Eigenvalues and eigenvectors of the 2 by 2 matrix 4 1 / 2 3, step by step

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 = \left[ \begin{array}{rr} 4 & 1 \\ 2 & 3 \end{array} \right], \qquad A - \lambda I = \begin{bmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{bmatrix}

Subtracting

\lambda I

takes

\lambda

off each diagonal entry and leaves the rest alone.

The matrix whose own directions we are hunting

4

1

2

3

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 $\det\left( A - \lambda I \right)$ with $\lambda$ left as a letter. The result is a polynomial in $\lambda$ of degree $2$ , 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 $A$ (the trace) and $\det A$ . They make a free check on the algebra.

\displaystyle \det\left( A - \lambda I \right)

=

\displaystyle \left( 4 - \lambda \right) \left( 3 - \lambda \right) - 1 \cdot 2

The

2 \times 2

rule: main diagonal product minus the other diagonal product. checked

=

\displaystyle 12 - 4 \lambda - 3 \lambda + \lambda^{2} - 2

Multiply out the two brackets, four products in all, and work out

1 \cdot 2

. checked

=

\displaystyle \lambda^{2} - 7 \lambda + 10

Collect the plain numbers and the

\lambda

terms, highest power first. checked

The matrix, before lambda is subtracted from its diagonal

4

1

2

3

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

Why this step

The eigenvalues are the roots of $\lambda^{2} - 7 \lambda + 10 = 0$ . It is a quadratic, so the discriminant decides everything in advance: positive means two different real eigenvalues, zero means one repeated eigenvalue, and negative means no real eigenvalue at all, which has a vivid meaning we will come to.

\displaystyle D

=

\displaystyle 7^{2} - 4 \cdot 10

The discriminant of

\lambda^2 - (\text{trace}) \lambda + \det A

is

(\text{trace})^2 - 4 \det A

. checked

=

\displaystyle 9

Work it out. checked

$D = 9$ is a perfect square, so the quadratic factorises. We need two numbers that add to the trace $7$ and multiply to $\det A = 10$ : they are $5$ and $2$ .

\lambda^{2} - 7 \lambda + 10 = \left( \lambda - 5 \right) \left( \lambda - 2 \right)

Check by multiplying the brackets back out. checked

The matrix whose characteristic equation this is

4

1

2

3

The roots of the characteristic equation are the values of lambda that flatten this matrix. Each one is an eigenvalue, and each brings its own directions that the matrix only stretches.

4

Eigenvectors for lambda = 2: set up

Why this step

With $\lambda = 2$ fixed, the eigenvectors are the non-zero solutions of $\left( A - \lambda I \right) v = 0$ . That is an ordinary system of linear equations, and we solve it the ordinary way, by row reduction. We already know it MUST have a non-zero solution, because this $\lambda$ was chosen to make the determinant zero. If row reduction produced a pivot in every column, that would prove we had the wrong eigenvalue.

A - \left( 2 \right) I = \left[ \begin{array}{rr} 2 & 1 \\ 2 & 1 \end{array} \right]

Take

2

off each diagonal entry of

A

. The right-hand side of the system is all zeros and stays all zeros under every row operation, so we do not write it.

The matrix this eigenvalue belongs to

4

1

2

3

Putting this value of lambda into A minus lambda times the identity flattens the matrix, and the directions it flattens to nothing are exactly the eigenvectors for that lambda.

5

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

Why this step

We work one column at a time, left to right. The goal for a column is a single $1$ (the pivot) with zeros everywhere else in it. A column in that state says something very plain: its variable appears in exactly one equation, with nothing in front of it. Each cleared column is one variable pinned down.

Multiply row 1 by $\frac{1}{2}$ , which turns the pivot $2$ into $1$ . Multiplying an equation through by a number that is not zero keeps it true for exactly the same values.

column	working	new entry
$1$	$\frac{1}{2} \cdot 2$	$1$
$2$	$\frac{1}{2} \cdot 1$	$\frac{1}{2}$

Row 1, entry by entry.

\left[ \begin{array}{rr} \boxed{1} & \frac{1}{2} \\ 2 & 1 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Replace row 2 by $R_{2} - 2 R_{1}$ . The multiplier is chosen so that the entry under attack, $2$ , is cancelled exactly by the pivot: $2 - 2 \cdot 1 = 0$ .

column	working	new entry
$1$	$2 - 2 \cdot 1$	$0$
$2$	$1 - 2 \cdot \frac{1}{2}$	$0$

Row 2, entry by entry: old entry minus the multiple of row 1.

\left[ \begin{array}{rr} \boxed{1} & \frac{1}{2} \\ 0 & 0 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Column 1: the pivot, and what it has to clear

2

1

2

1

the pivot: the entry being used the entries it is driving to zero the row the pivot lives in

The matrix before this column is dealt with. The boxed entry is the pivot, the entry this step works with; the marked entries below it are the ones being driven to zero. Every other number is carried along unchanged by the moves.

