Characteristic Equation

When analyzing linear transformations, eigenvalues ($\lambda$) and eigenvectors ($v$) reveal the fundamental directions and scaling factors of a matrix. An eigenvector $v$ of a matrix $A$ is a non-zero vector that, when $A$ acts upon it, only scales the vector without changing its direction. This relationship is captured by the equation $Av = \lambda v$. To find these special values $\lambda$, we need a systematic algebraic approach, which leads us directly to the characteristic equation.

The core definition of an eigenvector and eigenvalue is $Av = \lambda v$. Our goal is to solve for $\lambda$. To do this, we first move all terms to one side of the equation, setting it to zero. This transforms the equation into $Av - \lambda v = 0$. This step is crucial as it sets up a homogeneous system, which is a system of linear equations where all constant terms are zero.

We cannot directly factor out the vector $v$ from $Av - \lambda v = 0$ because $A$ is a matrix and $\lambda$ is a scalar. To perform matrix subtraction, $\lambda$ must be converted into a matrix of the same dimensions as $A$. We achieve this by multiplying $\lambda$ by the identity matrix $I$, which has ones on the main diagonal and zeros elsewhere. This allows us to write $\lambda v$ as $\lambda I v$, maintaining the vector $v$ and enabling the factorization $(A - \lambda I)v = 0$.

The equation $(A - \lambda I)v = 0$ represents a homogeneous system of linear equations. If the matrix $(A - \lambda I)$ were invertible, the only solution for $v$ would be the trivial solution $v = 0$. However, eigenvectors are defined as non-zero vectors. Therefore, for a non-trivial solution $v \neq 0$ to exist, the matrix $(A - \lambda I)$ must be singular (non-invertible). A fundamental property of singular matrices is that their determinant is zero.

Combining these insights, the condition for the existence of non-zero eigenvectors $v$ is that the matrix $(A - \lambda I)$ must be singular. This directly implies that its determinant must be equal to zero. This crucial relationship, $det(A - \lambda I) = 0$, is known as the characteristic equation. Solving this equation for $\lambda$ will yield all the eigenvalues of the matrix $A$.

Before calculating the determinant, we first need to explicitly construct the matrix $(A - \lambda I)$. This involves subtracting $\lambda$ from each element on the main diagonal of matrix $A$, while all off-diagonal elements remain unchanged. For a $2 \times 2$ matrix

Once we have the matrix $(A - \lambda I)$, the next step is to calculate its determinant and set it to zero. For a $2 \times 2$ matrix

The expression $det(A - \lambda I)$ is not just an equation; it's a polynomial in $\lambda$. This polynomial is called the characteristic polynomial of matrix $A$. For an $n \times n$ matrix, the characteristic polynomial will always be of degree $n$. The roots of this polynomial are precisely the eigenvalues of the matrix. Finding these roots is the primary method for determining eigenvalues.

Sometimes, an eigenvalue can be a repeated root of the characteristic polynomial. The number of times an eigenvalue $\lambda_i$ appears as a root of the characteristic polynomial is called its algebraic multiplicity. For example, if the characteristic polynomial is $(\lambda - 2)^3 (\lambda - 5) = 0$, then $\lambda = 2$ has an algebraic multiplicity of 3, and $\lambda = 5$ has an algebraic multiplicity of 1. This concept is important for understanding the full set of eigenvalues and their properties.

The entire derivation of the characteristic equation hinges on the requirement that eigenvectors must be non-zero. If we allowed $v = 0$, then $Av = \lambda v$ would always be true for any $\lambda$ and any matrix $A$, rendering the concept of eigenvalues meaningless. The condition $det(A - \lambda I) = 0$ specifically ensures that we are looking for values of $\lambda$ that allow for non-trivial (non-zero) solutions to the homogeneous system, which are the true eigenvectors.