Find $a_0$ such that the matrix has repeated real eigenvectors

Consider the 2x2 matrix $$A=\begin{bmatrix}a & 1\\ 0 & 1\end{bmatrix}$$

Find the value $a_0$ of the parameter $a$ for which A has repeated real eigenvalues. What happens to the eigenvectors of this matrix as $a$ approaches $a_0$?

Trivially, it is easy to see that if $a=1$ then the system has a repeated real eigenvalue of 1. Is this the answer then for $a_0$? I also tried to expand to solve the system $$det\begin{bmatrix}a-\lambda & 1\\ 0 & 1-\lambda\end{bmatrix}=0$$ and got the equation $\lambda^2+a-\lambda(1+a)=0$ but I got stuck from there on finding $a$.

As far as the second part of the question (what happens as $a\to a_0$) I'm not sure what would happen or what I need to do to even see what would happen... plug in values of $a$ close to $a_0$ and look at the eigenvectors?

• Did you try the Quadratic Formula for the roots? – Moo Sep 6 '16 at 14:44
• Hint: The quadratic immediately finds $a_0$. For the second part, do you notice that you go from two linearly independent eigenvectors to need a generalized one? Have you learned about algebraic and geometric multiplicity or defective matrices? – Moo Sep 6 '16 at 14:51

1 Answer

As you showed, the characteristic polynomial of the matrix is $\lambda^2 -(1+a)\lambda+a=(a-\lambda)(1-\lambda),$ which has multiple root $\lambda=1$ iff $a=1$. So $a_{0}=1$.

As for the second part, find the eigenspaces for the eigenvalues $a$ and $1$. These are $span\{(1,1-a)\}$ and $span\{(1,0)\}$ respectively. so if $a\neq1$, they are distinct. But the eigenspaces coincide when $a\rightarrow 1$. So the set of eigenvectors collapses to the line $span\{(1,0)\}$. In other words, you cannot find $2$ linearly independent eigenvectors anymore.