# Linear Transformation or not? [closed]

$$T(x) = \left[ \begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right] \cdot \left[ \begin{array}{c} x \\ y \\ 1 \end{array} \right]$$

You have to solve it by definition: $$T(w_{1}) + T(w_{2}) = T(w_{1} + w_{2}),$$ and $$\alpha T(x) = T(\alpha x).$$

I am new here, I do not know the format, I’m sorry for any inconvenience that this may cause. Just wondering if the exercise above is a linear transformation or not.

## closed as unclear what you're asking by Cameron Williams, Jyrki Lahtonen, Mostafa Ayaz, Namaste, JMPJun 27 '18 at 13:43

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• Hint: see if $T(x+y)$ is... (do the same for the other part of the definition) – Sean Roberson Jun 11 '18 at 15:09
• Is $\theta$ constant, or does it depend on $x$ and $y$? Is $y$ constant, or does it depend on $x$ and $\theta$? – GFauxPas Jun 11 '18 at 15:30
• θ Is constant, π/4 to be precise – Juan Antonio Ayala Rascón Jun 11 '18 at 15:37
• Have you checked if it meets the definition of a linear transformation? – Morgan Rodgers Jun 11 '18 at 16:07
• (Also, is it supposed to be $T(x,y)$?) – Morgan Rodgers Jun 11 '18 at 16:08

$L : \mathbb{R}^n \rightarrow \mathbb{R}^m$ is a linear transformation if and only if there is a matrix, $A$, such that $L( z ) = A z$... What you have here is sort of like that... BUT...
But here $L$ is a bit tricky: it is a function of two variables. Another thing you know about a linear transformation is that it should map the zero vector to the zero vector. What happens if you pick $x = y = 0$?