Eigenvectors and Eigenvalues

Definition

A complex $\lambda$ and a vector $\vec{v}$ is an eigenvalue and eigenvector of a matrix $A$ if

$$A\vec{v}=\lambda \vec{v}$$

We also say that $\vec{v}$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$.

Note: that $\vec{v}$ cannot be the zero vector

Arising from the note above, we find that there are two cases for $\lambda$:

• If $\lambda =0$, then $A\vec{v}=\vec{0}$, meaning that $\vec{v}$ is in the null space of $A$.
• If $\lambda \ne 0$, then $\vec{v}$ is in the column space of $A$.

Geometric Interpretation

As matrices are linear transformations which simply transform some vector $\vec{v}$ into a new vector $A\vec{v}$. We note that an eigenvector is some special vector that, when transformed by the matrix $A$ simply results in a scaled version of itself. The scaling factor is the eigenvalue.

Connection to Characteristic Equation

In general,

$$A\vec{x}=\lambda \vec{x}$$

can be rewritten as

$$A\vec{x}-\lambda I\vec{x}=\vec{0}\Rightarrow (A-\lambda I)\vec{x}=\vec{0}$$

This equation is the precursor to the characteristic equation.

Existence

Not all matrices have eigenvalues and eigenvectors. A matrix $A$ has an eigenvalue $\lambda$ iff the equation

$$(A-\lambda I)\vec{x}=\vec{0}$$

has a non-trivial solution. Which also translates to, if $(A-\lambda I)$ does not have full rank. Meaning:

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

This is the characteristic equation.

Examples

Example 1

Let $A$ be the matrix

$$A=\left[\begin{array}{cc}1& 6\\ 5& 2\end{array}\right]$$

Is $7$ an eigenvalue of $A$?

Human: this also means, is there a vector $\vec{x}$ such that the below equation is satisfied?

$$\left[\begin{array}{cc}1& 6\\ 5& 2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}7x_1\\ 7x_2\end{array}\right]$$

This creates the linear system

$$\{\begin{array}{l}{x}_1+6x_2=7x_1\\ 5x_1+2x_2=7x_2\end{array}\Rightarrow \{\begin{array}{l}-6x_1+6x_2=0\\ 5x_1-5x_2=0\end{array}$$

constructing an augmented matrix gives

$$\left[\begin{array}{ccc}-6& 6& 0\\ 5& -5& 0\end{array}\right]\to \left[\begin{array}{ccc}1& -1& 0\\ 0& 0& 0\end{array}\right]$$

Reassigning free variables, we let $x_2=s$, then: $x_1=s$ and

$$\vec{x}=\left[\begin{array}{c}s\\ s\end{array}\right]=s\cdot \left[\begin{array}{c}1\\ 1\end{array}\right],s\in \mathbb{R}$$

Therefore, any scalar multiple of $\left[\begin{array}{c}1\\ 1\end{array}\right]$ is an eigenvector for the eigenvalue $\lambda =7$.

Example 2

let:

$$A=\left[\begin{array}{cc}3& -2\\ 1& 0\end{array}\right],\vec{v}=\left[\begin{array}{c}2\\ 1\end{array}\right]$$ $$A\cdot \vec{v}=\left[\begin{array}{c}4\\ 2\end{array}\right]=2\cdot \vec{v}$$

This result is specific to this vector.

Contrary, let:

$$\vec{u}=\left[\begin{array}{c}-1\\ 1\end{array}\right]\Rightarrow A\cdot \vec{u}=\left[\begin{array}{c}-5\\ -1\end{array}\right]$$

which is not a multiple of $\vec{u}$

Example 3

$$A=\left[\begin{array}{ccc}4& -1& 6\\ 2& 1& 6\\ 2& -1& 8\end{array}\right]$$

given $\lambda =2$ is an eigenvalue of $A$, find a basis for the corresponding eigenspace.

$$(A-\lambda I)\vec{x}=0\Rightarrow (A-2I)\vec{x}=0$$ $$A-2I=A-\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]=\left[\begin{array}{ccc}2& -1& 6\\ 2& -1& 6\\ 2& -1& 6\end{array}\right]$$

wow! All rows are the same, who could’ve predicted this !!

$$\left[\begin{array}{ccc}2& -1& 6\\ 2& -1& 6\\ 2& -1& 6\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$$

using this equation, we can construct the augmented matrix and row reduction.

$$\left[\begin{array}{cccc}2& -1& 6& 0\\ 2& -1& 6& 0\\ 2& -1& 6& 0\end{array}\right]\to \left[\begin{array}{cccc}1& \frac{-1}{2}& 3& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$$

This matrix has 2 free variables and only 1 Pivot Position. Now, we let: $x_2=s,x_3=t$ and,

$$x_1=\frac{1}{2}s-3t\Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}\frac{1}{2}\\ 1\\ 0\end{array}\right]s+\left[\begin{array}{c}-3\\ 0\\ 1\end{array}\right]t$$

Therefore,

$$\{\left[\begin{array}{c}\frac{1}{2}\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 0\\ 1\end{array}\right]\}$$

form a basis for the eigenspace of $A$ for $\lambda =2$.