1. ## Question about solving eigenvector problems

I understand how to get eigenvalues, but I have a problem understanding eigenvectors.
I'll demonstrate with an example:

I am given A = $\begin{bmatrix}3 & 1.5 \\1.5 & 3 \end{bmatrix}$

Solving for lambda, I get lambda = 1.5 and 4.5

Now my matrix equations become:

$-1.5x_1+1.5x_2=0$ and $1.5x_1-1.5x_2=0$

dividing by 1.5 I get: $x_1-x_2=0$
so I get $x_1=x_2$

Now, the answer is $\begin{bmatrix}1 & 1 \end{bmatrix}$ because if you set x1=1, then x2=1

If x1=x2, then why isn't the answer [ℝ], where ℝ is the set of all real numbers?

2. ## Re: Question about solving eigenvector problems

Originally Posted by yaro99
I understand how to get eigenvalues, but I have a problem understanding eigenvectors.
I'll demonstrate with an example:

I am given A = $\begin{bmatrix}3 & 1.5 \\1.5 & 3 \end{bmatrix}$

Solving for lambda, I get lambda = 1.5 and 4.5

Now my matrix equations become:

$-1.5x_1+1.5x_2=0$ and $1.5x_1-1.5x_2=0$

dividing by 1.5 I get: $x_1-x_2=0$
so I get $x_1=x_2$

Now, the answer is $\begin{bmatrix}1 & 1 \end{bmatrix}$ because if you set x1=1, then x2=1

If x1=x2, then why isn't the answer [ℝ], where ℝ is the set of all real numbers?

There are actually two eigenvectors 1/sqrt(2) {1,1} and 1/sqrt(2) {1,-1}, corresonding to eigenvalues 1.5 and 4.5.

{1,1} doesn't span R2. It spans the line y = x. Likewise {1,-1} spans the line y = -x

together these two vectors span R2

3. ## Re: Question about solving eigenvector problems

Originally Posted by romsek

{1,1} doesn't span R2. It spans the line y = x.
In that case, my question is, why aren't there infinite values for x and y that satisfy the eigenvector?
for example, {2,2}, {5,5}, {1000,1000}, etc.

Originally Posted by romsek

There are actually two eigenvectors 1/sqrt(2) {1,1} and 1/sqrt(2) {1,-1}, corresonding to eigenvalues 1.5 and 4.5.
How did you get this? My book has this answer:
1.5, [1 1], 45°; 4.5, [1 1], 45°

Does that make sense?

4. ## Re: Question about solving eigenvector problems

Originally Posted by yaro99
so I get $x_1=x_2$

Now, the answer is $\begin{bmatrix}1 & 1 \end{bmatrix}$ because if you set x1=1, then x2=1

If x1=x2, then why isn't the answer [ℝ], where ℝ is the set of all real numbers?
First, the notation [ℝ ℝ], where the objects inside brackets are sets rather than numbers, is not universally accepted. Even if it were, [A B] would probably mean {[x y] | x ∈ A, y ∈ B}, i.e., A × B. There is no coordination between the first and second coordinate here. It is clear that you mean {[x x] | x ∈ ℝ}, but I don't see an easier way to write this. By the way, even if you delimit matrices with square brackets, I would use parentheses for vectors (i.e., 1 × n matrices): e.g., (x, x), because square brackets are also used to denote closed segments.

Second, you are right, any (x, x) is an eigenvector of A as is (x, -x). Indeed, you solve the equation det(A -λI) = 0 for λ, so if λ is an eigenvalue, the matrix A - λI is singular by definition. Therefore, the system of equations (A - λI)x = 0 has infinitely many solutions.

Edit: If x is an eigenvector with value λ, then so is (ax) for any a ∈ ℝ such that a ≠ 0. Indeed, A(ax) = a(Ax) = a(λx) = λ(ax).

5. ## Re: Question about solving eigenvector problems

Originally Posted by yaro99
In that case, my question is, why aren't there infinite values for x and y that satisfy the eigenvector?
for example, {2,2}, {5,5}, {1000,1000}, etc.

How did you get this? My book has this answer:
1.5, [1 1], 45°; 4.5, [1 1], 45°

Does that make sense?
there are infinite points. That's why you say this space, in this case the 1 dimensional line y=x, is "spanned" by the vector 1/sqrt(2) {1,1}. Multiply this vector by any real scalar and you remain in the space, but yes there are infinitely many points.

Your book is missing a minus sign somewhere. The answer I got was

1.5 [1,-1]/sqrt(2)

4.5 [1,1]/sqrt(2)

6. ## Re: Question about solving eigenvector problems

Originally Posted by romsek
there are infinite points. That's why you say this space, in this case the 1 dimensional line y=x, is "spanned" by the vector 1/sqrt(2) {1,1}. Multiply this vector by any real scalar and you remain in the space, but yes there are infinitely many points.

Your book is missing a minus sign somewhere. The answer I got was

1.5 [1,-1]/sqrt(2)

4.5 [1,1]/sqrt(2)
Whoops, that was my bad, there was a minus sign.
where do you get the 1/sqrt(2) from?
is it a scalar that you can multiply throughout the matrix?

7. ## Re: Question about solving eigenvector problems

Originally Posted by yaro99
Whoops, that was my bad, there was a minus sign.
where do you get the 1/sqrt(2) from?
is it a scalar that you can multiply throughout the matrix?
it just normalizes the vector to unit length. You tend to want the unit eigenvectors