Finding the eigenvectors for T()

[davemoosehead]

Homework Statement

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

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
$$T \left(<br /> \left[ {\begin{array}{cc}<br /> -2 \\<br /> -2 \\<br /> \end{array} } \right] \right) = <br /> \left[ {\begin{array}{cc}<br /> -4 \\<br /> -4 \\<br /> \end{array} } \right]<br /> \left[ {\begin{array}{cc}<br /> -2 \\<br /> -2 \\<br /> \end{array} } \right]<br /> =$$

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 $$\left[ {\begin{array}{cc}<br /> 1&1 \\<br /> 1&1 \\<br /> \end{array} } \right]$$
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.
$$T \left(<br /> \left[ {\begin{array}{cc}<br /> -2 \\<br /> -2 \\<br /> \end{array} } \right] \right) = <br /> \left[ {\begin{array}{cc}<br /> 1&1 \\<br /> 1&1 \\<br /> \end{array} } \right]<br /> \left[ {\begin{array}{cc}<br /> -2 \\<br /> -2 \\<br /> \end{array} } \right] = <br /> \left[ {\begin{array}{cc}<br /> -4 \\<br /> -4 \\<br /> \end{array} } \right] = 2<br /> \left[ {\begin{array}{cc}<br /> -2 \\<br /> -2 \\<br /> \end{array} } \right]$$
$$T \left(<br /> \left[ {\begin{array}{cc}<br /> 1 \\<br /> -1 \\<br /> \end{array} } \right] \right) = <br /> \left[ {\begin{array}{cc}<br /> 1&1 \\<br /> 1&1 \\<br /> \end{array} } \right]<br /> \left[ {\begin{array}{cc}<br /> 1 \\<br /> -1 \\<br /> \end{array} } \right] = <br /> \left[ {\begin{array}{cc}<br /> 0 \\<br /> 0 \\<br /> \end{array} } \right] = 0<br /> \left[ {\begin{array}{cc}<br /> 1 \\<br /> -1 \\<br /> \end{array} } \right]$$
$$T \left(<br /> \left[ {\begin{array}{cc}<br /> 1 \\<br /> 2 \\<br /> \end{array} } \right] \right) = <br /> \left[ {\begin{array}{cc}<br /> 1&1 \\<br /> 1&1 \\<br /> \end{array} } \right]<br /> \left[ {\begin{array}{cc}<br /> 1 \\<br /> 2 \\<br /> \end{array} } \right] = <br /> \left[ {\begin{array}{cc}<br /> 3 \\<br /> 3 \\<br /> \end{array} } \right] = ?$$

 

[Mark44]

Homework Statement

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

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
$$T \left(<br /> \left[ {\begin{array}{cc}<br /> -2 \\<br /> -2 \\<br /> \end{array} } \right] \right) = <br /> \left[ {\begin{array}{cc}<br /> -4 \\<br /> -4 \\<br /> \end{array} } \right]<br /> \left[ {\begin{array}{cc}<br /> -2 \\<br /> -2 \\<br /> \end{array} } \right]<br /> =$$

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 $$\left[ {\begin{array}{cc}<br /> 1&1 \\<br /> 1&1 \\<br /> \end{array} } \right]$$
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.
$$T \left(<br /> \left[ {\begin{array}{cc}<br /> -2 \\<br /> -2 \\<br /> \end{array} } \right] \right) = <br /> \left[ {\begin{array}{cc}<br /> 1&1 \\<br /> 1&1 \\<br /> \end{array} } \right]<br /> \left[ {\begin{array}{cc}<br /> -2 \\<br /> -2 \\<br /> \end{array} } \right] = <br /> \left[ {\begin{array}{cc}<br /> -4 \\<br /> -4 \\<br /> \end{array} } \right] = 2<br /> \left[ {\begin{array}{cc}<br /> -2 \\<br /> -2 \\<br /> \end{array} } \right]$$
$$T \left(<br /> \left[ {\begin{array}{cc}<br /> 1 \\<br /> -1 \\<br /> \end{array} } \right] \right) = <br /> \left[ {\begin{array}{cc}<br /> 1&1 \\<br /> 1&1 \\<br /> \end{array} } \right]<br /> \left[ {\begin{array}{cc}<br /> 1 \\<br /> -1 \\<br /> \end{array} } \right] = <br /> \left[ {\begin{array}{cc}<br /> 0 \\<br /> 0 \\<br /> \end{array} } \right] = 0<br /> \left[ {\begin{array}{cc}<br /> 1 \\<br /> -1 \\<br /> \end{array} } \right]$$
$$T \left(<br /> \left[ {\begin{array}{cc}<br /> 1 \\<br /> 2 \\<br /> \end{array} } \right] \right) = <br /> \left[ {\begin{array}{cc}<br /> 1&1 \\<br /> 1&1 \\<br /> \end{array} } \right]<br /> \left[ {\begin{array}{cc}<br /> 1 \\<br /> 2 \\<br /> \end{array} } \right] = <br /> \left[ {\begin{array}{cc}<br /> 3 \\<br /> 3 \\<br /> \end{array} } \right] = ?$$

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).