# Eigenvectors and Eigenvalues

For a given square matrix \(A\), the vector \(x\) is an eigenvector of \(A\) with corresponding
eigenvalue \(\lambda \in \mathbb{R}\) if

\(A x = \lambda x\).

Geometrically, this means that the linear transformation \(A\) applied to \(x\) doesn’t change
the direction of \(x\) i.e. \(A\) either stretches, compresses or reflects \(x\) in the opposite
direction.

Properties:

- If \(M\) is an invertible linear map, the eigenvalues of \(A\) are the same eigenvalues of \(M^{-1} A
M\). To see why, suppose \(y\) is an eigenvector of \(M^{-1} A M\) with eigenvalue \(\mu\):

\[M^{-1} A M y = \mu y \rightarrow A M y = \mu M y \rightarrow My\]
The implication is that a change of variables from \(y\) to \(M y\) doesn’t affect the eigenvalues
of \(A\) but changes the eigenvectors from \(y\) to \(My\) (so long as M is invertible).

### Eigenvalues of Special Matrices

- If square \(A\) is real and symmetric, then its eigenvalues are real. To see why, consider
an eigenvalue-eigenvector pair \((\lambda, v)\):

\[v^T A \overline{v} = v^T \overline{A v} = \overline{\lambda} v^T \overline{v}\]
and we also have

\[v^T A \overline{v} = v^T A^T \overline{v} = \lambda v^T \overline{v}\]
Consequently, \(\lambda = \overline{\lambda}\) and thus \(\lambda\) must be real.