1. Definition:

• If $A$ is an $n\times n$ matrix, do nonzero vectors x in $R^n$ exist such that $A$x is a scalar multiple of x? The scalar, denoted by $\lambda$, is called an eigenvalue of matrix $A$, and nonzero vector x is called an eigenvector of $A$ corresponding to $\lambda$. So we have: $A$x$=$$\lambda$x.

  Let $A$ be an $n\times n$ matrix. The scalar $\lambda$ is called an eigenvalue of $A$ if there is a nonzero vector x such that $A$x$=$$\lambda$x.

  The vector x is called an eigenvector of $A$ corresponding to $\lambda$.

  • Note: x and $\lambda$ cannot be zero.

• Eigenspaces:

  • If $A$ is an $n\times n$ matrix with an eigenvalue $\lambda$, then the set of all eigenvectors of $\lambda$, together with the zero vector

    {0} $\cup$ {x: x is an eigenvector of $\lambda$}

    is a subspace of $R^n$. This subspace is called the eigenspace of $\lambda$.

2. Finding eigenvectors and eigenvalues:

• First, we writing equation $A$x$=$$\lambda$x in the form $A$x$=$$\lambda I$x with $I$ is the $n\times n$ identity matrix. Then we produce

  ($\lambda I-A$)x $=$ 0. $\tag1$

• Note: We can see that: $(1)$ has nonzero solution if and only if the coefficient matrix ($\lambda I-A$) is not invertible, it means determinant of ($\lambda I-A$) is zero. So we have next theorem.

• Let $A$ be an $n\times n$ matrix.

  1. An eigenvalue of $A$ is a scalar $\lambda$ such that

  ​ det($\lambda I-A$) $=0$.

  2. The eigenvectors of $A$ is corresponding to $\lambda$ are the nonzero solutions of

  ​ ($\lambda I-A$)x$=$0​.

  • The equation det($\lambda I-A$) $=0$ is called the characteristic equation if $A$. Moreover, when expanded to polynomial form, the polynomial

    $|\lambda I-A|=\lambda^n+c_{n-1}\lambda^{n-1}+...+c_1\lambda+c_0$

    is called the characteristic polynomial of $A$.

• Finding eigenvalues and eigenvectors:

  • Let A be an $n\times n$ matrix.

    1. Form the characteristic equation $|\lambda I-A|=0$. It will be polynomial equation of degree $n$ in the variable $\lambda$.
    2. Find the real roots of the characteristic equation. These are the eigenvalues of $A$.
    3. For each value $\lambda_i$, find the eigenvectors corresponding to $\lambda_i$, by solving ($\lambda I-A$)x$=$0.

• Example: Finding the eigenvalues and corresponding eigenvectors of $A=\begin{bmatrix}2&1&0\\0&2&0\\0&0&2\end{bmatrix}$. What is the dimension of the eigenspace of each eigenvalue?

  • The characteristic polynomial is

    $$|\lambda I-A|= \begin{vmatrix}\lambda-2&-1&0\\0&\lambda-2&0\\0&0&\lambda-2\end{vmatrix}=(\lambda-2)^3$$

  • So, characteristic equation is $(\lambda-2)^3=0$. So we have only eigenvalue is $\lambda=2$.

  • To find eigenvectors corresponding of $\lambda=2$, we find x such that $(2I-A)$x$=$0$\tag2$.

    $$2I-A=\begin{bmatrix}0&-1&0\\0&0&0\\0&0&0\end{bmatrix}$$
    • Let x $=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$, $(2I-A)$x$=\begin{bmatrix}-x_2\\0\\0\end{bmatrix}$. Equation $(2)$ satisfied when $x_2=0$. Therefore, x $=\begin{bmatrix}x_1\\0\\x_3\end{bmatrix}\begin{bmatrix}s\\0\\t\end{bmatrix}=s\begin{bmatrix}1\\0\\x\end{bmatrix}+t\begin{bmatrix}0\\0\\1\end{bmatrix}$, $s$ and $t$ not both zero.

  • Because $\lambda=2$ has two linearly independent eigenvectors, the dimension of degree of its eigenspace is $2$.

• Eigenvalues of triangular matrix:

  • If $A$ is an $n\times n$ triangular matrix, then its eigenvalues are the entries on its main diagonal.

Reference

1. Elementary Linear Algebra.