UNIT 4 • LINEAR ALGEBRA & MATRIX METHODS

Power Method
Dominant Eigenvalue

The power method finds the largest eigenvalue of a matrix: multiply a vector by A again and again, rescaling each time so its biggest entry is 1. The scale factor settles on the eigenvalue $\lambda$ and the vector on its eigenvector.

ONE ITERATIONLARGEST |λ|

y^{(k)} = A\,x^{(k-1)}, \qquad \lambda^{(k)} = \text{largest entry of } y^{(k)}, \qquad x^{(k)} = \frac{y^{(k)}}{\lambda^{(k)}}

1. MultiplyWork out

y = Ax

2. ScaleTake the entry of y with the biggest size as λ, and divide y by it.

3. RepeatStop when λ stops changing; the vector is the eigenvector.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Find the dominant eigenvalue and eigenvector of $A = \begin{bmatrix} 1 & 6 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ , starting from $x^{(0)} = [1, 0, 0]^T$ .

STEP 01 • FIRST TWO ITERATIONS

A\,x^{(0)} = [1, 1, 0]^T \Rightarrow \lambda = 1, \quad x^{(1)} = [1, 1, 0]^T

A\,x^{(1)} = [7, 3, 0]^T \Rightarrow \lambda = 7, \quad x^{(2)} = [1, 0.4286, 0]^T

STEP 02 • KEEP GOING

A\,x^{(2)} = [3.5714, 1.8571, 0]^T \Rightarrow \lambda = 3.5714, \quad x^{(3)} = [1, 0.52, 0]^T

A\,x^{(3)} = [4.12, 2.04, 0]^T \Rightarrow \lambda = 4.12

The estimates swing around 4 and settle there after about ten iterations.

STEP 03 • CHECK

A\begin{bmatrix} 1 \\ 0.5 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 + 3 \\ 1 + 1 \\ 0 \end{bmatrix} = 4\begin{bmatrix} 1 \\ 0.5 \\ 0 \end{bmatrix} \ \checkmark

FINAL CONCLUSION

λ = 4, eigenvector [1, 0.5, 0]

\lambda_{\max} = 4, \qquad x = [1,\ 0.5,\ 0]^T

The calculator below opens on this matrix. Switch to "smallest" for the inverse power method.

LABORATORY

Interactive Power Method Calculator

LABORATORY

Power Method Engine

Size of A:

A =

Find

Starting vector x₀ (blank = all 1s)ToleranceMax iterations

Common Questions

What does the power method find?

The dominant eigenvalue (the one with the largest size) of a square matrix, and its eigenvector. Repeatedly multiplying any starting vector by A makes the dominant direction take over.

How do you find the smallest eigenvalue?

Use the inverse power method: run the power method on

A^{-1}

. Its largest eigenvalue is

1/\lambda_{\min}

of A, so take the reciprocal.

When does the power method fail or go slowly?

When two eigenvalues have the same size (like 3 and −3) it can't settle, and when the top two are close it converges slowly. A starting vector with no part in the dominant direction can also cause trouble; the default of all 1s usually works.

PreviousJacobi Method

Unit 4•30 of 63

NextTaylor's Series Method

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 4 • LINEAR ALGEBRA & MATRIX METHODS

Power Method
Dominant Eigenvalue

ONE ITERATIONLARGEST |λ|

y^{(k)} = A\,x^{(k-1)}, \qquad \lambda^{(k)} = \text{largest entry of } y^{(k)}, \qquad x^{(k)} = \frac{y^{(k)}}{\lambda^{(k)}}

1. MultiplyWork out

y = Ax

2. ScaleTake the entry of y with the biggest size as λ, and divide y by it.

3. RepeatStop when λ stops changing; the vector is the eigenvector.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Find the dominant eigenvalue and eigenvector of $A = \begin{bmatrix} 1 & 6 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ , starting from $x^{(0)} = [1, 0, 0]^T$ .

STEP 01 • FIRST TWO ITERATIONS

A\,x^{(0)} = [1, 1, 0]^T \Rightarrow \lambda = 1, \quad x^{(1)} = [1, 1, 0]^T

A\,x^{(1)} = [7, 3, 0]^T \Rightarrow \lambda = 7, \quad x^{(2)} = [1, 0.4286, 0]^T

STEP 02 • KEEP GOING

A\,x^{(2)} = [3.5714, 1.8571, 0]^T \Rightarrow \lambda = 3.5714, \quad x^{(3)} = [1, 0.52, 0]^T

A\,x^{(3)} = [4.12, 2.04, 0]^T \Rightarrow \lambda = 4.12

The estimates swing around 4 and settle there after about ten iterations.

STEP 03 • CHECK

A\begin{bmatrix} 1 \\ 0.5 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 + 3 \\ 1 + 1 \\ 0 \end{bmatrix} = 4\begin{bmatrix} 1 \\ 0.5 \\ 0 \end{bmatrix} \ \checkmark

FINAL CONCLUSION

λ = 4, eigenvector [1, 0.5, 0]

\lambda_{\max} = 4, \qquad x = [1,\ 0.5,\ 0]^T

The calculator below opens on this matrix. Switch to "smallest" for the inverse power method.

LABORATORY

Interactive Power Method Calculator

LABORATORY

Power Method Engine

Size of A:

A =

Find

Starting vector x₀ (blank = all 1s)ToleranceMax iterations

Common Questions

What does the power method find?

The dominant eigenvalue (the one with the largest size) of a square matrix, and its eigenvector. Repeatedly multiplying any starting vector by A makes the dominant direction take over.

How do you find the smallest eigenvalue?

Use the inverse power method: run the power method on

A^{-1}

. Its largest eigenvalue is

1/\lambda_{\min}

of A, so take the reciprocal.

When does the power method fail or go slowly?

PreviousJacobi Method

Unit 4•30 of 63

NextTaylor's Series Method

Power Method Dominant Eigenvalue

Step-by-Step Worked Example

λ = 4, eigenvector [1, 0.5, 0]

Interactive Power Method Calculator

Power Method Engine

Common Questions

NETZ Engine

Power Method Dominant Eigenvalue

Step-by-Step Worked Example

λ = 4, eigenvector [1, 0.5, 0]

Interactive Power Method Calculator

Power Method Engine

Common Questions

Power Method
Dominant Eigenvalue

Power Method
Dominant Eigenvalue