Finding the eigenvectors for T()

[davemoosehead]

Homework Statement

Which of the following is not an eigenvector for $$T \left( \left[ {\begin{array}{cc} x \\ y \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} x + y \\ x+ y \\ \end{array} } \right]$$ ?

A) v = [-2 -2]^T
B) v = [1 -1]^T
C) v = [1 2]^T
D) All are eigenvectors

Homework Equations

Ax = $$\lambda$$x

The Attempt at a Solution

My problem is that all eigenvectors I've computed have come from 2x2 matrices. My best guess on starting is
$$T \left( \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} -4 \\ -4 \\ \end{array} } \right] \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] =$$

but this obviously doesn't work because of the size. How do I find an eigenvalue of a 2x1 matrix? Is it possible? Am I even looking at this correctly?

Edit: I think I've figured it out. First, I constructed the standard matrix for T and got $$\left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right]$$
Then I used Ax = $$\lambda$$x with the vectors given to find the eigenvalues. Letter C didn't have an eigenvalue, so that is the answer.
$$T \left( \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] = \left[ {\begin{array}{cc} -4 \\ -4 \\ \end{array} } \right] = 2 \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right]$$
$$T \left( \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right] = \left[ {\begin{array}{cc} 0 \\ 0 \\ \end{array} } \right] = 0 \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right]$$
$$T \left( \left[ {\begin{array}{cc} 1 \\ 2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} 1 \\ 2 \\ \end{array} } \right] = \left[ {\begin{array}{cc} 3 \\ 3 \\ \end{array} } \right] = ?$$

 

[Mark44]

Homework Statement

Which of the following is not an eigenvector for $$T \left( \left[ {\begin{array}{cc} x \\ y \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} x + y \\ x+ y \\ \end{array} } \right]$$ ?

A) v = [-2 -2]^T
B) v = [1 -1]^T
C) v = [1 2]^T
D) All are eigenvectors

Homework Equations

Ax = $$\lambda$$x

The Attempt at a Solution

My problem is that all eigenvectors I've computed have come from 2x2 matrices. My best guess on starting is
$$T \left( \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} -4 \\ -4 \\ \end{array} } \right] \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] =$$

but this obviously doesn't work because of the size. How do I find an eigenvalue of a 2x1 matrix? Is it possible? Am I even looking at this correctly?

Edit: I think I've figured it out. First, I constructed the standard matrix for T and got $$\left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right]$$
Then I used Ax = $$\lambda$$x with the vectors given to find the eigenvalues. Letter C didn't have an eigenvalue, so that is the answer.
$$T \left( \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] = \left[ {\begin{array}{cc} -4 \\ -4 \\ \end{array} } \right] = 2 \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right]$$
$$T \left( \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right] = \left[ {\begin{array}{cc} 0 \\ 0 \\ \end{array} } \right] = 0 \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right]$$
$$T \left( \left[ {\begin{array}{cc} 1 \\ 2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} 1 \\ 2 \\ \end{array} } \right] = \left[ {\begin{array}{cc} 3 \\ 3 \\ \end{array} } \right] = ?$$

Your answer is correct, but you have made extra work for yourself in this problem. You didn't need to find a matrix representation for this transformation. All you needed to do was determine whether T(x) is equal to a scalar multiple of x.
For example, T(-2, -2) = (-4, -4) = 2*(-2, -2).