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Section 3.9 Eigenspaces

Given \(\mtx{A} = \left[ \begin{array}{rr} 3 \amp 0 \\ 8 \amp -1 \end{array} \right]\) and \(\vect{v} = \op{v_1}{v_2}\text{.}\) Compute \(\mtx{A} \vect{v} - 3\vect{v}\) and \((\mtx{A} - 3 \mtx{I})\vect{v}\text{.}\) What do you notice? What does it mean when \((\mtx{A} - 3 \mtx{I})\vect{v} = \vect{0}\text{?}\) Find at least two vectors \(\vect{v}\) such that \(\mtx{A}\vect{v} = 3\vect{v}\text{.}\)

Definition 3.9.2.

Given a square matrix \(\mtx{A}\text{,}\) when there exists a non-zero vector \(\vect{v}\) such that \(\mtx{A}\vect{v} = \lambda \vect{v}\) for some real number \(\lambda\) that vector is called an eigenvector for \(\mtx{A}\text{,}\) and \(\lambda\) its corresponding eigenvalue.

Example. The effect of multiplication by a matrix on a representative set of vectors is illustrated below. Shown is a set of vectors and then those same vectors with the result of multiplying each of them by \(\mtx{A} = \left[ \begin{array}{rr} 3 \amp 0 \\ 8 \amp -1 \end{array} \right]\text{.}\)

Set of vectors.
Vectors moved by A.
Figure 3.9.4.
The vectors \(\op{0}{1}\) and \(\op{1}{2}\) are eigenvectors. Any scalar multiple of these vectors is also an eigenvector. The eigenvalue corresponding to \(\op{0}{1}\) is \(\lambda= -1\) and the eigenvalue corresponding to \(\op{1}{2}\) is \(\lambda = 3\text{.}\)

Given \(\mtx{A} = \left[ \begin{array}{cc} 1 \amp 3 \\ 4 \amp 2 \end{array} \right]\text{,}\) find all values for \(\lambda\) such that the matrix equation

\begin{equation*} (\mtx{A} - \lambda \mtx{I})\vect{v} = \vect{0} \end{equation*}

has a solution other that other than \(\vect{v} = \vect{0}\text{.}\) These are the eigenvalues for \(\mtx{A}\text{.}\)

Given \(\mtx{A} = \left[ \begin{array}{rrr} 2 \amp 0 \amp 0 \\ 3 \amp -1 \amp 0 \\ 3 \amp 0 \amp -1 \end{array} \right]\) has eigenvalues \(\lambda_\vect{v} = 2\) and \(\lambda_\vect{w} = -1\text{.}\)

  1. Describe the set of vectors \(\vect{v}\) such that \(\mtx{A}\)\(\vect{v}\) = 2\(\vect{v}\)

  2. Describe the set of vectors \(\vect{w}\) such that \(\mtx{A}\)\(\vect{w}\) = -1\(\vect{w}\)

If an eigenvector could be zero what would the corresponding eigenvalue be?