6

Eigenvectors for lambda = 2: read off and check

Why this step

Each column without a pivot is a free variable, and each free variable gives one independent eigenvector: set it to $1$ , set any other free variables to $0$ , and read the pivot variables off the rows. An eigenvector is only a direction, so any non-zero multiple is equally good; we scale to whole numbers.

\left[ \begin{array}{rr} 1 & \frac{1}{2} \\ 0 & 0 \end{array} \right]

The reduced matrix. It has 1 pivot and 1 free column. checked

v = \begin{bmatrix} 1 \\ -2 \end{bmatrix}

An eigenvector for

\lambda = 2

. Now the check,

A v = \lambda v

, one row at a time:

4 \cdot 1 + 1 \cdot \left( -2 \right) = 2 \cdot 1

Row 1 of

A

times

v

, against

\lambda

times entry 1 of

v

. Both sides are

2

. checked

2 \cdot 1 + 3 \cdot \left( -2 \right) = 2 \cdot \left( -2 \right)

Row 2 of

A

times

v

, against

\lambda

times entry 2 of

v

. Both sides are

-4

. checked

Back to A, to check the eigenvector

4

1

2

3

The original matrix. An eigenvector is only an eigenvector if A really does send it back along its own line, so the check multiplies it by A and compares with lambda times the vector.

7

Eigenvectors for lambda = 5: set up

Why this step

With $\lambda = 5$ fixed, the eigenvectors are the non-zero solutions of $\left( A - \lambda I \right) v = 0$ . That is an ordinary system of linear equations, and we solve it the ordinary way, by row reduction. We already know it MUST have a non-zero solution, because this $\lambda$ was chosen to make the determinant zero. If row reduction produced a pivot in every column, that would prove we had the wrong eigenvalue.

A - \left( 5 \right) I = \left[ \begin{array}{rr} -1 & 1 \\ 2 & -2 \end{array} \right]

Take

5

off each diagonal entry of

A

. The right-hand side of the system is all zeros and stays all zeros under every row operation, so we do not write it.

The matrix this eigenvalue belongs to

4

1

2

3

Putting this value of lambda into A minus lambda times the identity flattens the matrix, and the directions it flattens to nothing are exactly the eigenvectors for that lambda.

8

lambda = 5, pivot 1: column 1, row 1

Why this step

We work one column at a time, left to right. The goal for a column is a single $1$ (the pivot) with zeros everywhere else in it. A column in that state says something very plain: its variable appears in exactly one equation, with nothing in front of it. Each cleared column is one variable pinned down.

Multiply row 1 by $-1$ , which turns the pivot $-1$ into $1$ . Multiplying an equation through by a number that is not zero keeps it true for exactly the same values.

column	working	new entry
$1$	$-1 \cdot \left( -1 \right)$	$1$
$2$	$-1 \cdot 1$	$-1$

Row 1, entry by entry.

\left[ \begin{array}{rr} \boxed{1} & -1 \\ 2 & -2 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Replace row 2 by $R_{2} - 2 R_{1}$ . The multiplier is chosen so that the entry under attack, $2$ , is cancelled exactly by the pivot: $2 - 2 \cdot 1 = 0$ .

column	working	new entry
$1$	$2 - 2 \cdot 1$	$0$
$2$	$-2 - 2 \cdot \left( -1 \right)$	$0$

Row 2, entry by entry: old entry minus the multiple of row 1.

\left[ \begin{array}{rr} \boxed{1} & -1 \\ 0 & 0 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Column 1: the pivot, and what it has to clear

-1

1

2

-2

the pivot: the entry being used the entries it is driving to zero the row the pivot lives in

The matrix before this column is dealt with. The boxed entry is the pivot, the entry this step works with; the marked entries below it are the ones being driven to zero. Every other number is carried along unchanged by the moves.

9

Eigenvectors for lambda = 5: read off and check

Why this step

Each column without a pivot is a free variable, and each free variable gives one independent eigenvector: set it to $1$ , set any other free variables to $0$ , and read the pivot variables off the rows. An eigenvector is only a direction, so any non-zero multiple is equally good; we scale to whole numbers.

\left[ \begin{array}{rr} 1 & -1 \\ 0 & 0 \end{array} \right]

The reduced matrix. It has 1 pivot and 1 free column. checked

v = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

An eigenvector for

\lambda = 5

. Now the check,

A v = \lambda v

, one row at a time:

4 \cdot 1 + 1 \cdot 1 = 5 \cdot 1

Row 1 of

A

times

v

, against

\lambda

times entry 1 of

v

. Both sides are

5

. checked

2 \cdot 1 + 3 \cdot 1 = 5 \cdot 1

Row 2 of

A

times

v

, against

\lambda

times entry 2 of

v

. Both sides are

5

. checked

Back to A, to check the eigenvector

4

1

2

3

The original matrix. An eigenvector is only an eigenvector if A really does send it back along its own line, so the check multiplies it by A and compares with lambda times the vector.

Each dashed line is an eigen-direction. The solid arrow on it is an eigenvector v and the faint arrow is A v: it lies on the SAME line, only longer, shorter or reversed. The grey pair shows an ordinary vector u and A u, which has been turned off its line, as almost every vector is.