Introduction to Eigenvectors & Eigenvalues

What Are Eigenvectors and Eigenvalues?

An eigenvector is a non-zero vector that only changes in magnitude when a matrix is applied to it. The corresponding scalar value that describes this change is the eigenvalue.

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

Where:

• $A$ is a square matrix;
• $\lambda$ is the eigenvalue;
• $\vec{v}$ is the eigenvector.

Example Matrix and Setup

Suppose:

$$A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$$

We want to find values of $\lambda$ and vectors $\vec{v}$ such that:

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

Characteristic Equation

To find $\lambda$, solve the characteristic equation:

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

Substitute:

$$\det \begin{bmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{bmatrix} = 0$$

Compute determinant:

$$(4-\lambda)(3-\lambda) - 2 = 0$$

Solve:

$$\lambda^2 - 7\lambda + 10 = 0 \\ \lambda = 5, \; \lambda = 2$$

Find Eigenvectors

Now solve for each $\lambda$.

For $\lambda = 5$:

Subtract:

$$(A - 5I)\vec{v} = 0$$ $$\begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix} \vec{v} = 0$$

Solve:

$$v_1 = v_2$$

So:

$$\vec{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$

For $\lambda = 2$:

Subtract:

$$(A - 2I)\vec{v} = 0$$ $$\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} \vec{v} = 0$$

Solve:

$$v_1 = -\tfrac{1}{2} v_2$$

So:

$$\vec{v} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$$

Confirm the Eigenpair

Once you have an eigenvalue $\lambda$ and an eigenvector $\vec{v}$, verify that:

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

Example:

$$A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 5 \end{bmatrix} = 5 \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$

Example:

$$\begin{bmatrix} 2 \\ 2 \end{bmatrix}$$

is also an eigenvector for $\lambda = 5$.

Diagonalization (Advanced)

If a matrix $A$ has $n$ linearly independent eigenvectors, then it can be diagonalized:

$$A = PDP^{-1}$$

Where:

• $P$ is the matrix of eigenvectors as columns;
• $D$ is a diagonal matrix of eigenvalues;
• $P^{-1}$ is the inverse of $P$.

You can confirm diagonalization by checking $A = PDP^{-1}$.
This is useful for computing powers of $A$:

Example

Let:

$$A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix}$$

Find eigenvalues:

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

Solve:

$$\lambda = 3, \; \lambda = 2$$

Find eigenvectors:

For $\lambda = 3$:

$$\vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

For $\lambda = 2$:

$$\vec{v} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$$

Construct $P, D$ and $P^{-1}$:

$$P = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}, \quad D = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}, \quad P^{-1} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$

Compute:

$$PDP^{-1} = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix} = A$$

Confirmed.

Why this matters:

To compute powers of $A$, like $A^k$. Since $D$ is diagonal:

$$A^k = P D^k P^{-1}$$

This makes calculating matrix powers much faster.

Important Notes

• Eigenvalues and eigenvectors are directions that remain unchanged under transformation;
• $\lambda$ stretches $\vec{v}$;
• $\lambda = 1$ keeps $\vec{v}$ unchanged in magnitude.