### Invariant subspaces of linear maps: Eigenvalues and eigenvectors

### Eigenspace

We now focus on finding eigenvectors for a given eigenvalue.

Eigenspace

Let #V# be a vector space and # L :V\rightarrow V# a linear map. For every number #\lambda#,

\[

E_\lambda=\ker{ L -\lambda {I_V}}

\] is a linear subspace of #V#.

The subspace #E_\lambda# is called the **eigenspace** of # L # corresponding to #\lambda#. This space consists of the zero vector and all eigenvectors with eigenvalue #\lambda#.

The process of finding eigenvectors consists of first determining the values of #\lambda# such that #\dim {E_\lambda }\gt0#. Then #\lambda# is an eigenvalue and the vectors in #E_\lambda# distinct from #\vec{0}# are the eigenvectors corresponding to the eigenvalue.

Let # L :V\rightarrow V# be a linear map. We recall that the equation \(\det (L_\alpha-\lambda I)=0\) is the *characteristic equation* of # L # and that the left side of this equation, #\det (L_\alpha-\lambda \,I)#, is the *characteristic polynomial* of # L#.

Characterization of eigenvalues and eigenvectors Let #L:V\to V# be a linear map, where #V# is a vector space of finite dimension #n#.

- A number #\lambda# is an eigenvalue of #L# if and only if #\det (L-\lambda \cdot I_V)=0#.
- The eigenvectors corresponding to the eigenvalue #\lambda# are the solutions #\vec{v}# distinct from the zero vector of the linear equation #L(\vec{v}) = \lambda\, \vec{v}#.

Let #\alpha# be a basis of #V#. Then the #\alpha#-coordinates #v_1,\ldots,v_n# of the eigenvectors of #L# corresponding to the eigenvalue #\lambda# may be calculated as the solutions distinct from the zero vector of the system of linear equations \[ \left(\,\begin{array}{cccc}

a_{11}-\lambda & a_{12} & \ldots & a_{1n}\\

a_{21} & a_{22}-\lambda & & \vdots\\

\vdots & & \ddots & \vdots \\

a_{n1} & \ldots & \ldots & a_{nn}-\lambda

\end{array}\,\right)\ \ \left(\,\begin{array}{c}

v_1\\ v_2\\ \vdots\\ v_n

\end{array}\,\right)\ =\ \left(\,\begin{array}{r}

0 \\ 0\\ \vdots\\ 0

\end{array}\,\right) \] where #a_{ij}# is the #(i,j)#-entry of #L_\alpha#.

Vectors on the line \(x -3 y = 0\) are mapped onto themselves, so \(1\) is an eigenvalue of #L#. Furthermore, since the normal vector to this line gets mapped to its negative, \(-1\) is an eigenvalue as well. A map from \(\mathbb R^2\) to \(\mathbb R^2\) has at most two different eigenvalues, so the answer is \(\{-1,1\}\).

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