1. ## [SOLVED] Practice test

I have my final coming just around the corner and I got this practice test. Any help would be great.

1. Define $\displaystyle T: R^2 \rightarrow R \ by \ T \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} = det \begin{bmatrix}x_{1}&1\\x_{2}&1\end{bmatrix}$.

Is T a linear transformation? If not, explain why not and provide a counterexample. If so, what is the geometric interpretation of the kernel of T?

2. Diagonalize $\displaystyle \begin{bmatrix}1&1\\-2&4\end{bmatrix}$
(I got Eigenvalues 2 and 3 and Eigenvectors $\displaystyle \begin{bmatrix}1\\1\end{bmatrix} and \begin{bmatrix}1\\2\end{bmatrix}$)

3. Given $\displaystyle T: R^2 \rightarrow R^2$ is linear and T(1,2) = (3,4) and T(3,4) = (1,2), then T(4,3) = ?
(I got (-8, -9))

2. 1)

$\displaystyle T \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} = det \begin{bmatrix}x_{1}&1\\x_{2}&1\end{bmatrix} = x_1 - x_2 = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\end{bmat rix}$
T is indeed linear!

Ker T = $\displaystyle \{{\bf x} = (x_1,x_2) \in \mathbb{R}^2| T{\bf x} = x_1 - x_2 = 0\}$
Ker T is the line passing through origin with a slope of $\displaystyle 45^{\circ}$ with the $\displaystyle x_2$ (or $\displaystyle x_1$) axis on the $\displaystyle x_1-x_2$ plane.

3. Originally Posted by Isomorphism
1)

$\displaystyle T \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} = det \begin{bmatrix}x_{1}&1\\x_{2}&1\end{bmatrix} = x_1 - x_2 = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\end{bmat rix}$
T is indeed linear!

Ker T = $\displaystyle \{{\bf x} = (x_1,x_2) \in \mathbb{R}^2| T{\bf x} = x_1 - x_2 = 0\}$
Ker T is the line passing through origin with a slope of $\displaystyle 45^{\circ}$ with the $\displaystyle x_2$ (or $\displaystyle x_1$) axis on the $\displaystyle x_1-x_2$ plane.
Thanks. I'm kind of struggling in the class. Could you please explain to me how you got $\displaystyle x_1 - x_2 = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\end{bmat rix}$? I'm confused as to how you got that 1, -1 on the diagonal.

4. Originally Posted by chickeneaterguy
Thanks. I'm kind of struggling in the class. Could you please explain to me how you got $\displaystyle x_1 - x_2 = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\end{bmat rix}$? I'm confused as to how you got that 1, -1 on the diagonal.
Just use matrix multiplication and show right hand side of the equation is equal to left hand side

5. Originally Posted by Isomorphism
Just use matrix multiplication and show right hand side of the equation is equal to left hand side
Oh, okay. I see now. Thanks again