## 2.5 Eigenvalues and eigenvectors

Definition 2.16 Let A be a square matrix.

• an eigenvalue of A is a number l such that there exists a nonzero vector $$\mathbf{v}$$ such that $$A\mathbf{v} = l\mathbf{v}$$.
• a vector $$\mathbf{v}$$ with this property is called an l-eigenvector of A.
• a vector $$\mathbf{v}$$ which is an l-eigenvector for some l is called an eigenvector of A.

Lemma 2.8 l is an eigenvalue of the nn matrix A if and only if $$\det (A-l I_n)=0$$.
Proof. $$\det (A-l I_n)=0$$ if and only if $$A-l I_n$$ is not invertible (Theorem 2.3), if and only if there is a nonzero vector $$\mathbf{v}$$ with $$(A-l I_n)\mathbf{v}=\mathbf{0}$$ (Corollary 2.2). But $\begin{equation*} (A-l I_n)\mathbf{v}= A \mathbf{v}-l \mathbf{v} \end{equation*}$ so this holds if and only if $$\mathbf{v}$$ is an l-eigenvector.

If we think of x as an unknown, equation $$\det (A-x I_n)=0$$ is a polynomial of degree n in the variable x. This is called the characteristic polynomial $$\chi_A(x)$$ of A, and the last lemma shows that the roots of $$\chi_A$$ are exactly the eigenvalues of A.

Example 2.8 Let $$A= \begin{pmatrix} 1&2 \\ 3&4 \end{pmatrix}$$. The characteristic polynomial is $\begin{equation*} \chi_A(x)= \det(A-xI_2)= \det \begin{pmatrix} 1-x&2 \\ 3&4-x \end{pmatrix}=x^2-5x-2. \end{equation*}$ The eigenvalues of A are the roots of the characteristic polynomial, which are $$\frac{5 \pm \sqrt{33}}{2}$$.

Once we know that a certain number l is an eigenvalue of A, we can find the eigenvectors corresponding to that eigenvalue by solving the matrix equation $\begin{equation*} (A-l I_n)\mathbf{x}= \mathbf{0} \end{equation*}$ by putting the augmented matrix into RRE form (or otherwise).

Example 2.9 The characteristic polynomial of $$A= \begin{pmatrix} 1&1\\1&1 \end{pmatrix}$$ is $\begin{equation*} \det (A-x I_2)= (1-x)(1-x)-1=x^2-2x=x(x-2) \end{equation*}$ and so the eigenvalues of A are 0 and 2. We can find the 0-eigenvectors by looking for nonzero solutions of the matrix equation $\begin{equation*} (A-0I_2) \mathbf{x} = \mathbf{0}, \end{equation*}$ that is, $$A \mathbf{x}= \mathbf{0}$$. By putting this into RRE form or just solving directly, you will find that the 0-eigenvectors are the vectors $$\begin{pmatrix} a\\-a \end{pmatrix}$$ for $$a \neq 0$$.

To find the 2-eigenvectors we look for nonzero solutions of the matrix equation $\begin{equation*} (A-2I_2) \mathbf{x}=\mathbf{0} \end{equation*}$ which is $\begin{equation*} \begin{pmatrix} -1&1\\1&-1 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \mathbf{0}. \end{equation*}$ Again by putting the augmented matrix into RRE form, or otherwise, you’ll find that the 2-eigenvectors are the vectors $$\begin{pmatrix} b\\b \end{pmatrix}$$ for $$b\neq 0$$.