# 2.4: Eigenvalues and eigenvectors

Every $$M\times M$$ square matrix $$\underset{\sim}{A} has \(M$$ eigenvalue-eigenvector pairs:

$A \vec{v}^{(m)}=\lambda^{(m)} \vec{v}^{(m)} \quad(\text { no sum on } m) \label{eq:1}$

The superscript ($$m$$) is the “mode number" an index running from 1 to $$M$$ that labels the eigenvalue-eigenvector pairs. To find $$\lambda^{(m)$$ and $$\vec{v}^{(m)}$$, $$m=1,2,...,M$$, rewrite Equation $$\ref{eq:1}$$ as:

$\underset{\sim}{A}\vec{v}^{(m)}=\lambda^{(m)}\underset{\sim}{\delta}\vec{v}^{(m)}\Rightarrow\quad\left[\underset{\sim}{A}-\lambda^{(m)}\underset{\sim}{\delta}\right] \vec{v}^{(m)}=0\label{eq:2}$

where $$\underset{\sim}{\delta}$$ is the $$M\times M$$ identity matrix. This is a homogeneous set of equations, and can therefore only have a nontrivial solution if the determinant is zero:

$\operatorname{det}\left(\underset{\sim}{A}-\lambda^{(m)}\underset{\sim}{\delta}\right)=0\label{eq:3}$

Because $$\underset{\sim}{A}$$ is an $$M\times M$$ matrix, Equation $$\ref{eq:3}$$ can be written as an $$M^{th}$$ order polynomial for the eigenvalues $$\lambda$$, referred to as the characteristic polynomial. It has $$M$$ solutions for $$\lambda$$. These may be complex, and they are not necessarily distinct. Each particular eigenvalue solution $$\lambda^{(m)}$$ has an associated non-zero eigenvector $$\vec{v}^{(m)}$$ that is obtained by substituting the value of $$\vec{v}^{(m)}$$ into Equation $$\ref{eq:2}$$ and solving the resulting set of $$M$$ equations for the $$M$$ components of $$\vec{v}^{(m)}$$.

An important property of Equation $$\ref{eq:2}$$ is that both sides of the equation can be multiplied by an arbitrary constant $$c$$ and the equation still holds. So, if $$\vec{v}^{(m)}$$ is an eigenvector of $$\underset{\sim}{A}$$ with eigenvalue $$\lambda^{(m)}$$, then $$c\vec{v}^{(m)}$$ is also an eigenvector of $$A$$ with the same eigenvalue $$\lambda^{(m})$$. The eigenvectors are therefore defined only to within an arbitrary scalar multiple. In many cases, we choose the eigenvectors to be unit vectors by making the appropriate choice for $$c$$.

Interesting tidbits:

• The sum of the eigenvalues equals the trace.

• The product of the eigenvalues equals the determinant.

You will prove these in due course.