eigenvalues and eigenvectors?

Question

Define what is meant by saying that a column vector v is an eigenvector for the square matrix A and that (h) is the corresponding eigenvalue. Verify that v = [2, 5]T is an eigenvector for the matrix

A = −38 ...14
......−105 .39

and determine the corresponding eigenvalue.

Answer 1

If v is an eigenvector of A, it means that the linear transformation Av is a scalar multiple of v. Indeed, Av = [-6, -15]T, which is equal to -3 times v.

Answer 2

Any linear combination of eigenvectors with the same eigenvalue is an eigenvector respect to the same eigenvalue. Let l be lambda and let a, b any two numbers (in some field K: R, C, ...). u is an eigenvector of [A] means that [A] u = l u. v is an eigenvector of [A] means that [A] v = l v. You can multiply the two equations respectively for a and b: you get [A] a u= l a u and [A] b v = l b v. You can move a and b inside since a works as [I] a (where [I] is the identity matrix). Summing equations and using linearity we get [A] (a u + bv) = l (a u + b v). Now let a=b=1 and you get [A] (u+v)=l (u+v), which is what you asked.

Answer 3

The eigenvector of a matrix is a vector that remains the same (multiplied by a constant) after being transformed by that matrix. For example any vector is an eigenvector of the identity matrix with eigenvalue 1.

In your example you'll see that
A v = lambda v

You just multiply A times v and you should get an integer times v in this case -3. So v is an eigenvector of a and the eigenvalue is -3.

Answer 4

if v is an eigenvector and h is the corresponding eigenvalue for matrix A then:

A*v = h*v

(*=multiply)
in your case

A*v = [-38*2 + 14*5, -105*2 + 39*5]T
= [-6, -15]T

comparing this to v we see that the corresponding eigenvalue must be -3

Answer 5

http://www.sosmath.com/matrix/eigen0/eigen0.html
http://www.math.hmc.edu/calculus/tutorials/eigenstuff/
http://www.cs.ut.ee/~toomas_l/linalg/lin1/node16.html
http://ceee.rice.edu/Books/LA/eigen/
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