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What kind of matrix is it?

Suppose that I want to compute $$\begin{bmatrix}a & 2 \\ 0 & 1\end{bmatrix}^n \vec v$$ or solve a (inhomogenous) recurrence, $v^2 = av+2$. I must eigendecompose everything.

The determinant $$\begin{vmatrix}a-\lambda & 2 \\ 0 & 1-\lambda\end{vmatrix} = 0$$ yeilds characteristic equation $(a-\lambda)(1-\lambda) = 0$ with roots $\lambda_1=a$ and $\lambda_2=1$.

The eigenvalues are distinct and everything should be ok. However, when I plug into

$$\begin{bmatrix}a & 2 \\ 0 & 1\end{bmatrix} \begin{bmatrix}e_1 \\ e_2\end{bmatrix} = \lambda\begin{bmatrix}e_1 \\ e_2\end{bmatrix},$$ that is,

$$ \begin{cases} a e_1 + 2e_2 &= \lambda e_1 \\ e_2 = \lambda e_2 \end{cases}$$

implies that only $\lambda = 1$ is a valid eigenvalue. Otherwise, second equality breaks down. What happens to the first eigenvalue, $a$?

Now, all eigenvectors are identical up to the constant factor. This means that we can choose one component(or coordinate?) of eigenvector arbitrarily. Let's assign $e_2 = 1$. This yeilds $a e_1 + 2 = 1 \cdot e_1$ or, $e_1 = 2/(1-a)$ and the eigendecomposition is

$$\begin{bmatrix}a & 2 \\ 0 & 1\end{bmatrix} \begin{bmatrix} 2/(1-a)\\ 1\end{bmatrix} = 1 \begin{bmatrix}2/(1-a) \\ 1\end{bmatrix}.$$

However, because there is only one eigencouple, I have a difficulty decomposing arbitrary 2-dimensional vector $\vec v$ into basis $\begin{bmatrix}e_1 & e_2\end{bmatrix}^T = \begin{bmatrix}2/(a-1) & 1\end{bmatrix}^T$ because there is only one vector in the basis.

So, I conclude the matrix of this kind is special, despite it has two distinct eigenvalues. Therefore, I ask, what kind of matrix is it? Might be you can say why eigendecomposition cannot be used to solve nonhomogenous recurrences?

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    $\begingroup$ "Otherwise, second equality breaks down." Only if you're assuming $e_2\neq 0$... $\endgroup$ – Casteels Dec 11 '13 at 12:14
  • $\begingroup$ Yes, that was a false implicit assumption! $\endgroup$ – Val Dec 11 '13 at 12:58
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Your eigenvalues are correct.

The eigenvectors are:

  • $\lambda_1 = 1, ~ v_2 = \left( -\dfrac{2}{a-1}, 1\right)$
  • $\lambda_2 = a, ~ v_1 = (1, 0)$

We can now write the Jordan Normal Form as:

$$A = P J P^{-1} = \begin{bmatrix} -\dfrac{2}{a-1} & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & a\end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & \dfrac{2}{a-1} \end{bmatrix}$$

From this we can now easily write:

$$A^n = P J^n P^{-1}$$

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