Understanding Eigenvalues, Eigenvectors, and Eigenspaces: Examples & Concepts - Prof. Gera, Study notes of Linear Algebra

An introduction to the concepts of eigenvalues, eigenvectors, and eigenspaces for a square matrix. It explains how to find eigenvalues and eigenvectors using examples and discusses the significance of these concepts. The document also touches upon the characteristic equation and its role in determining the number of eigenvalues for a matrix.

Typology: Study notes

Pre 2010

Uploaded on 09/17/2009

koofers-user-t51
koofers-user-t51 🇺🇸

10 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
1
5.1–5.2. EIGENVALUES, EIGENVECTORS, EIGENSPACES,
CHARACTERISTIC EQUATION
If A is a square matrix, a scalar λ is called an eigenvalue of A if there
exists a column vector x 0 such that A x = λ x . If this is true, then each such
non-zero x is called an eigenvector for A associated with λ. There are other
standard phrases that are used for these concepts. An eigenvalue is also called a
characteristic value or a latent root. An eigenvector is also called a characteristic
vector. Since (for a specific λ) the set of all solutions of A x = λ x is a subspace,
we can call the set of all x, including the zero vector, which satisfy A x = λ x the
eigenspace associated with λ.
Notice that if λ = 0, then λ is an eigenvalue of A if and only if the
homogeneous system with A as coefficient matrix has at least one non-trivial
solution. Thus 0 is an eigenvalue of A if and only if A is singular (i.e., not
invertible).
Let’s consider some examples.
Example 1. Let A =
A x = λ x means that . So the eigenvalue
condition is x1 = λ x1 and 2x2 =λ x2.
If x1 0, then it follows from the first requirement that we must have λ =
1 and then from the second requirement, 2x2 = x2, so that x2 = 0. So λ = 1 is an
eigenvalue of A and the associated eigenspace consists of all vectors in R2 of the
form (So is a basis for this eigenspace.) But if
x1 =0, then since x2 can’t also be zero (after all, x is assumed not to be the zero
vector.), it follows from the second requirement that λ = 2. So λ = 2 is the other
eigenvalue for A and the associated eigenspace consists of all vectors in R2 of the
form (And is a basis for this eigenspace.)
In the same way, one can show that for every n by n diagonal matrix, the
eigenvalues are the diagonal entries and for each k from 1 through n, the column
vector ek is an eigenvector associated with the kth entry of the matrix.
Example 2. Let A =
A x = λ x means that . So the
eigenvalue condition is x1 + x2 = λ x1 and 2x2 =λ x2. Now the second of the two
pf3

Partial preview of the text

Download Understanding Eigenvalues, Eigenvectors, and Eigenspaces: Examples & Concepts - Prof. Gera and more Study notes Linear Algebra in PDF only on Docsity!

5.1–5.2. EIGENVALUES, EIGENVECTORS, EIGENSPACES,

CHARACTERISTIC EQUATION

If A is a square matrix, a scalar λ is called an eigenvalue of A if there exists a column vector x0 such that A x = λ x. If this is true, then each such non-zero x is called an eigenvector for A associated with λ. There are other standard phrases that are used for these concepts. An eigenvalue is also called a characteristic value or a latent root. An eigenvector is also called a characteristic vector. Since (for a specific λ) the set of all solutions of A x = λ x is a subspace, we can call the set of all x , including the zero vector, which satisfy A x = λ x the eigenspace associated with λ. Notice that if λ = 0, then λ is an eigenvalue of A if and only if the homogeneous system with A as coefficient matrix has at least one non-trivial solution. Thus 0 is an eigenvalue of A if and only if A is singular (i.e., not invertible). Let’s consider some examples. Example 1. Let A = A x = λ x means that. So the eigenvalue condition is x 1 = λ x 1 and 2 x 2 = λ x 2. If x 1 ≠ 0, then it follows from the first requirement that we must have λ = 1 and then from the second requirement, 2 x 2 = x 2 , so that x 2 = 0. So λ = 1 is an eigenvalue of A and the associated eigenspace consists of all vectors in R 2 of the form (So is a basis for this eigenspace.) But if x 1 =0, then since x 2 can’t also be zero (after all, x is assumed not to be the zero vector.), it follows from the second requirement that λ = 2. So λ = 2 is the other eigenvalue for A and the associated eigenspace consists of all vectors in R 2 of the form (And is a basis for this eigenspace.) In the same way, one can show that for every n by n diagonal matrix, the eigenvalues are the diagonal entries and for each k from 1 through n , the column vector e k is an eigenvector associated with the k th entry of the matrix. Example 2. Let A = A x = λ x means that. So the eigenvalue condition is x 1 + x 2 = λ x 1 and 2 x 2 = λ x 2. Now the second of the two

requirements is simpler than the first one. Either x 2 = 0 or x 2 ≠ 0. In the first case, the first requirement becomes x 1 = λ x 1 and since x 1 ≠ 0 , we must have λ = 1. So λ = 1 is an eigenvalue of A and the associated eigenspace consists of all vectors in R 2 of the form (So is a basis for this eigenspace.) But if x 2 ≠ 0, then it follows from the second requirement that λ = 2. So λ = 2 is the other eigenvalue for A. The requirement that x 1 + x 2 = λ x 1 is now x 1 + x 2 = 2 x 1. This simplifies: x 2 = x 1. Thus the eigenspace associated with λ = 2 consists of all vectors in R^2 of the form (And is a basis for this eigenspace.) It seems reasonable to guess that the diagonal entries of every upper triangular square matrix are eigenvalues of the matrix. This turns out to be true, as we shall soon see. But the next example shows that a repeated entry on the diagonal doesn’t necessarily get repeated as an eigenvalue. Example 3. Let A = A x = λ x means that. So the eigenvalue condition is (a) 2 x 1 + x 2 = λ x 1 and (b) 2 x 2 = λ x 2. If x 2 = 0, requirement (a) becomes 2 x 1 = λ x 1 and since x 1 ≠ 0 , we must have λ = 2. So λ = 2 is an eigenvalue of A. If x 2 ≠ 0, then it follows from requirement (b) that λ = 2. So whether x 2 = 0 or x 2 ≠ 0, λ = 2. So this matrix has just the one eigenvalue, λ = 2. In order to find the eigenvectors, replace λ by 2 in (a) and (b). They become (a) 2 x 1 + x 2 = 2 x 1 and (b) 2 x 2 = 2 x 2. Which says exactly that x 2 = 0 and that x 1 is a free variable. So for this matrix, λ = 2 occurs twice algebraically but has only a one-dimensional eigenspace, namely Span. Example 4. Let A = It turns out that this matrix has no eigenvalues in the real number system. In fact, A x = λ x means that (a) – x 2 = λ x 1 and (b) x 1 = λ x 2. If λ is to be an eigenvalue, at least one of the x i’s isn’t zero. But if either one were 0, (a) and (b) imply that both would be zero. So neither is. These (a) and (b) are very different from the (a) and (b) in the previous examples because of the way they are coupled. To unscramble them, we can take the expression in (b) and use it to express (a) in terms of just x 2. Namely, – x 2 = λ(λ x 2 )= λ^2 x 2. Since x 2 ≠ 0, canceling is justified; so λ^2 = – 1. Since the square of a real number cannot be negative, this matrix has no eigenvalues and eigenvectors. If a matrix is given, is there a systematic way to determine how many eigenvalues it has? Is there an upper bound that limits the number of eigenvalues that an