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Problem type

Eigenvalues and eigenvectors

What it solves
Find the eigenvalues and eigenvectors of a 2 by 2 or 3 by 3 matrix: build the characteristic polynomial, solve it, find the directions the matrix does not turn, and check every one by multiplying.
Subject
Linear Algebra
Topic
Eigenvalues and eigenvectors
Also called
eigenvalue, eigenvector, characteristic polynomial, diagonalise
Your problem

Find the eigenvalues and eigenvectors of this matrix.

A=[1−304]A = \left[ \begin{array}{rr} 1 & -3 \\ 0 & 4 \end{array} \right]
The square matrix
Size2
Whole numbers or fractions for 2 by 2; whole numbers for 3 by 3.
Invent a new one:
Femtobytes
Eigenvalues and eigenvectors of the 2 by 2 matrix 1 -3 / 0 4, step by step
femtobytes.com/solve/eigenvalues/1c-3r0c4
The problem

Find the eigenvalues and eigenvectors of this matrix.

A=[1−304]A = \left[ \begin{array}{rr} 1 & -3 \\ 0 & 4 \end{array} \right]
The result
λ1=1:  [10],λ2=4:  [1−1]\lambda_{1} = 1: \; \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad \lambda_{2} = 4: \; \begin{bmatrix} 1 \\ -1 \end{bmatrix}

Two different real eigenvalues, each with its own direction.

checked 22 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.
Picture 1 of 2Step 1: Say what we are looking for
The matrix whose own directions we are hunting
11−3-30044
A matrix turns most vectors off their own line. The question this page answers is which lines it does NOT turn: on those, the matrix only stretches or shrinks, and the amount it stretches by is the eigenvalue.
1

Say what we are looking for

Why this step

A matrix moves almost every vector to a new direction. The exceptions are rare and precious: directions the matrix does NOT turn, only stretches. Along such a direction the whole complicated matrix behaves like multiplication by a single number. Find those directions and the matrix stops being a table of numbers and becomes something you can see.

A=[1−304],A−λI=[1−λ−304−λ]A = \left[ \begin{array}{rr} 1 & -3 \\ 0 & 4 \end{array} \right], \qquad A - \lambda I = \begin{bmatrix} 1 - \lambda & -3 \\ 0 & 4 - \lambda \end{bmatrix}
Subtracting λI\lambda I takes λ\lambda off each diagonal entry and leaves the rest alone.
The matrix whose own directions we are hunting
11−3-30044
A matrix turns most vectors off their own line. The question this page answers is which lines it does NOT turn: on those, the matrix only stretches or shrinks, and the amount it stretches by is the eigenvalue.
2

Build the characteristic polynomial

Why this step

We expand det⁡(A−λI)\det\left( A - \lambda I \right) with λ\lambda left as a letter. The result is a polynomial in λ\lambda of degree 22, called the characteristic polynomial, and its roots are the eigenvalues. Two of its coefficients can be predicted before any work: the sum of the diagonal of AA (the trace) and det⁡A\det A. They make a free check on the algebra.

det⁡(A−λI)\displaystyle \det\left( A - \lambda I \right)==(1−λ)(4−λ)\displaystyle \left( 1 - \lambda \right) \left( 4 - \lambda \right)
The 2×22 \times 2 rule: main diagonal product minus the other diagonal product. checked
==4−λ−4λ+λ2\displaystyle 4 - \lambda - 4 \lambda + \lambda^{2}
Multiply out the two brackets, four products in all, and work out −3⋅(0)-3 \cdot \left( 0 \right). checked
==λ2−5λ+4\displaystyle \lambda^{2} - 5 \lambda + 4
Collect the plain numbers and the λ\lambda terms, highest power first. checked
The matrix, before lambda is subtracted from its diagonal
11−3-30044
Subtracting lambda down the diagonal is what turns the question into one about a determinant: we want the values of lambda that flatten this matrix, because only a flattened matrix can send a non-zero vector to zero.
3

Solve the characteristic equation

4

Eigenvectors for lambda = 1: set up

5

lambda = 1, pivot 1: column 2, row 1

6

Eigenvectors for lambda = 1: read off and check

7

Eigenvectors for lambda = 4: set up

8

lambda = 4, pivot 1: column 1, row 1

9

Eigenvectors for lambda = 4: read off and check