Eigenvalues And Eigenvectors

Here is the most important definition in this text.

Definition

Let A be an n × n matrix.

1. An eigenvector of A is a nonzero vector v in R n such that Av = λ v , for some scalar λ .
2. An eigenvalue of A is a scalar λ such that the equation Av = λ v has a nontrivial solution.

If Av = λ v for v A = 0, we say that λ is the eigenvalue for v , and that v is an eigenvector for λ .

The German prefix “eigen” roughly translates to “self” or “own”. An eigenvector of A is a vector that is taken to a multiple of itself by the matrix transformation T ( x )= Ax , which perhaps explains the terminology. On the other hand, “eigen” is often translated as “characteristic”; we may think of an eigenvector as describing an intrinsic, or characteristic, property of A .

Note

Eigenvalues and eigenvectors are only for square matrices.

Eigenvectors are by definition nonzero. Eigenvalues may be equal to zero.

We do not consider the zero vector to be an eigenvector: since A 0 = 0 = λ 0 for every scalar λ , the associated eigenvalue would be undefined.

If someone hands you a matrix A and a vector v , it is easy to check if v is an eigenvector of A : simply multiply v by A and see if Av is a scalar multiple of v . On the other hand, given just the matrix A , it is not obvious at all how to find the eigenvectors. We will learn how to do this in Section 5.2.

Example(Verifying eigenvectors)

Example(Verifying eigenvectors)

Example(An eigenvector with eigenvalue 0 )

To say that Av = λ v means that Av and λ v are collinear with the origin. So, an eigenvector of A is a nonzero vector v such that Av and v lie on the same line through the origin. In this case, Av is a scalar multiple of v ; the eigenvalue is the scaling factor.

v Av w Aw 0 v isaneigenvector w isnotaneigenvector

For matrices that arise as the standard matrix of a linear transformation, it is often best to draw a picture, then find the eigenvectors and eigenvalues geometrically by studying which vectors are not moved off of their line. For a transformation that is defined geometrically, it is not necessary even to compute its matrix to find the eigenvectors and eigenvalues.

Example(Reflection)

Here is an example of this. Let T : R 2 → R 2 be the linear transformation that reflects over the line L defined by y = − x , and let A be the matrix for T . We will find the eigenvalues and eigenvectors of A without doing any computations.

This transformation is defined geometrically, so we draw a picture.

L u Au 0

The vector u is not an eigenvector, because Au is not collinear with u and the origin.

L z Az 0

The vector z is not an eigenvector either.

L v Av 0

The vector v is an eigenvector because Av is collinear with v and the origin. The vector Av has the same length as v , but the opposite direction, so the associated eigenvalue is − 1.

L w Aw 0

The vector w is an eigenvector because Aw is collinear with w and the origin: indeed, Aw is equal to w ! This means that w is an eigenvector with eigenvalue 1.

It appears that all eigenvectors lie either on L , or on the line perpendicular to L . The vectors on L have eigenvalue 1, and the vectors perpendicular to L have eigenvalue − 1.

 

We will now give five more examples of this nature

Example(Projection)

Example(Identity)

Example(Dilation)

Example(Shear)

Example(Rotation)

Here we mention one basic fact about eigenvectors.

Fact(Eigenvectors with distinct eigenvalues are linearly independent)

Let v 1 , v 2 ,..., v k be eigenvectors of a matrix A , and suppose that the corresponding eigenvalues λ 1 , λ 2 ,..., λ k are distinct (all different from each other). Then { v 1 , v 2 ,..., v k } is linearly independent.

Proof

Suppose that { v 1 , v 2 ,..., v k } were linearly dependent. According to the increasing span criterion in Section 2.5, this means that for some j , the vector v j is in Span { v 1 , v 2 ,..., v j − 1 } . If we choose the first such j , then { v 1 , v 2 ,..., v j − 1 } is linearly independent. Note that j > 1 since v 1 A = 0.

Since v j is in Span { v 1 , v 2 ,..., v j − 1 } ,, we can write

v j = c 1 v 1 + c 2 v 2 + ··· + c j − 1 v j − 1

for some scalars c 1 , c 2 ,..., c j − 1 . Multiplying both sides of the above equation by A gives

λ j v j = Av j = A A c 1 v 1 + c 2 v 2 + ··· + c j − 1 v j − 1 B = c 1 Av 1 + c 2 Av 2 + ··· + c j − 1 Av j − 1 = c 1 λ 1 v 1 + c 2 λ 2 v 2 + ··· + c j − 1 λ j − 1 v j − 1 .

Subtracting λ j times the first equation from the second gives

0 = λ j v j − λ j v j = c 1 ( λ 1 − λ j ) v 1 + c 2 ( λ 2 − λ j ) v 2 + ··· + c j − 1 ( λ j − 1 − λ j ) v j − 1 .

Since λ i A = λ j for i < j , this is an equation of linear dependence among v 1 , v 2 ,..., v j − 1 , which is impossible because those vectors are linearly independent. Therefore, { v 1 , v 2 ,..., v k } must have been linearly independent after all.

When k = 2, this says that if v 1 , v 2 are eigenvectors with eigenvalues λ 1 A = λ 2 , then v 2 is not a multiple of v 1 . In fact, any nonzero multiple cv 1 of v 1 is also an eigenvector with eigenvalue λ 1 :

A ( cv 1 )= cAv 1 = c ( λ 1 v 1 )= λ 1 ( cv 1 ) .

As a consequence of the above fact, we have the following.

An n × n matrix A has at most n eigenvalues.