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!

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.

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.

1. To find a simple expression for (M^n)v, we need to understand how matrix multiplication works with eigenvectors and eigenvalues.

Given that v is an eigenvector of M with eigenvalue k, we have Mv = kv. Let's multiply both sides of this equation by M:

M(Mv) = M(kv)
(MM)v = k(Mv)

Since Mv = kv, we can substitute kv for Mv:

(MM)v = k(kv)
(M^2)v = k^2v

By applying the same logic repeatedly, we can find an expression for (M^n)v:

(M^n)v = k^n v

Therefore, (M^n)v is equal to k^n times the eigenvector v.

2. To show that v1 = {1 ; 1 ; 1}, v2 = {-1 ; 1 ; 1}, and v3 = {11 ; 1 ; -14} are eigenvectors of matrix A, we need to demonstrate that for each vector, A multiplied by the vector equals a scalar multiple of the vector.

For v1, we have:
A * v1 =
{3 , -2 , 3 ;
1 , 2 , 1 ;
1 , 3 , 0} * {1 ; 1 ; 1} =
{4 ; 4 ; 4} =
4 * {1 ; 1 ; 1} = 4v1

Therefore, v1 is an eigenvector of A with eigenvalue 4.

For v2, we have:
A * v2 =
{3 , -2 , 3 ;
1 , 2 , 1 ;
1 , 3 , 0} * {-1 ; 1 ; 1} =
{-4 ; 2 ; -2} =
-2 * {-1 ; 1 ; 1} = -2v2

Therefore, v2 is an eigenvector of A with eigenvalue -2.

For v3, we have:
A * v3 =
{3 , -2 , 3 ;
1 , 2 , 1 ;
1 , 3 , 0} * {11 ; 1 ; -14} =
{90 ; 18 ; -90} =
18 * {5 ; 1 ; -5} = 18v3

Therefore, v3 is an eigenvector of A with eigenvalue 18.

3. To show that any three-dimensional vector v can be expressed as a linear combination of v1, v2, and v3, we need to find the coefficients that give the linear combination. Let's call these coefficients a, b, and c.

We want to find:
v = a * v1 + b * v2 + c * v3

Substituting the values of v1, v2, and v3:
v = a * {1 ; 1 ; 1} + b * {-1 ; 1 ; 1} + c * {11 ; 1 ; -14}
v = {a - b + 11c ; a + b + c ; a + b - 14c}

Now, we set this equal to a general three-dimensional vector:
v = {x ; y ; z}

Equating the corresponding components:
x = a - b + 11c
y = a + b + c
z = a + b - 14c

Now we solve this system of equations for a, b, and c. This can be done through various methods, such as Gaussian elimination or matrix operations, to find the values of a, b, and c that satisfy the equations. Once we have these values, we can express any three-dimensional vector as a linear combination of v1, v2, and v3 using the coefficients a, b, and c.