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6.1 Eigenvalues and Eigenvectors

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

When multiply a vector by this matrix, the vector usually changes direction and magnitude.

For example,

\[ \vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad A\vec{x} = \begin{bmatrix} 5 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \end{bmatrix} \]
A 2D coordinate graph showing vector [1, 0] transformed to vector [5, 2] by matrix A.
\[ \vec{x} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \quad A\vec{x} = \begin{bmatrix} 5 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 3 \end{bmatrix} \]
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Most matrices have special vectors that preserve their directions after transformation.

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

direction preserved
same magnitude

\[ \vec{x} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \quad A\vec{x} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} = 5 \begin{bmatrix} 2 \\ 1 \end{bmatrix} \]

same direction
magnitude increased by factor of 5

These vectors that keep their directions are called the eigenvectors of the matrix.

The factors by which their magnitudes change are called the eigenvalues. So, in the above example,

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  • \[ \begin{bmatrix} 0 \\ 1 \end{bmatrix} \] is an eigenvector w/ the associated eigenvalue of 1
  • \[ \begin{bmatrix} 2 \\ 1 \end{bmatrix} \] is an eigenvector w/ the associated eigenvalue of 5

How to find them?

if \( \vec{v} \) is an eigenvector, then \( A\vec{v} \) keeps the same direction but changes magnitude by a factor of \( \lambda \) (lambda).

\[ A\vec{v} = \lambda \vec{v} \]\[ A\vec{v} - \lambda \vec{v} = \vec{0} \]
\[ (A - \lambda I)\vec{v} = \vec{0} \]

\( I \): identity matrix

this is a homogeneous eq, which always has solution \( \vec{v} = \vec{0} \) (trivial solution) but we don't want that. to force nontrivial solution, we force multiple solutions

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which implies \[ |A - \lambda I| = \det(A - \lambda I) = 0 \]

So, to find eigenvalues and eigenvectors, we solve

\[ \det(A - \lambda I) = 0 \quad \text{for } \lambda\text{'s} \]

then use those \( \lambda \)'s to solve for \( \vec{v} \)'s from

\[ (A - \lambda I)\vec{v} = \vec{0} \]
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Example: Finding Eigenvalues

\[ A = \begin{bmatrix} 7 & 4 \\ -3 & -1 \end{bmatrix} \]
\[ A - \lambda I = \begin{bmatrix} 7 & 4 \\ -3 & -1 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

First, find eigenvalues:

\[ \det(A - \lambda I) = \begin{vmatrix} 7 - \lambda & 4 \\ -3 & -1 - \lambda \end{vmatrix} = 0 \]

\[ (7 - \lambda)(-1 - \lambda) - (-3)(4) = 0 \]

\[ \lambda^2 - 6\lambda + 5 = 0 \]

characteristic eq.

This is an \( n^{th} \)-order polynomial for an \( n \times n \) matrix.

\[ (\lambda - 5)(\lambda - 1) = 0 \]

\[ \lambda = 1, \quad \lambda = 5 \]

eigenvalues of A

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Now find the associated eigenvectors:

Solve \( (A - \lambda I)\vec{v} = \vec{0} \) for each \( \lambda \).

\( \lambda = 1 \)

\[ (A - \lambda I)\vec{v} = \vec{0} \]\[ \begin{bmatrix} 6 & 4 \\ -3 & -2 \end{bmatrix} \vec{v} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]
\[ \begin{bmatrix} 6 & 4 & 0 \\ -3 & -2 & 0 \end{bmatrix} \rightarrow \dots \rightarrow \begin{bmatrix} 3 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]
There is ALWAYS at least one zero row.

\[ \vec{v} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \quad \begin{aligned} x_2 &= r \text{ (free)} \\ x_1 &\mapsto \text{row 1: } 3x_1 + 2x_2 = 0 \\ x_1 &= -\frac{2}{3}x_2 = -\frac{2}{3}r \end{aligned} \]

\[ \text{so, } \vec{v} = \begin{bmatrix} -\frac{2}{3}r \\ r \end{bmatrix} = r \begin{bmatrix} -\frac{2}{3} \\ 1 \end{bmatrix} \]

Choose any convenient \( r \neq 0 \).

