

In the $\alpha=\beta\ne1$ case, the equations for the eigenvalue $\lambda_1=\alpha$ become: $$a\alpha=a\alpha$$ $$b\alpha= b\alpha $$ $$a(1-\alpha)+b(1-\alpha)+c= c\alpha$$

Note I just used the equations from before replaced all occurrences of $\beta$ with $\alpha$ since they are the same.

Now the first two equations are trivial and the last equation gives us $(a+b)(1-\alpha)=c(\alpha-1)\hspace{3mm}\implies\hspace{3mm} a+b=-c$. Hence setting $a=t$ and $b=s$ we get $\left[\begin{matrix}t\\s\\-(s+t)\end{matrix}\right]$ where $t,s\in\mathbb{R}$ as our eigenvectors for $\lambda_1=\lambda_2=\alpha$.

For $\lambda_3=1$, we get the system of equations $$a\alpha=a$$ $$b\alpha= b$$ $$a(1-\alpha)+b(1-\alpha)+c= c$$ and simplification gives us $a=b=0$ and $c=c$. Hence setting $c=t$ gives us the eigenvectors for this eigenvalue as $\left[\begin{matrix}0\\0\\t\end{matrix}\right]$ for $t\in\mathbb{R}$.

I leave the rest up to you.

Eigen vectors for matrix with unknown constants?

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