Eigenvector
Definition
Eigenvector of a matrix is a nonzero vector and becomes scalar times of the original vector after the transformation by a square matrix.
By definition, eigenvectors cannot be zero vectors.
\[\tag{1} A\vec x = \lambda \vec x\]Where $A$ is a square matrix in $\mathbb{R}^{n \times n}$ and $\lambda$ is a scalar.
There are exactly $n$ eigenvectors and associated eigenvalues in a $n$ by $n$ matrix, including duplicate values.
$A$, $\lambda$, and the eigenvector $\vec x$ can include complex numbers.
Characteristic equation
By rearranging the equation (1) above, we can get
\[A\vec x - \lambda \vec x = 0\]Using the identity matrix, we can rewrite the equation.
\[A\vec x - \lambda I \vec x = 0\]And finally, we can factor out $\vec x$.
\[(A - \lambda I)\vec x = 0\]Since $\vec x$ should be nonzero, it means $(A - \lambda I)$ should be singular, and thus the determinant should be zero.
\[\det(A-\lambda I) = 0\]By solving the determeinant, we get $n$ th degree polynomial called characteristic equation.
Also, this is why $n \times n$ matrix has exactly $n$ roots including multiplicity by the fundamental threorem of algebra.
Example of finding eigenvalues using determinant
Let $A$ be 2 by 2 matrix. $$ A = \begin{bmatrix}1&2 \\ -1&4\end{bmatrix} $$ Solving the determinant $\det(A - \lambda I) = 0$ $$ \begin{vmatrix}1-\lambda &2 \\ -1&4-\lambda \end{vmatrix} = 0 $$ $$ (1-\lambda )(4-\lambda ) + 2 = 0 $$ $$ \lambda^2 - 5\lambda + 6 = 0\\ $$ Finally, we can get two eigenvalues of $$ \lambda = 2, \lambda = 3 $$Example of finding associated eigenvalues
Using the example above with $$ A = \begin{bmatrix}1&2 \\ -1&4\end{bmatrix}, \lambda_1 = 2, \lambda_2 = 3 $$ We can find associated eigenvectors with each $\lambda_1$ and $\lambda_2$ using Gauss Elimination with augemented matrix. Starting with $\lambda_1$, $$ (A - \lambda I)\vec x = \vec 0 $$ $$ \left(\begin{bmatrix}1&2 \\ -1&4\end{bmatrix} - 2\begin{bmatrix}1&0\\0&1\end{bmatrix}\right)\vec x = \vec 0\\ \begin{bmatrix}1-2&2 \\ -1&4-2\end{bmatrix}\vec x = \vec 0 $$ Now we can make it augmented and apply Gauss Elimination $$ \begin{bmatrix}-1&2&|&0\\-1&2&|&0\end{bmatrix} \to \begin{bmatrix}-1&2&|&0\\0&0&|&0\end{bmatrix} $$ $$ -1x_1 +2x_2 = 0\\ x_1 = 2x_2 $$ Therefore, we can solve for $\vec x$ $$ \vec x = \begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}x_1\\x_1/2\end{bmatrix} = \begin{bmatrix}1\\1/2\end{bmatrix}x_1 $$ Therefore the eigenvector associated with $\lambda = 2$ is $\begin{bmatrix}1\\1/2\end{bmatrix}$Relation to Trace
trace is sum of diagonal entries of a matirix, and it is equal to the sum of eigenvalues of the matrix.
\[\text{trace}(A) = \sum_{i=1}^n \lambda_i\]Complex Eigenvectors/Eigenvalues
If a matrix is a real matrix, $A\in \mathbb{R}^{n\times n}$, then the complex conjugate of its eigenvectors and eigenvalues are also the eigenvector and eigenvalues of the matrix $A$.
Which implies that if there a matrix have odd number of eigenvalues, at least one of the eigenvalues and associated eigenvector is real.
Having only complex numbers in eigenvalues/eigenvectors
If a matrix has only complex eigenvectors and eigenvaues, it means any real vectors will be rotated by the matrix.
Eigenvalues of bipartite graph
Eigenvalues of bipartite graph are plus minus pairs.
It means the sum of eigenvalues are zero, thus trace of matrix is zero.