Furthermore, a square matrix M is invertible if and only if \det(M)\not=0\text{.}
Observation 5.3.3
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{.}
Figure1.Transformation of the unit square by the linear transformation A
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{.}
Figure2.The map A stretches out the eigenvector \left[\begin{array}{c}2 \\ 1 \end{array}\right] by a factor of 3 (the corresponding eigenvalue).
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 5.3.5 (~5 min)
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?
The kernel of the transformation with standard matrix A-\lambda I must contain the zero vector, so A-\lambda I is invertible.
The kernel of the transformation with standard matrix A-\lambda I must contain a non-zero vector, so A-\lambda I is not invertible.
The image of the transformation with standard matrix A-\lambda I must contain the zero vector, so A-\lambda I is invertible.
The image of the transformation with standard matrix A-\lambda I must contain a non-zero vector, so A-\lambda I is not invertible.
Fact 5.3.6
The eigenvalues \lambda for a matrix A are the values that make A-\lambda I non-invertible.
Thus the eigenvalues \lambda for a matrix A are the solutions to the equation
and its eigenvalues are the solutions to \lambda^2-5\lambda-2=0\text{.}
Activity 5.3.8 (~10 min)
Let A = \left[\begin{array}{cc} 5 & 2 \\ -3 & -2 \end{array}\right]\text{.}
Part 1.
Compute \det (A-\lambda I) to determine the characteristic polynomial of A\text{.}
Activity 5.3.8 (~10 min)
Let A = \left[\begin{array}{cc} 5 & 2 \\ -3 & -2 \end{array}\right]\text{.}
Part 2.
Set this characteristic polynomial equal to zero and factor to determine the eigenvalues of A\text{.}
Activity 5.3.9 (~5 min)
Find all the eigenvalues for the matrix A=\left[\begin{array}{cc} 3 & -3 \\ 2 & -4 \end{array}\right]\text{.}
Activity 5.3.10 (~5 min)
Find all the eigenvalues for the matrix A=\left[\begin{array}{cc} 1 & -4 \\ 0 & 5 \end{array}\right]\text{.}
Activity 5.3.11 (~10 min)
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{.}
Linear Algebra for Team-Based Inquiry Learning 2023 Early Edition Steven Clontz Drew Lewis University of South Alabama University of South Alabama December 22, 2022