Eigenvalues and Eigenvectors: The Characteristic Polynomial

1. From a Condition to a Calculation: The Characteristic Polynomial

In the last lesson, we established that for an eigenvalue $\lambda$ to exist, the matrix $(A-\lambda I)$ must have a non-trivial null space. This is equivalent to saying its determinant must be zero. Let's formalize this.

The expression $\det (A-\lambda I)$ is not just a number; it's a polynomial in the variable $\lambda$. This is called the characteristic polynomial of matrix $A$. The roots of the characteristic equation, $\det (A-\lambda I)=0$, are the eigenvalues of $A$.

As the video demonstrated, the process for finding eigenvalues is:

1. Construct the matrix $A-\lambda I$ by subtracting $\lambda$ from each diagonal entry of $A$.
2. Compute its determinant, $\det (A-\lambda I)$. This will give you the characteristic polynomial in $\lambda$.
3. Solve the equation $\det (A-\lambda I)=0$ for $\lambda$. The solutions are your eigenvalues.

Let's see this in action with a concrete example.

2. Step 1: Solving for Eigenvalues ($\lambda$)

Consider the matrix $A=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)$.

1. Construct $A-\lambda I$:

2. Compute the determinant:

This is the characteristic polynomial.

3. Solve $\det (A-\lambda I)=0$:

Factoring the polynomial gives:

The roots, and therefore the eigenvalues, are ${\lambda }_1=4$ and ${\lambda }_2=2$.

An interesting side note for 2x2 matrices: notice that the coefficient of $\lambda$ (-6) is the negative of the sum of the diagonal elements of A (3+3=6), which is called the trace. The constant term (8) is the determinant of A (33 - 11 = 8). This is a useful shortcut for checking your work!

3. Step 2: Solving for Eigenvectors ($\mathbf{x}$)

Finding the eigenvalues is only half the problem. For each eigenvalue, we now need to find its corresponding eigenvector(s).

Remember our starting point: $(A-\lambda I)\mathbf{x}=0$. An eigenvector $\mathbf{x}$ is simply a non-zero vector in the null space of the matrix $(A-\lambda I)$. Now that we have the values for $\lambda$, we can plug them in and solve this system for $\mathbf{x}$.

Let's continue with our example matrix $A=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)$ and its eigenvalues ${\lambda }_1=4$ and ${\lambda }_2=2$.

Case 1: Find the eigenvector for ${\lambda }_1=4$.
We need to solve $(A-4I)\mathbf{x}=0$.
First, find the matrix $(A-4I)$:

Now, solve the system $\left(\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$.
This gives the equation $-x_1+x_2=0$, or $x_1=x_2$. (The second row gives $x_1-x_2=0$, which is the same equation, as expected for a singular matrix).

Any non-zero vector where $x_1=x_2$ is a valid eigenvector. The simplest choice is ${\mathbf{x}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$. The entire set of eigenvectors for ${\lambda }_1=4$ is the line spanned by this vector.

Case 2: Find the eigenvector for ${\lambda }_2=2$.
We need to solve $(A-2I)\mathbf{x}=0$.
First, find the matrix $(A-2I)$:

Now, solve the system $\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$.
This gives the equation $x_1+x_2=0$, or $x_1=-x_2$.

A simple choice for the eigenvector is ${\mathbf{x}}_2=\left(\begin{array}{c}1\\ -1\end{array}\right)$.

We have successfully found the eigenvalues and a basis for their corresponding eigenspaces!

4. A Note on Complex Eigenvalues

What happens if the characteristic polynomial has no real roots? For example, a 90-degree rotation matrix $R=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$.

Its characteristic polynomial is:
$\det (R-\lambda I)=\det \left(\begin{array}{cc}-\lambda & -1\\ 1& -\lambda \end{array}\right)={\lambda }^2+1$
Setting ${\lambda }^2+1=0$ gives the solutions $\lambda =i$ and $\lambda =-i$.

This is a profound result. The absence of real eigenvectors, which we observed geometrically in the last lesson, corresponds algebraically to the characteristic polynomial having complex roots. These complex eigenvalues and their corresponding complex eigenvectors are absolutely essential in quantum mechanics, where they are related to the dynamics of quantum systems. We will explore complex vector spaces more formally later in the course.

Conclusion

Today we established the core computational algorithm for finding eigenvalues and eigenvectors. You now have a reliable, two-step process to analyze any square matrix.

Key Takeaways:

• The Algorithm:
  1. Find Eigenvalues: Calculate the characteristic polynomial $\det (A-\lambda I)$ and solve the equation $\det (A-\lambda I)=0$ to find the eigenvalues $\lambda$.
  2. Find Eigenvectors: For each eigenvalue $\lambda$, find the non-zero vectors $\mathbf{x}$ in the null space of $(A-\lambda I)$ by solving the system $(A-\lambda I)\mathbf{x}=0$.
• The set of all eigenvectors for a given $\lambda$, plus the zero vector, forms a subspace called the eigenspace.
• The existence of complex eigenvalues is not a failure but an important feature, corresponding to transformations (like rotations) that have no real invariant directions.

Preview of the Next Lesson:

In our examples today, each distinct eigenvalue corresponded to one "dimension" or direction of eigenvectors. But what happens if the characteristic polynomial has repeated roots, for example, $(\lambda -3{)}^2=0$? Does the eigenvalue $\lambda =3$ give us two independent eigenvectors or only one? In the next lesson, we will explore this question by distinguishing between the algebraic and geometric multiplicity of an eigenvalue, which is the key to understanding when a matrix can be "diagonalized".