How to Find Eigenvalues and Eigenvectors?

Before beginning to see how to find eigenvalues and eigenvectors of a matrix, let's see what they actually are. First, eigenvalues are a set of scalars that can satisfy $A \textbf{v} = \lambda \textbf{v}$. Let's take a look at this equation. On the left-hand side, we multiply vector $\textbf{v}$ with matrix $A$. As mentioned in the last section, multiplying a matrix with a vector is a linear transformation of the vector. On the right-hand side, we multiply a scalar $\lambda$ with the vector $\textbf{v}$. Multiplying a scalar is also a linear transformation; however, it does not change the 'direction' of the vector; rather, it changes the magnitude of the vector, as its name describes. So, the goal of finding scalar $\lambda$ is that $\lambda$ must satisfy the linear transformation of vector $\textbf{v}$. Eigenvectors are vectors that give a scaled multiple of itself after a given linear operator. In this document, we are going to see how to find eigenvalues and eigenvectors.

Eigenvalues

First, let's derive an equation called the characteristic equation. From the matrix equation given above, we can derive a characteristic equation by following these steps. We are assuming that $\textbf{v} \neq \textbf{0}$.

$$ \begin{align*} A \textbf{v} &= \lambda \textbf{v} \ A \textbf{v} - \lambda \textbf{v} &= [0] \ (A - \lambda I) \textbf{v} &= [0] \ A - \lambda I &= [0] \ \det(A - \lambda I) &= \det([0]) \ \det(A - \lambda I) &= 0 \end{align*} $$

Now, we see that the characteristic equation is in terms of the determinant of $A - \lambda I$. If we find a determinant of $A - \lambda I$, there would be a polynomial equation of $\lambda$. If we solve that equation, we obtain the eigenvalues of matrix $A$. Let's find the eigenvalues of $A$.

$$ A = \begin{bmatrix} 2&10 \ 1&3 \end{bmatrix} $$

By using the formula above, we can get a characteristic equation.

$$ \begin{align*} \det(A - \lambda I) &= \begin{vmatrix} 2-\lambda & 10 \ 1 & 3 - \lambda \end{vmatrix} \ &= (2-\lambda)(3-\lambda) - 10 = 0 \end{align*} $$

We solve for $\lambda$, then we obtain the eigenvalues of $A$.

$$ \begin{align*} (2-\lambda)(3-\lambda) - 10 = \lambda^2 -5\lambda - 4 \Rightarrow (\lambda+1)(\lambda-5) = 0 \end{align*} $$

Eigenvectors

Useful Properties

• The sum of diagonal entries in the matrix $A$ is the sum of the eigenvalues of $A$: $\text{tr}(A) = \sum (\lambda)$.
• The determinant of the matrix $A$ is the product of the eigenvalues of $A$: $\det(A) = \prod (\lambda)$.