Computing eigenvalues

Section 6.2 Computing eigenvalues

An easy but important fact is used to compute eigenvalues.

Proposition 6.2.1. Eigenvalues \(x\) satisfy \((A-\lambda I)\vec x=\vec0\).

\begin{equation*} A\vec x=\lambda \vec x \textrm{ if and only if } (A-\lambda I)\vec x=\vec0 \end{equation*}

Proof.

If \(A\vec x=\lambda\vec x\text{,}\) then

\begin{equation*} \vec0=A\vec x -\lambda\vec x=A\vec x -\lambda I\vec x = (A - \lambda I)\vec x. \end{equation*}

Corollary 6.2.2.

\(\lambda\) is an eigenvalue of \(A\) if and only if \(A-\lambda I\) is singular.

Proof.

One of the equivalent conditions of singularity given in Theorem 4.4.21 is that \(B\) is singular if \(B\vec x=\vec 0\) for some \(\vec x\not=\vec 0\text{.}\) The matrix \(A-\lambda I\) plays the role of \(B\text{.}\)

Corollary 6.2.3.

\(\lambda\) is an eigenvalue of \(A\) if and only if \(\det (A-\lambda I)=0\)

Proof.

\(B\) is singular if and only if \(\det B=0\text{.}\)

We now can see how these corollaries allow us to compute eigenvalues. We look at the matrix from Example 6.1.3.

Example 6.2.4.

We use

\begin{equation*} A-\lambda I= \begin{bmatrix} 5-\lambda\amp -1\amp -2\\ 1\amp 3-\lambda\amp -2\\ -1\amp -1\amp 4-\lambda \end{bmatrix} \end{equation*}

and

\begin{equation*} \det(A-\lambda I)=0. \end{equation*}

Evaluating the determinant gives

\begin{align*} \det\amp \begin{bmatrix} 5-\lambda\amp -1\amp -2\\ 1\amp 3-\lambda\amp -2\\ -1\amp -1\amp 4-\lambda \end{bmatrix}\\ \amp=(5-\lambda)(3-\lambda)(4-\lambda)-2+2 -2(5-\lambda) -2(3-\lambda)+(4-\lambda)\\ \amp = -\lambda^3 +12\lambda^2 -44\lambda +48\\ \amp = -(\lambda-2)(\lambda-4)(\lambda-6) \end{align*}

So we see that \(\det(A-\lambda I)=0\) only if \(\lambda=2\text{,}\) \(\lambda=4\) or \(\lambda=6\text{.}\) The evaluation of \(\det(A-\lambda I)=0\) gave us a polynomial \(p(\lambda)=-(\lambda-2)(\lambda-4)(\lambda-6)\) whose roots are the eigenvalues.

Definition 6.2.5. The characteristic polynomial.

The characteristic polynomial of a square matrix \(A\) is the polynomial

\begin{equation*} p(\lambda)=\det(\lambda I-A). \end{equation*}

Note 6.2.6.

Some sources define the characteristic polynomial as \(\det(A-\lambda I)\text{.}\) Since \(\lambda I-A=-(A-\lambda I)\text{,}\) it follows that for any square matrix \(A\) of size \(n\text{,}\)

\begin{equation*} \det(\lambda I-A)=(-1)^n\det(A-\lambda I) \end{equation*}

and so both polynomials have the same roots.

We look at the matrix from Example 6.1.4.

Example 6.2.7.

\(A=\begin{bmatrix}2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5 \end{bmatrix}\text{.}\)

\begin{equation*} \begin{array}{rl} \det(A-\lambda I) \amp =\det\begin{bmatrix}2-\lambda\amp 1\amp 4\\ 0\amp 3-\lambda\amp 0\\ 2\amp -2\amp -5-\lambda \end{bmatrix}\\ \amp =(2-\lambda)(3-\lambda)(-5-\lambda)-8(3-\lambda)\\ \amp = (\lambda+6)(\lambda-3)^2 \end{array} \end{equation*}

and so the eigenvalues are \(\lambda=-6\) and \(\lambda=3\)