Here, let's use \( r = -3 \).

\[ \text{so, } \vec{v} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}, \quad \lambda = 1 \]

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Next Pair

λ = 5

solve \( (A - \lambda I) \vec{v} = \vec{0} \)

\[ \begin{bmatrix} 2 & 4 & 0 \\ -3 & -6 & 0 \end{bmatrix} \rightarrow \dots \rightarrow \begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]
\[ \vec{v} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \]

\( x_2 = r \)

\( x_1 + 2x_2 = 0 \implies x_1 = -2x_2 = -2r \)

\[ \vec{v} = \begin{bmatrix} -2r \\ r \end{bmatrix} \]

choose \( r = 1 \)

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

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Example

\[ A = \begin{bmatrix} 4 & 0 & 1 \\ -2 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix} \]

solve \( \det(A - \lambda I) = 0 \)

\[ \begin{vmatrix} 4 - \lambda & 0 & 1 \\ -2 & 1 - \lambda & 0 \\ -2 & 0 & 1 - \lambda \end{vmatrix} = 0 \]

Cofactor expansion along column 2

\[ (1 - \lambda) \begin{vmatrix} 4 - \lambda & 1 \\ -2 & 1 - \lambda \end{vmatrix} = 0 \]

\[ (1 - \lambda) [ (4 - \lambda)(1 - \lambda) - (-2)(1) ] = 0 \]

\[ (1 - \lambda) (\lambda^2 - 5\lambda + 6) = 0 \]

\[ (1 - \lambda) (\lambda - 2) (\lambda - 3) = 0 \]

\( \lambda = 1, 2, 3 \)

\( 3 \times 3 \rightarrow 3 \text{ } \lambda \text{'s} \)

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Solving for Eigenvectors

Solve \((A - \lambda I)\vec{v} = \vec{0}\) for each \(\lambda\)

Case \(\lambda = 1\)

\((A - \lambda I)\vec{v} = \vec{0}\)

\[\begin{bmatrix} 3 & 0 & 1 & 0 \\ -2 & 0 & 0 & 0 \\ -2 & 0 & 0 & 0 \end{bmatrix}\]
\[\rightarrow \dots \rightarrow \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \quad \vec{v} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\]

\(x_2 = r\)

row 2: \(x_3 = 0\)

row 1: \(x_1 = 0\)

\(\vec{v} = \begin{bmatrix} 0 \\ r \\ 0 \end{bmatrix}\)

Choose \(r = 1\)

\(\vec{v} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \lambda = 1\)

this eigenvector spans an eigenspace of dimension 1

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\(\lambda = 2 \quad (A - \lambda I)\vec{v} = \vec{0} \dots \vec{v} = \begin{bmatrix} -1 \\ 2 \\ 2 \end{bmatrix}\)

\(\lambda = 3 \quad \dots \vec{v} = \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}\)

if \(A\) is triangular or diagonal, its eigenvalues are the main diagonal elements

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

triangular

\[\det(A - \lambda I) = 0 \implies \begin{vmatrix} 1 - \lambda & 2 & 3 \\ 0 & 4 - \lambda & 5 \\ 0 & 0 & 6 - \lambda \end{vmatrix} = 0\]
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\[ (1-\lambda) \begin{vmatrix} 4-\lambda & 5 \\ 0 & 6-\lambda \end{vmatrix} = 0 \]
\[ (1-\lambda)(4-\lambda)(6-\lambda) = 0 \rightarrow \lambda = 1, 4, 6 \]

(main diagonal elements)

Eigenvalues can be complex → come in conjugate pairs

\[ A = \begin{bmatrix} 0 & 1 \\ -4 & 0 \end{bmatrix} \quad \begin{vmatrix} 0-\lambda & 1 \\ -4 & 0-\lambda \end{vmatrix} = 0 \]
\[ \lambda^2 + 4 = 0 \]

\[ \lambda = 2i, -2i \]

Eigenvectors are also complex conjugate pairs

\[ \vec{v} = \begin{bmatrix} -1 \\ -2i \end{bmatrix}, \lambda = 2i \]
\[ \vec{v} = \begin{bmatrix} -1 \\ 2i \end{bmatrix}, \lambda = -2i \]