# Solving for eigenvectors (Introduction to Linear Algebra by Serge Lang, example 2 page 240)

In "Introduction to Linear Algebra" by Serge Lang, example 2 page 240, there is the matrix $\begin{pmatrix}1&4\\2&3\end{pmatrix}$. The characteristic polynomial is $(t-5)(t+1)$. The system to solve for eigenvectors is $\begin{equation*}(1-\lambda)x+4y=0\\2x+(3-\lambda)y=0\end{equation*}$.

The author gives $x$ the value 1 and solve for $y$ from the second equation to get $y=-2/(3-\lambda)$. Why does this work ? Why can he assume the value 1 ? How does the author know that $x=1$ and $y=-2/(3-\lambda)$ is consistent with the first equation $(1-\lambda)x+4y=0$ ? It is actually true but how does he know it ?

Recall that the way to solve for eigenvalue is to let the characteristic polynomial $$|\lambda I-A|=0$$ and solve for $\lambda$. Hence the solution $\lambda$ is calculated to make the matrix $\lambda I- A$ singular. In this case the matrix is of dimension 2, singularity means it has rank 1 (since it obviously is not a zero matrix). Hence in the Gaussian elimination only one row is going to survive, and we only need to consider one row, which is exactly to solve only one equation of the system $$(\lambda I-A)v=0,$$ i.e. one of \begin{equation*}(1-\lambda)x+4y=0\\2x+(3-\lambda)y=0\end{equation*}.
The reason is because if you try multiplying the matrix by a vector of the form $(0, y)$, you can see immediately that it is not an eigenvector so we can assume that the $x$ component is non-zero. Because we don't care about scaling (with a non-zero constant) when it comes to eigenvectors, we can scale the $x$ component so that it is $1$.
The reason we know it's consistent with the first equation is because we always expect to get infinitely many solutions when it comes to eigenvectors (due to the invariance under scaling). That means that these rows will end up cancelling out when it comes to Gaussian elimination (because if they didn't, we'd have a rank $2$ matrix with $2$ variables giving us only the trivial solution).
• I don't think this answer is correct or, at least, not completely correct: the fact that the system's rank is 1 because we're inputting one of the values that makes the matrix $\,\lambda I-A\,$ singular seems to be, at least, undermentioned. – DonAntonio Apr 3 '13 at 13:05
• Yes you are right, I left it assumed that a solution for $\lambda$ to $|\lambda I - A| = 0$ implies there is a non-zero $x$ such that $A x = \lambda x$. – muzzlator Apr 3 '13 at 13:12