Defining Eigenvalues and Eigenvectors

When a matrix acts on a vector, it usually changes the vector's direction. But for most square matrices, certain non-zero vectors are exceptional. The transformation only stretches or shrinks them, leaving their original direction unchanged. These special vectors are called eigenvectors, and the factors by which they are scaled are their corresponding eigenvalues.

This unique relationship is captured in a simple but powerful equation:

$$Av = \lambda v$$

This equation states that when the matrix $A$ is applied to its eigenvector $v$, the result is the same as multiplying $v$ by a simple scalar, the eigenvalue $\lambda$.

Let's break down each component:

• $A$ is the square matrix representing the linear transformation.
• $v$ is the eigenvector, a non-zero vector whose direction is preserved by the transformation.
• $\lambda$ (the Greek letter lambda) is the eigenvalue, a scalar that tells us the factor by which the eigenvector $v$ is stretched, shrunk, or flipped.

Essentially, eigenvectors define the "axes" of a transformation. While other vectors are rotated and scaled in complex ways, eigenvectors point along directions where the transformation acts simply as a scaling operation.

For a general vector, a matrix transformation changes its direction. For an eigenvector, the transformation only scales it along the same line (its "eigenspace").

A Concrete Example

Let's make this tangible with numbers. Consider the following matrix $A$:

$$A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$$

Now, let's test if the vector $v = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ is an eigenvector of $A$. We do this by applying the transformation (multiplying $A$ by $v$) and seeing if the result is a scaled version of $v$.

First, calculate the left side of the equation, $Av$:

$$Av = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 4(2) + (-2)(1) \\ 1(2) + 1(1) \end{pmatrix} = \begin{pmatrix} 8 - 2 \\ 2 + 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \end{pmatrix}$$

The result of the transformation is the vector $\begin{pmatrix} 6 \\ 3 \end{pmatrix}$. Now, we check if this result is a multiple of our original vector $v = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

Is there a scalar $\lambda$ such that $\begin{pmatrix} 6 \\ 3 \end{pmatrix} = \lambda \begin{pmatrix} 2 \\ 1 \end{pmatrix}$?

Yes, there is. We can see that:

$$\begin{pmatrix} 6 \\ 3 \end{pmatrix} = 3 \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$

Because we found such a scalar, we have confirmed that $v = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ is an eigenvector of matrix $A$. The corresponding eigenvalue is $\lambda = 3$. The transformation $A$ stretches this specific vector by a factor of 3 without changing its direction.

Interpreting Eigenvalues

The value of $\lambda$ gives us important information about what the transformation does to the corresponding eigenvector:

• If $\lambda > 1$, the eigenvector is stretched.
• If $0 < \lambda < 1$, the eigenvector is shrunk.
• If $\lambda = 1$, the eigenvector is unchanged by the transformation.
• If $\lambda < 0$, the eigenvector is flipped to point in the opposite direction and then scaled.
• If $\lambda = 0$, the eigenvector is collapsed into the origin (the zero vector).

The Non-Zero Condition

It is important that eigenvectors must be non-zero vectors. Why? If we allowed $v$ to be the zero vector, the equation $A v = \lambda v$ would become $A\mathbf{0} = \lambda\mathbf{0}$, which simplifies to $\mathbf{0} = \mathbf{0}$. This equation is true for any matrix $A$ and any scalar $\lambda$. It provides no useful information about the transformation, so we exclude the zero vector by definition.