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Section 5.3 Eigenvalues and Characteristic Polynomials (G3)

Activity 5.3.1.

An invertible matrix \(M\) and its inverse \(M^{-1}\) are given below:

\begin{equation*} M=\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \hspace{2em} M^{-1}=\left[\begin{array}{cc}-2&1\\3/2&-1/2\end{array}\right] \end{equation*}

Which of the following is equal to \(\det(M)\det(M^{-1})\text{?}\)

  1. \(\displaystyle -1\)

  2. \(\displaystyle 0\)

  3. \(\displaystyle 1\)

  4. \(\displaystyle 4\)

Observation 5.3.2.

Consider the linear transformation \(A : \IR^2 \rightarrow \IR^2\) given by the matrix \(A = \left[\begin{array}{cc} 2 & 2 \\ 0 & 3 \end{array}\right]\text{.}\)

\begin{tikzpicture}[scale=0.75]
\fill[red!50] (0,0) rectangle (1,1);
\draw[thin,gray,<->] (-1,0)-- (4,0);
\draw[thin,gray,<->] (0,-1)-- (0,4);
\draw[thick,blue,->] (0,0) -- node[below right] {\(A \vec{e}_1\)}++ (2,0);
\draw[thick,red,->] (0,0) -- node[below] {\(\vec{e}_1\)}++ (1,0);
\draw[thick,blue,->] (0,0) -- node[above left] {\(A \vec{e}_2\)}++(2,3);
\draw[thick,red,->] (0,0) -- node[left] {\(\vec{e}_2\)}++ (0,1);
\draw[blue,dashed] (2,0) -- (4,3) -- (2,3);
\draw[purple!50!blue,thick,->] (0,0) -- (6,3);
\draw[purple!50!red,thick,->] (0,0) -- (2,1);
\end{tikzpicture}

It is easy to see geometrically that

\begin{equation*} A\left[\begin{array}{c}1 \\ 0 \end{array}\right] = \left[\begin{array}{cc} 2 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{c}1 \\ 0 \end{array}\right]= \left[\begin{array}{c}2 \\ 0 \end{array}\right]= 2 \left[\begin{array}{c}1 \\ 0 \end{array}\right]\text{.} \end{equation*}

It is less obvious (but easily checked once you find it) that

\begin{equation*} A\left[\begin{array}{c} 2 \\ 1 \end{array}\right] = \left[\begin{array}{cc} 2 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{c}2 \\ 1 \end{array}\right]= \left[\begin{array}{c} 6 \\ 3 \end{array}\right] = 3\left[\begin{array}{c} 2 \\ 1 \end{array}\right]\text{.} \end{equation*}
Definition 5.3.3.

Let \(A \in M_{n,n}\text{.}\) An eigenvector for \(A\) is a vector \(\vec{x} \in \IR^n\) such that \(A\vec{x}\) is parallel to \(\vec{x}\text{.}\)

\begin{tikzpicture}[scale=0.75]
\fill[gray!50] (0,0) rectangle (1,1);
\draw[thin,gray,<->] (-1,0)-- (4,0);
\draw[thin,gray,<->] (0,-1)-- (0,4);
\draw[thick,blue,->] (0,0) -- node[below right] {\(A \vec{e}_1=2\vec e_1\)}++ (2,0);
\draw[thick,red,->] (0,0) -- node[below] {\(\vec{e}_1\)}++ (1,0);
\draw[thick,gray,->] (0,0) -- node[above left] {\(A \vec{e}_2\)}++(2,3);
\draw[thick,gray,->] (0,0) -- node[left] {\(\vec{e}_2\)}++ (0,1);
\draw[gray,dashed] (2,0) -- (4,3) -- (2,3);
\draw[purple!50!blue,thick,->] (0,0) -- (6,3) 
  node [below right] {\(
   A\left[\begin{array}{c}2\\1\end{array}\right]
     =
   3\left[\begin{array}{c}2\\1\end{array}\right]
  \)};
\draw[purple!50!red,thick,->] (0,0) -- (2,1)
  node [above] {\(\left[\begin{array}{c}2\\1\end{array}\right]\)};
\end{tikzpicture}

In other words, \(A\vec{x}=\lambda \vec{x}\) for some scalar \(\lambda\text{.}\) If \(\vec x\not=\vec 0\text{,}\) then we say \(\vec x\) is a nontrivial eigenvector and we call this \(\lambda\) an eigenvalue of \(A\text{.}\)

Activity 5.3.2.

Finding the eigenvalues \(\lambda\) that satisfy

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

for some nontrivial eigenvector \(\vec x\) is equivalent to finding nonzero solutions for the matrix equation

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

Which of the following must be true for any eigenvalue?

  1. The kernel of the transformation with standard matrix \(A-\lambda I\) must contain the zero vector, so \(A-\lambda I\) is invertible.

  2. The kernel of the transformation with standard matrix \(A-\lambda I\) must contain a non-zero vector, so \(A-\lambda I\) is not invertible.

  3. The image of the transformation with standard matrix \(A-\lambda I\) must contain the zero vector, so \(A-\lambda I\) is invertible.

