# How can I find the eigenvectors for this matrix?

Here is the matrix A:

$$\begin{pmatrix} a & b\\ 0 & d \\ \end{pmatrix}$$

I've been able to find the eigenvalues ($$a$$ and $$d$$), however when you put these eigenvalues into the matrix $$|A - \lambda I|$$

$$\begin{pmatrix} a - \lambda & b\\ 0 & d - \lambda \\ \end{pmatrix}$$

the matrix reduces to either a single row or single column. How can I get around this problem?

• For $\lambda = a$, it reduces to $S = \pmatrix{0 & b \\ 0 & d-a}$. That's not a "problem". Can you solve $S\pmatrix{x \\y} = \pmatrix{0 \\ 0}$? Because the solution is an eigenvector for the eigenvalue $a$. – John Hughes Feb 4 at 19:23
• To find eigenvectors $v_1$ and $v_2$, solve $Av_1=av_1$ and $Av_2=dv_2$ – J. W. Tanner Feb 4 at 19:27
• Do you want left or right eigenvectors? – robjohn Feb 4 at 19:46
• You should be able to find an eigenvector of $a$ at a glance. Recall the meaning of the columns of a matrix. – amd Feb 5 at 0:56

$$\lambda_1 = a$$:

$$\begin{pmatrix} 0 & b \\ 0 & d-a \end{pmatrix}\cdot \begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}$$

$$by=0 \quad\wedge\quad (d-a)y=0$$

which is solved by $$y=0$$ given that $$b\neq 0$$ or $$a\neq d$$. Because $$x$$ is not present in the above system of equations at all, it means it can be any number. So the eigenvector $$v_1$$ corresponding to $$\lambda_1$$ is

$$v_1 = \begin{pmatrix} x\\ 0 \end{pmatrix},$$

which after normalisation gives

$$v_1 = \begin{pmatrix} 1\\ 0 \end{pmatrix}.$$

For $$\lambda_2=d$$ you'll get one equation involving both $$x$$ and $$y$$, meaning that the solution $$x$$ will be given in terms of $$y$$ (or vice versa). Then you insert it to $$v_2 = \begin{pmatrix} x\\ y \end{pmatrix}$$, take out the common factor (i.e., $$x$$ or $$y$$) and normalise to get the second eigenvector.

when $$d \neq a$$ we do get two eigenvectors, which I put as the columns of $$E = \left( \begin{array}{cc} 1 & b \\ 0 & d-a \end{array} \right)$$ Indeed, we get $$E^{-1} = \frac{1}{d-a} \left( \begin{array}{cc} d-a & -b \\ 0 & 1 \end{array} \right)$$ and $$E^{-1}AE = .....$$

Yes, of course that matrix reduces- that's the whole point of an "eigenvector". That's because the set of all eigenvectors corresponding to a given eigenvalue form a subspace. There are necessarily an infinite number of vectors.

The definition of "eigenvector" of matrix A corresponding to eigenvalue $$\lambda$$ is a vector v such that $$Av= \lambda v$$. Here, $$A= \begin{bmatrix}a & b \\ 0 & d \end{bmatrix}$$. The eigenvalues are a and d. Any eigenvector, $$v= \begin{bmatrix}x \\ y \end{bmatrix}$$, corresponding to eigenvalue a, must satisfy $$Av= \begin{bmatrix}a & b \\ 0 & d \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}ax+ by\\ dy\end{bmatrix}= \begin{bmatrix}ax \\ ay\end{bmatrix}$$. I would write that as the pair of equations ax+ by= ax and dy= ay. The first of those reduces to by= 0 so y= 0 which also satisfies dy= ay. But that does not say anything about x. In fact x can be anything. The set of all eigenvectors corresponding to eigenvalue a is the set of all eigenvectors of the form $$\begin{bmatrix} x \\ 0\end{bmatrix}$$.

Similarly the set of all eigenvectors corresponding to eigenvalue b is the set of all vectors of the for $$\begin{bmatrix}0 \\ y \end{bmatrix}$$.

Boy, I really bollixed that up, didn't I? I should have said that the eigenvalues are a and d, not a and b. And an eigenvector, $$\begin{bmatrix}x \\ y \end{bmatrix}$$ corresponding to eigenvalue d must satisfy $$\begin{bmatrix}a & b \\ 0 & d \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}ax+ by \\ dy\end{bmatrix}= \begin{bmatrix}dx \\ dy \end{bmatrix}$$ so we have equations ax+ by= dx and dy= dy. The second equation is true for any y and we can solve the first equation for x= by/(d- a). Every eigenvector corresponding to eigenvalue d is of the form $$\begin{bmatrix} by/(d-a) \\ y\end{bmatrix}$$ so is a multiple of $$\begin{bmatrix} b/(d-a) \\ 1\end{bmatrix}$$. But my basic concept is still true- there is not a single "eigenvector" but an infinite number of them- an entire subspace.
• Neither is $b$ an eigenvalue nor is $\begin{bmatrix}0\\y\end{bmatrix}$ an eigenvector (in general). – Christoph Feb 4 at 19:51
Assuming you want right eigenvectors, we are looking for vectors perpendicular to the rows of $$A-I\lambda$$. A row of the cofactor matrix is perpendicular to all the other rows of the original matrix. Since the rows are dependent, a row of the cofactor matrix $$\operatorname{cof}\begin{bmatrix}0&b\\0&d-a\end{bmatrix}=\begin{bmatrix}\color{#C00}{d-a}&\color{#C00}{0}\\-b&0\end{bmatrix}$$ $$\operatorname{cof}\begin{bmatrix}a-d&b\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\\color{#C00}{-b}&\color{#C00}{a-d}\end{bmatrix}$$ Therefore, $$\begin{bmatrix}d-a\\0\end{bmatrix}$$ and $$\begin{bmatrix}-b\\a-d\end{bmatrix}$$ are right eigenvectors.