there are many eigenvalue calculators online.
As for #3, there should be a proof in your text. The discussion about eigenvectors will show that they are a basis for the vector space.
Given a square matrix M, we say that a nonzero vector v is an eigenvector of M if Mv=kv for some real number k. The real number k is called the eigenvalue of v with respect to M.
1. Let v be an eigenvector of the matrix M with eigenvalue k. Find a simple expression for (M^n)v, where n is a positive integer.
2. Let A = {3 , -2 , 3 ; 1 , 2 , 1 ; 1 , 3 , 0}.
Show that
v1 = {1 ; 1 ; 1}, v2 = {-1 ; 1 ; 1}, v3 = {11 ; 1 ; -14} are eigenvectors of A, and for each of these eigenvectors, find the corresponding eigenvalue.
3. Show that any three-dimensional vector v can be expressed as a linear combination of v1, v2, and v3.
Even just one part helps! Thanks!
3 answers
I don't see how an eigenvalue calculator helps me with #1 and the first part of #2. I also don't know what you mean by "there should be a proof in your text. The discussion about eigenvectors will show that they are a basis for the vector space." Thanks!~
Since v is an eigenvector,
Mv = λv
so, M^2 v = M Mv = M λv = λ Mv = λ^2 v
So, M^n v = λ^n v
a ={1,1,1} is an eigenvector because Ma = {4,4,4} = λa where λ=4
So, 4 is the eigenvalue of a
If we call the eigenvectors a,b,c, then you can show that
a/2 - b/2 = {1,0,0}
a/2 + b/2 = {0,1,0}
a/3 + 3b/5 - c/15 = {0,0,1}
Clearly any 3-vector can be formed using combinations of those combinations.
Mv = λv
so, M^2 v = M Mv = M λv = λ Mv = λ^2 v
So, M^n v = λ^n v
a ={1,1,1} is an eigenvector because Ma = {4,4,4} = λa where λ=4
So, 4 is the eigenvalue of a
If we call the eigenvectors a,b,c, then you can show that
a/2 - b/2 = {1,0,0}
a/2 + b/2 = {0,1,0}
a/3 + 3b/5 - c/15 = {0,0,1}
Clearly any 3-vector can be formed using combinations of those combinations.
0 answers