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5.1 + 5.2 Eigenvectors, eigenvalues, and Characteristic equations

HW 21 + HW 22 due together

Let \( A = \begin{bmatrix} 5 & 0 \\ 2 & 1 \end{bmatrix} \). \( A\vec{x} \) is a transformation of \( \vec{x} \).

\[ A \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \end{bmatrix} \]

More than likely both magnitude and direction will change.

Figure: Graph showing vector x=(1,0) transformed to Ax=(5,2) on Cartesian axes.

However, some do not change direction:

\[ \begin{bmatrix} 5 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

Figure: Graph showing vector x=(0,1) and its transformation Ax=(0,1) overlapping on the vertical axis.

\[ \begin{bmatrix} 5 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} = 5 \begin{bmatrix} 2 \\ 1 \end{bmatrix} \]

Figure: Graph showing vector x=(2,1) and its transformation Ax=(10,5) as a longer vector in the same direction.

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These are called eigenvectors.

The scaling factors after transformation are called eigenvalues.

e.g. \( \begin{bmatrix} 2 \\ 1 \end{bmatrix} \) is an eigenvector of \( \begin{bmatrix} 5 & 0 \\ 2 & 1 \end{bmatrix} \) with the corresponding eigenvalue of 5.

This means if \( \vec{x} \) is an eigenvector of \( A \) w/ corresponding eigenvalue \( \lambda \), then:

\[ A\vec{x} = \lambda\vec{x} \]

Example

Is \( \vec{x} = \begin{bmatrix} -2 \\ 1 \end{bmatrix} \) an eigenvector of \( A = \begin{bmatrix} -1 & 4 \\ 3 & 3 \end{bmatrix} \)?

\[ A\vec{x} = \begin{bmatrix} -1 & 4 \\ 3 & 3 \end{bmatrix} \begin{bmatrix} -2 \\ 1 \end{bmatrix} = \begin{bmatrix} 6 \\ -3 \end{bmatrix} = -3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} \]

Result: Yes, with \( \lambda = -3 \) and \( \vec{x} = \begin{bmatrix} -2 \\ 1 \end{bmatrix} \).

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How to find eigenvector if we know eigenvalue?

Example

Given matrix:

\[ A = \begin{bmatrix} 7 & 4 \\ -3 & -1 \end{bmatrix} \quad \text{eigenvalues: } 1, 5 \]

Eigenvector for \( \lambda = 1 \):

\[ A\vec{x} = \lambda\vec{x} \]\[ A\vec{x} - \lambda\vec{x} = \vec{0} \]\[ (A - \lambda I)\vec{x} = \vec{0} \implies \text{homogeneous eq.} \]

Substituting \( \lambda = 1 \):

\[ \begin{bmatrix} 7-1 & 4 & 0 \\ -3 & -1-1 & 0 \end{bmatrix} \]\[ \begin{bmatrix} 6 & 4 & 0 \\ -3 & -2 & 0 \end{bmatrix} \sim \begin{bmatrix} 6 & 4 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]

eigenvector \( \neq \vec{0} \)

must have nontrivial solution

From the reduced matrix:

\( x_2 \) is free
\( x_1 = -\frac{2}{3}x_2 \)

\[ \vec{x} = \begin{bmatrix} -2/3 \\ 1 \end{bmatrix} x_2 \]

make this ANY convenient \( \neq 0 \)

Let \( x_2 = 3 \):

\[ \vec{x} = \begin{bmatrix} -2 \\ 3 \end{bmatrix}, \lambda = 1 \]

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\( \lambda = 5 \)

\[ (A - \lambda I)\vec{x} = \vec{0} \]\[ \begin{bmatrix} 2 & 4 & 0 \\ -3 & -6 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]

\( x_2 \) is free, \( x_1 = -2x_2 \)

\[ \vec{x} = x_2 \begin{bmatrix} -2 \\ 1 \end{bmatrix} \]

\[ \vec{x} = \begin{bmatrix} -2 \\ 1 \end{bmatrix}, \lambda = 5 \]

is a vector space with basis \( \begin{bmatrix} -2 \\ 1 \end{bmatrix} \)

called eigenspace

all multiples of eigenvector and the zero vector

How to find eigenvalues?

