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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?

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    $\begingroup$ 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 $a$. $\endgroup$ – John Hughes Feb 4 at 19:23
  • $\begingroup$ To find eigenvectors $v_1$ and $v_2$, solve $Av_1=av_1$ and $Av_2=dv_2$ $\endgroup$ – J. W. Tanner Feb 4 at 19:27
  • $\begingroup$ Do you want left or right eigenvectors? $\endgroup$ – robjohn Feb 4 at 19:46
  • $\begingroup$ You should be able to find an eigenvector of $a$ at a glance. Recall the meaning of the columns of a matrix. $\endgroup$ – amd Feb 5 at 0:56
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$\lambda_1 = a$:

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

leads to

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

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.

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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 $E^{-1}AE = .....$

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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 $\begin{bmatrix}0 \\ y \end{bmatrix}$.

Added after Christoph's comment:

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.

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  • $\begingroup$ Neither is $b$ an eigenvalue nor is $\begin{bmatrix}0\\y\end{bmatrix}$ an eigenvector (in general). $\endgroup$ – Christoph Feb 4 at 19:51
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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.

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