The Characteristic Equation

We have established that for an eigenvector $v$, the transformation $A$ simply scales it by its corresponding eigenvalue $\lambda$. This relationship is captured by the equation $Av = \lambda v$. While this equation perfectly describes the property of eigenvalues and eigenvectors, it doesn't immediately show us how to find them for a given matrix $A$. To do that, we need an algebraic method, which starts by rearranging this fundamental equation.

Our goal is to find the values of $\lambda$ (the eigenvalues) that make this equation true for some non-zero vector $v$. Let's begin by moving all terms to one side of the equation:

$Av - \lambda v = 0$

It might seem like we can factor out the vector $v$, but there's a small hurdle. In the term $Av$, $v$ is multiplied by a matrix $A$, while in $\lambda v$, it's multiplied by a scalar $\lambda$. We cannot subtract a scalar from a matrix. To resolve this, we can use the identity matrix, $I$. Recall that multiplying any vector $v$ by the identity matrix $I$ leaves it unchanged, so $v = Iv$. We can substitute this into our equation:

$Av - \lambda(Iv) = 0$

Now, both terms involve a matrix multiplying the vector $v$, which allows us to factor $v$ out:

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

This single equation is very informative. It tells us that when we multiply the vector $v$ by the matrix $(A - \lambda I)$, the result is the zero vector.

Finding Non-Trivial Solutions

One obvious solution to the equation $(A - \lambda I)v = 0$ is the trivial solution, where $v$ is the zero vector. However, the definition of an eigenvector requires it to be a non-zero vector. After all, the zero vector's direction is undefined, so the idea of its direction being unchanged by a transformation isn't very useful.

This means we are looking for a non-zero vector $v$ that satisfies the equation. A system of linear equations of the form $Mx = 0$ has a non-zero solution for $x$ only if the matrix $M$ is singular. A singular matrix is one that does not have an inverse. A convenient way to check if a square matrix is singular is to calculate its determinant. If the determinant is zero, the matrix is singular.

Applying this to our situation, for the equation $(A - \lambda I)v = 0$ to have a non-zero solution for $v$, the matrix $(A - \lambda I)$ must be singular. This leads us directly to the condition we need:

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

This equation is called the characteristic equation of the matrix $A$.

The process of deriving the characteristic equation from the initial eigenvalue definition.

Example: Calculating Eigenvalues

Let's find the eigenvalues for a simple 2x2 matrix to see the characteristic equation in action. Suppose we have the matrix $A$:

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

Step 1: Set up the matrix $(A - \lambda I)$

First, we subtract $\lambda I$ from $A$:

$$A - \lambda I = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{pmatrix}$$

Step 2: Calculate the determinant and set it to zero

Next, we find the determinant of this new matrix. For a 2x2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $ad - bc$.

$$\det(A - \lambda I) = (2-\lambda)(2-\lambda) - (1)(1) = 0$$

Step 3: Solve the characteristic equation for $\lambda$

Expanding the equation gives us a polynomial in $\lambda$, which is called the characteristic polynomial.

$$(4 - 4\lambda + \lambda^2) - 1 = 0$$ $$\lambda^2 - 4\lambda + 3 = 0$$

We can solve this quadratic equation by factoring it:

$$(\lambda - 3)(\lambda - 1) = 0$$

This gives us two solutions for $\lambda$. These solutions are the eigenvalues of the matrix $A$:

$$\lambda_1 = 3, \quad \lambda_2 = 1$$

By solving the characteristic equation, we have successfully found the eigenvalues for matrix $A$. These are the two special scaling factors for the linear transformation defined by $A$. The next step, which we will address later, would be to substitute each of these eigenvalues back into the equation $(A - \lambda I)v = 0$ to find their corresponding eigenvectors.