Eigenvectors and Eigenvalues

Diagnalizing a Matrix Given a linear transformation $T: V \to V$, a non-zero vector $v \in V$ is called an eigenvector of $T$ if there exists a scalar $\lambda$ such that:

$$ T(v) = \lambda v $$

The scalar $\lambda$ is called the eigenvalue corresponding to the eigenvector $v$.

#Finding Eigenvectors and Eigenvalues

#Eigenvalues

To find the eigenvectors and eigenvalues of a linear transformation $T$ represented by a matrix $A$, we need to solve the equation:

$$ Ax = \lambda x $$ This equation can be rewritten as: $$ (A - \lambda I)x = 0 $$

where $I$ is the identity matrix.

#Characteristic Equation

The equation:

$$ \det(A - \lambda I) = 0 $$

is called the characteristic equation. The solutions to this equation are the eigenvalues of $A$.

#Example

Suppose we have a matrix:

$$ A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} $$ To find the eigenvalues of $A$, we need to solve the characteristic equation: $$ \det(A - \lambda I) = \det\begin{pmatrix} 2 - \lambda & 1 \\ 1 & 1 - \lambda \end{pmatrix} = (2 - \lambda)(1 - \lambda) - 1 = 0 $$ Solving this equation, we get: $$ \lambda^2 - 3\lambda + 1 = 0 $$ Using the quadratic formula, we get: $$ \lambda = \frac{3 \pm \sqrt{5}}{2} $$

So, the eigenvalues of $A$ are $\lambda_1 = \frac{3 + \sqrt{5}}{2}$ and $\lambda_2 = \frac{3 - \sqrt{5}}{2}$.

#Eigenvectors

Once we have determined the eigenvalues, we can find the corresponding eigenvectors. For each eigenvalue $\lambda$, we solve the equation:

$$ (A - \lambda I)x = 0 $$

#Finding Eigenvectors for $\lambda_1 = \frac{3 + \sqrt{5}}{2}$

Substitute $\lambda_1$ into the equation:

$$ A - \lambda_1 I = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} - \begin{pmatrix} \frac{3 + \sqrt{5}}{2} & 0 \\ 0 & \frac{3 + \sqrt{5}}{2} \end{pmatrix} = \begin{pmatrix} 2 - \frac{3 + \sqrt{5}}{2} & 1 \\ 1 & 1 - \frac{3 + \sqrt{5}}{2} \end{pmatrix} $$ Simplifying the matrix: $$ = \begin{pmatrix} \frac{4 - \sqrt{5}}{2} & 1 \\ 1 & \frac{2 - \sqrt{5}}{2} \end{pmatrix} $$ We solve the system: $$ \begin{pmatrix} \frac{4 - \sqrt{5}}{2} & 1 \\ 1 & \frac{2 - \sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ From the first row: $$ \left(\frac{4 - \sqrt{5}}{2}\right)x_1 + x_2 = 0 \quad \Rightarrow \quad x_2 = -\left(\frac{4 - \sqrt{5}}{2}\right)x_1 $$ Thus, the eigenvector corresponding to $\lambda_1$ is: $$ x = \begin{pmatrix} x_1 \\ -\left(\frac{4 - \sqrt{5}}{2}\right)x_1 \end{pmatrix} = x_1 \begin{pmatrix} 1 \\ -\left(\frac{4 - \sqrt{5}}{2}\right) \end{pmatrix} $$

#Finding Eigenvectors for $\lambda_2 = \frac{3 - \sqrt{5}}{2}$

Substitute $\lambda_2$ into the equation:

$$ A - \lambda_2 I = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} - \begin{pmatrix} \frac{3 - \sqrt{5}}{2} & 0 \\ 0 & \frac{3 - \sqrt{5}}{2} \end{pmatrix} = \begin{pmatrix} 2 - \frac{3 - \sqrt{5}}{2} & 1 \\ 1 & 1 - \frac{3 - \sqrt{5}}{2} \end{pmatrix} $$ Simplifying the matrix: $$ = \begin{pmatrix} \frac{4 + \sqrt{5}}{2} & 1 \\ 1 & \frac{2 + \sqrt{5}}{2} \end{pmatrix} $$ We solve the system: $$ \begin{pmatrix} \frac{4 + \sqrt{5}}{2} & 1 \\ 1 & \frac{2 + \sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ From the first row: $$ \left(\frac{4 + \sqrt{5}}{2}\right)x_1 + x_2 = 0 \quad \Rightarrow \quad x_2 = -\left(\frac{4 + \sqrt{5}}{2}\right)x_1 $$ Thus, the eigenvector corresponding to $\lambda_2$ is: $$ x = \begin{pmatrix} x_1 \\ -\left(\frac{4 + \sqrt{5}}{2}\right)x_1 \end{pmatrix} = x_1 \begin{pmatrix} 1 \\ -\left(\frac{4 + \sqrt{5}}{2}\right) \end{pmatrix} $$

#Summary

The eigenvalues of matrix $A$ are $\lambda_1 = \frac{3 + \sqrt{5}}{2}$ and $\lambda_2 = \frac{3 - \sqrt{5}}{2}$. The corresponding eigenvectors are:

For $\lambda_1$: $\begin{pmatrix} 1 \ -\left(\frac{4 - \sqrt{5}}{2}\right) \end{pmatrix}$
For $\lambda_2$: $\begin{pmatrix} 1 \ -\left(\frac{4 + \sqrt{5}}{2}\right) \end{pmatrix}$