Special case: if \( A \) is triangular, then the main diagonal elements are eigenvalues.

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Eigenvalues and Matrix Rank

Why?

\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \quad A - \lambda I = \begin{bmatrix} 1 - \lambda & 2 & 3 \\ 0 & 4 - \lambda & 5 \\ 0 & 0 & 6 - \lambda \end{bmatrix} \]

If \( \lambda \) is an eigenvalue, then \( (A - \lambda I) \vec{x} = \vec{0} \) has nontrivial solutions (eigenvectors).

So \( \lambda \) must be equal to 1, 4, or 6 in above \( A \) (to not have rank of 3).

eigenvector CANNOT be zero vector
eigenvalue CAN be zero.

if \( \lambda = 0 \), then \( A \vec{x} = \lambda \vec{x} \) becomes \( A \vec{x} = \vec{0} \) and \( (A - \lambda I) \vec{x} = 0 \) has nontrivial solutions.

\( \Rightarrow A^{-1} \) does NOT exist (because \( \det A = 0 \), columns of \( A \) are not independent, etc.)

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also, all eigenvectors corresponding to distinct eigenvalues are linearly independent.

How to find eigenvalues in general case

\[ A \vec{x} = \lambda \vec{x} \] \[ (A - \lambda I) \vec{x} = \vec{0} \quad \text{must have nontrivial solutions} \] \[ \Rightarrow \boxed{\det(A - \lambda I) = 0} \]

\( \hookrightarrow \) solve for \( \lambda \)

example

\[ A = \begin{bmatrix} 5 & 3 \\ 3 & 5 \end{bmatrix} \quad \lambda = ? \] \[ A - \lambda I = \begin{bmatrix} 5 - \lambda & 3 \\ 3 & 5 - \lambda \end{bmatrix} \] \[ \det(A - \lambda I) = \begin{vmatrix} 5 - \lambda & 3 \\ 3 & 5 - \lambda \end{vmatrix} = (5 - \lambda)^2 - 9 = 0 \]

characteristic eq. (\( n \times n \) matrix \( \to n^{\text{th}} \)-degree polynomial)

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\[ (5 - \lambda) = 3 \quad \text{or} \quad (5 - \lambda) = -3 \]\[ \lambda = 2 \quad \text{or} \quad \lambda = 8 \]

Note:

\[ 2 \times 2 \ A \Rightarrow 2 \text{, possibly repeated or complex } \lambda \]

then find eigenvectors by following earlier example.

Example

\[ A = \begin{bmatrix} -8 & -1 \\ 1 & -6 \end{bmatrix} \]\[ \det(A - \lambda I) = 0 \]\[ \begin{vmatrix} -8 - \lambda & -1 \\ 1 & -6 - \lambda \end{vmatrix} = 0 \]\[ (-8 - \lambda)(-6 - \lambda) + 1 = 0 \]\[ \lambda^2 + 14\lambda + 49 = 0 \]\[ \lambda = 7, 7 \]

algebraic multiplicity

7 is repeated, shows up twice

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Example

\[ A = \begin{bmatrix} 1 & 0 & 1 \\ -3 & 2 & -2 \\ 0 & 7 & 0 \end{bmatrix} \]

row reduction changes eigenvalues

do NOT do ERO's before finding \(\lambda\)'s

\[ \det(A - \lambda I) = 0 \]\[ \begin{vmatrix} 1 - \lambda & 0 & 1 \\ -3 & 2 - \lambda & -2 \\ 0 & 7 & -\lambda \end{vmatrix} = 0 \]\[ (1 - \lambda) \begin{vmatrix} 2 - \lambda & -2 \\ 7 & -\lambda \end{vmatrix} + (1) \begin{vmatrix} -3 & 2 - \lambda \\ 0 & 7 \end{vmatrix} = 0 \]\[ (1 - \lambda) [(2 - \lambda)(-\lambda) + 14] + (-3)(7) = 0 \]\[ \dots \quad -\lambda^3 + 3\lambda^2 - 16\lambda - 7 = 0 \]