# 2.4: Eigenvalues and eigenvectors

- Page ID
- 18056

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.