  4. The image of the transformation with standard matrix \(A-\lambda I\) must contain a non-zero vector, so \(A-\lambda I\) is not invertible.

Definition 5.3.5.

The expression \(\det(A-\lambda I)\) is called characteristic polynomial of \(A\text{.}\)

For example, when \(A=\left[\begin{array}{cc}1 & 2 \\ 3 & 4\end{array}\right]\text{,}\) we have

\begin{equation*} A-\lambda I= \left[\begin{array}{cc}1 & 2 \\ 3 & 4\end{array}\right]- \left[\begin{array}{cc}\lambda & 0 \\ 0 & \lambda\end{array}\right]= \left[\begin{array}{cc}1-\lambda & 2 \\ 3 & 4-\lambda\end{array}\right]\text{.} \end{equation*}

Thus the characteristic polynomial of \(A\) is

\begin{equation*} \det\left[\begin{array}{cc}1-\lambda & 2 \\ 3 & 4-\lambda\end{array}\right] = (1-\lambda)(4-\lambda)-(2)(3) = \lambda^2-5\lambda-2 \end{equation*}

and its eigenvalues are the solutions to \(\lambda^2-5\lambda-2=0\text{.}\)

Activity 5.3.3.

Let \(A = \left[\begin{array}{cc} 5 & 2 \\ -3 & -2 \end{array}\right]\text{.}\)

(a)

Compute \(\det (A-\lambda I)\) to determine the characteristic polynomial of \(A\text{.}\)

(b)

Set this characteristic polynomial equal to zero and factor to determine the eigenvalues of \(A\text{.}\)

Activity 5.3.4.

Find all the eigenvalues for the matrix \(A=\left[\begin{array}{cc} 3 & -3 \\ 2 & -4 \end{array}\right]\text{.}\)

Activity 5.3.5.

Find all the eigenvalues for the matrix \(A=\left[\begin{array}{cc} 1 & -4 \\ 0 & 5 \end{array}\right]\text{.}\)

Activity 5.3.6.

Find all the eigenvalues for the matrix \(A=\left[\begin{array}{ccc} 3 & -3 & 1 \\ 0 & -4 & 2 \\ 0 & 0 & 7 \end{array}\right]\text{.}\)

Exercises 5.3.1 Exercises

1.

Explain how to find the eigenvalues of the matrix \(\left[\begin{array}{cc} -2 & 1 \\ 18 & 1 \end{array}\right] \text{.}\)

Answer

The characteristic polynomial of \(\left[\begin{array}{cc} -2 & 1 \\ 18 & 1 \end{array}\right] \) is \(\lambda^{2} + \lambda - 20 \text{.}\)

The eigenvalues of \(\left[\begin{array}{cc} -2 & 1 \\ 18 & 1 \end{array}\right] \) are \(-5 \) and \(4 \text{.}\)

2.

Explain how to find the eigenvalues of the matrix \(\left[\begin{array}{cc} 3 & 2 \\ 2 & 3 \end{array}\right] \text{.}\)

Answer

The characteristic polynomial of \(\left[\begin{array}{cc} 3 & 2 \\ 2 & 3 \end{array}\right] \) is \(\lambda^{2} - 6 \lambda + 5 \text{.}\)

The eigenvalues of \(\left[\begin{array}{cc} 3 & 2 \\ 2 & 3 \end{array}\right] \) are \(1 \) and \(5 \text{.}\)

3.

Explain how to find the eigenvalues of the matrix \(\left[\begin{array}{cc} 5 & 2 \\ -10 & -7 \end{array}\right] \text{.}\)

Answer

The characteristic polynomial of \(\left[\begin{array}{cc} 5 & 2 \\ -10 & -7 \end{array}\right] \) is \(\lambda^{2} + 2 \lambda - 15 \text{.}\)

The eigenvalues of \(\left[\begin{array}{cc} 5 & 2 \\ -10 & -7 \end{array}\right] \) are \(3 \) and \(-5 \text{.}\)

4.

Explain how to find the eigenvalues of the matrix \(\left[\begin{array}{cc} 8 & 2 \\ -15 & -3 \end{array}\right] \text{.}\)

Answer

The characteristic polynomial of \(\left[\begin{array}{cc} 8 & 2 \\ -15 & -3 \end{array}\right] \) is \(\lambda^{2} - 5 \lambda + 6 \text{.}\)

The eigenvalues of \(\left[\begin{array}{cc} 8 & 2 \\ -15 & -3 \end{array}\right] \) are \(2 \) and \(3 \text{.}\)

5.

Explain how to find the eigenvalues of the matrix \(\left[\begin{array}{cc} 7 & 1 \\ -18 & -4 \end{array}\right] \text{.}\)

Answer

The characteristic polynomial of \(\left[\begin{array}{cc} 7 & 1 \\ -18 & -4 \end{array}\right] \) is \(\lambda^{2} - 3 \lambda - 10 \text{.}\)

The eigenvalues of \(\left[\begin{array}{cc} 7 & 1 \\ -18 & -4 \end{array}\right] \) are \(5 \) and \(-2 \text{.}\)

Additional exercises available at checkit.clontz.org.