TITLE: Does it make sense to talk about eigenvalues in a 1x1 matrix?
QUESTION [5 upvotes]: I am an Economics student and I am trying to apply the determinacy conditions as described in Blanchard, Kahn (1980) to check for the existence of a unique solution in a differential equation.
These conditions state that I need to have the number of eigenvalues of a matrix inside the unit circle equal to the number of jumpy variables in the system. 
Putting aside the economic meaning of that, I have a case in which I am dealing with a unique differential equation so that I do not have a matrix but a scalar, i.e.
\begin{gather}
\pi_t=\frac{1}{1+\phi}\pi_{t+1}+\epsilon_t
\end{gather}
Thus I am wondering if the eigenvalue in this context could be identified with the unique element of the 1x1 matrix made of $\frac{1}{1+\phi}$.

REPLY [6 votes]: You can think like that.
If we think about a linear transformation from $\Bbb R$ to $\Bbb R$ it will be represented by a scalar number $"a"$.
And following the definition of eigenvalue we have $aX=\lambda X$ for $X \in \Bbb R^*$.
Then we can conclude that $\lambda = a$ and $X$ can be any non null real number. So all non null real number will be a eigenvector and the eigenvalue will be $"a"$.