Calculating Eigenvectors

Hello! Welcome back to our study of eigenanalysis.

Introduction

In our last lesson, we developed a powerful method for finding the eigenvalues of a matrix $A$. By understanding that an eigenvalue $\lambda$ makes the matrix $(A-\lambda I)$ singular, we derived the characteristic equation $\det (A-\lambda I)=0$. Solving this polynomial equation gave us the set of all eigenvalues for $A$.

Today, we complete the picture. Our goal is to compute the eigenvectors corresponding to each eigenvalue by solving the system $(A-\lambda I)\mathbf{x}=0$. Now that we know how to find the specific values of $\lambda$ that make the matrix $(A-\lambda I)$ singular, we will leverage that singularity to find the non-zero vectors $\mathbf{x}$ that satisfy the equation. These vectors form the null space of $(A-\lambda I)$, a crucial concept we will explore in detail.

1. From Eigenvalues to Eigenvectors: The Procedure

The fundamental equation for an eigenvector $\mathbf{x}$ and its corresponding eigenvalue $\lambda$ is $A\mathbf{x}=\lambda \mathbf{x}$. As we've seen, this can be rewritten as:
$(A-\lambda I)\mathbf{x}=0$
This is a homogeneous system of linear equations. Since we specifically chose $\lambda$ to make the matrix $(A-\lambda I)$ singular (by making its determinant zero), we are guaranteed that this system has non-trivial (non-zero) solutions for $\mathbf{x}$. These non-zero solutions are precisely the eigenvectors corresponding to $\lambda$.

The set of all solutions to this equation, including the zero vector, forms a subspace.

The Eigenspace
For a given eigenvalue $\lambda$ of a matrix $A$, the set of all vectors $\mathbf{x}$ satisfying $(A-\lambda I)\mathbf{x}=0$ is the null space of the matrix $(A-\lambda I)$. This subspace is called the eigenspace of $A$ corresponding to $\lambda$. An eigenvector is any non-zero vector in the corresponding eigenspace.

So, our procedure for finding eigenvectors for a given $\lambda$ is:

1. Construct the matrix $B=A-\lambda I$.
2. Find the null space of $B$ by solving the homogeneous system $B\mathbf{x}=0$. This is a standard problem that you can solve using Gaussian elimination.
3. The basis vectors for this null space are a set of linearly independent eigenvectors for $\lambda$.

2. A Concrete 2x2 Example

Let's apply this procedure. In the previous lesson's video, Professor Strang used the matrix $A=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)$ and found its eigenvalues to be ${\lambda }_1=4$ and ${\lambda }_2=2$. Now, let's find the corresponding eigenvectors.

Let's summarize the steps for ${\lambda }_1=4$:

1. Construct the matrix:
   $A-4I=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)-4\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right)$
2. Solve the system $(A-4I)\mathbf{x}=0$:
   $\left(\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$
   Both equations simplify to $-x_1+x_2=0$, or $x_1=x_2$. This means any vector where the two components are equal is an eigenvector.
3. Find a basis for the eigenspace: The general solution is $\mathbf{x}=\left(\begin{array}{c}{x}_2\\ x_2\end{array}\right)=x_2\left(\begin{array}{c}1\\ 1\end{array}\right)$. The eigenspace is the line spanned by the vector $\left(\begin{array}{c}1\\ 1\end{array}\right)$. A basis for this eigenspace is simply $\left\{\left(\begin{array}{c}1\\ 1\end{array}\right)\right\}$.

So, ${\mathbf{x}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$ is an eigenvector for ${\lambda }_1=4$.

For a clear, step-by-step written example, you can also review the worked example from the textbook.

This image provides a great visual summary of the process for a reflection matrix, connecting the geometry to the algebraic computation.

3. Eigenspaces with Higher Dimensions

What happens if the null space of $(A-\lambda I)$ has a dimension greater than one? This is not only possible but common. In such cases, the eigenspace for a single eigenvalue $\lambda$ will be spanned by two or more linearly independent eigenvectors.

The dimension of the eigenspace corresponding to $\lambda$ is called its geometric multiplicity. Finding a basis for the eigenspace means finding a complete set of linearly independent eigenvectors for that eigenvalue.

The following resource provides a thorough walkthrough for a 3x3 matrix where one eigenvalue has a geometric multiplicity of 2.

The image below shows another example where an eigenvalue gives rise to a two-dimensional eigenspace.

4. When Eigenvectors Are in Short Supply

In the previous lesson, we noted that eigenvalues can be repeated. This is called algebraic multiplicity. For example, in the characteristic polynomial $(\lambda -3{)}^2(\lambda -1)=0$, the eigenvalue $\lambda =3$ has an algebraic multiplicity of 2.

A crucial point, which will become central to our next lesson on diagonalization, is that the geometric multiplicity of an eigenvalue is always less than or equal to its algebraic multiplicity.
$1\le \text{Geometric Multiplicity}\le \text{Algebraic Multiplicity}$
Sometimes, an eigenvalue with an algebraic multiplicity greater than 1 can have a geometric multiplicity of only 1. This results in a "shortage" of linearly independent eigenvectors.

This situation is sometimes called a "defective" or "degenerate" case, and it has important consequences for whether a matrix can be simplified in a particular way, which we will study next.

Conclusion

In this lesson, we completed the full eigenvalue-eigenvector computation. By taking the eigenvalues found from the characteristic equation, we can systematically find the corresponding eigenvectors.

Key Takeaways:

• To find the eigenvectors for a known eigenvalue $\lambda$, you must find the null space of the matrix $(A-\lambda I)$.
• The process involves solving the homogeneous system $(A-\lambda I)\mathbf{x}=0$ using methods like Gaussian elimination.
• The set of all solutions (including $0$) is a subspace called the eigenspace. Any non-zero vector in this space is an eigenvector.
• The number of linearly independent eigenvectors for an eigenvalue $\lambda$ is the dimension of its eigenspace, also known as its geometric multiplicity.
• A single eigenvalue can have multiple linearly independent eigenvectors associated with it if its eigenspace has a dimension greater than one.

Preview of the Next Lesson:
We now have all the tools to perform one of the most powerful operations in linear algebra: diagonalization. The eigenvectors we just learned to compute can be used as basis vectors for a new coordinate system in which the linear transformation $A$ acts in a very simple way—it just stretches or shrinks along the axes. In the next lesson, we will see how to construct this transformation and prove the conditions under which a matrix is diagonalizable, a property that hinges directly on having a sufficient number of linearly independent eigenvectors.