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Section5.1Eigenvalues and Eigenvectors

Objectives
  1. Learn the definition of eigenvector and eigenvalue.
  2. Learn to find eigenvectors and eigenvalues geometrically.
  3. Learn to decide if a number is an eigenvalue of a matrix, and if so, how to find an associated eigenvector.
  4. Recipe: find a basis for the λ -eigenspace.
  5. Pictures: whether or not a vector is an eigenvector, eigenvectors of standard matrix transformations.
  6. Theorem: the expanded invertible matrix theorem.
  7. Vocabulary word: eigenspace.
  8. Essential vocabulary words: eigenvector, eigenvalue.

In this section, we define eigenvalues and eigenvectors. These form the most important facet of the structure theory of square matrices. As such, eigenvalues and eigenvectors tend to play a key role in the real-life applications of linear algebra.

Subsection5.1.1Eigenvalues 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.

Figure8An eigenvector of A is a vector x such that Ax is collinear with x and the origin. Click and drag the head of x to convince yourself that all such vectors lie either on L , or on the line perpendicular to L .

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.

Proof

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.

Subsection5.1.2Eigenspaces

Suppose that A is a square matrix. We already know how to check if a given vector is an eigenvector of A and in that case to find the eigenvalue. Our next goal is to check if a given real number is an eigenvalue of A and in that case to find all of the corresponding eigenvectors. Again this will be straightforward, but more involved. The only missing piece, then, will be to find the eigenvalues of A ; this is the main content of Section 5.2.

Let A be an n × n matrix, and let λ be a scalar. The eigenvectors with eigenvalue λ , if any, are the nonzero solutions of the equation Av = λ v . We can rewrite this equation as follows:

Av = λ v ⇐⇒ Av − λ v = 0 ⇐⇒ Av − λ I n v = 0 ⇐⇒ ( A − λ I n ) v = 0.

Therefore, the eigenvectors of A with eigenvalue λ , if any, are the nontrivial solutions of the matrix equation ( A − λ I n ) v = 0, i.e., the nonzero vectors in Nul ( A − λ I n ) . If this equation has no nontrivial solutions, then λ is not an eigenvector of A .

The above observation is important because it says that finding the eigenvectors for a given eigenvalue means solving a homogeneous system of equations. For instance, if

A = C 713 − 32 − 3 − 3 − 2 − 1 D ,

then an eigenvector with eigenvalue λ is a nontrivial solution of the matrix equation

C 713 − 32 − 3 − 3 − 2 − 1 DC xyz D = λ C xyz D .

This translates to the system of equations

E 7 x + y + 3 z = λ x − 3 x + 2 y − 3 z = λ y − 3 x − 2 y − z = λ z −−−→ E ( 7 − λ ) x + y + 3 z = 0 − 3 x +( 2 − λ ) y − 3 z = 0 − 3 x − 2 y +( − 1 − λ ) z = 0.

This is the same as the homogeneous matrix equation

C 7 − λ 13 − 32 − λ − 3 − 3 − 2 − 1 − λ DC xyz D = 0,

i.e., ( A − λ I 3 ) v = 0.

Definition

Let A be an n × n matrix, and let λ be an eigenvalue of A . The λ -eigenspace of A is the solution set of ( A − λ I n ) v = 0, i.e., the subspace Nul ( A − λ I n ) .

The λ -eigenspace is a subspace because it is the null space of a matrix, namely, the matrix A − λ I n . This subspace consists of the zero vector and all eigenvectors of A with eigenvalue λ .

Note

Since a nonzero subspace is infinite, every eigenvalue has infinitely many eigenvectors. (For example, multiplying an eigenvector by a nonzero scalar gives another eigenvector.) On the other hand, there can be at most n linearly independent eigenvectors of an n × n matrix, since R n has dimension n .

Example(Computing eigenspaces)
Example(Computing eigenspaces)
Example(Reflection)
Recipes: Eigenspaces

Let A be an n × n matrix and let λ be a number.

  1. λ is an eigenvalue of A if and only if ( A − λ I n ) v = 0 has a nontrivial solution, if and only if Nul ( A − λ I n ) A = { 0 } .
  2. In this case, finding a basis for the λ -eigenspace of A means finding a basis for Nul ( A − λ I n ) , which can be done by finding the parametric vector form of the solutions of the homogeneous system of equations ( A − λ I n ) v = 0.
  3. The dimension of the λ -eigenspace of A is equal to the number of free variables in the system of equations ( A − λ I n ) v = 0, which is the number of columns of A − λ I n without pivots.
  4. The eigenvectors with eigenvalue λ are the nonzero vectors in Nul ( A − λ I n ) , or equivalently, the nontrivial solutions of ( A − λ I n ) v = 0.

We conclude with an observation about the 0 -eigenspace of a matrix.

Proof

We know that 0 is an eigenvalue of A if and only if Nul ( A − 0 I n )= Nul ( A ) is nonzero, which is equivalent to the noninvertibility of A by the invertible matrix theorem in Section 3.6. In this case, the 0 -eigenspace is by definition Nul ( A − 0 I n )= Nul ( A ) .

Concretely, an eigenvector with eigenvalue 0 is a nonzero vector v such that Av = 0 v , i.e., such that Av = 0. These are exactly the nonzero vectors in the null space of A .

Subsection5.1.3The Invertible Matrix Theorem: Addenda

We now have two new ways of saying that a matrix is invertible, so we add them to the invertible matrix theorem.