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A linear transformation acts on $R^2$ by first doubling the $x$-coordinate and then rotating the plane by an angle of $\pi/2$ ($90^{\circ}$) counter-clockwise. What is the corresponding matrix to this linear transformation?

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  • $\begingroup$ Well, can you at least figure out the matrix for the doubling transformation? We can take care of rotation later. $\endgroup$ – астон вілла олоф мэллбэрг Sep 29 '16 at 0:59
  • $\begingroup$ I can do the rotation not the doubling $\endgroup$ – Anonymous Sep 29 '16 at 1:00
  • $\begingroup$ is it 2*([0,-1] [1,0]) $\endgroup$ – Anonymous Sep 29 '16 at 1:00
  • $\begingroup$ where do your principle compenent vecors ($\hat i, \hat j$, or $(1,0), (0,1)$ depending on your notation preference) go in this transformation? $\endgroup$ – Doug M Sep 29 '16 at 1:03
  • $\begingroup$ what do you mean? $\endgroup$ – Anonymous Sep 29 '16 at 1:04
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See, the key to finding the matrix of a linear transformation is watching wher the basis elements go. To be precise : TO find the matrix representation of $T$, find $T(e_i)$, and write down the answer in column format.

So let's look at our basis vectors, which are $(1,0)$ and $(0,1)$.

For $(1,0)$, first we double the x-coordinate, which gives $(2,0)$, and then we rotate counter clockwise by $90^\circ$, which gives $(0,2)$.

For $(0,1)$, first we double the x-coordinate, which gives $(0,1)$, and then we rotate counter clockwise by $90^\circ$, which gives $(-1,0)$.

Now, the key is to write these as columns, and just join them together. $$ T = \begin{bmatrix}0 & -1\\ 2 & 0\end{bmatrix} $$

We can see that this transformation does exactly what you want.

Don't look only at the answer, but also note the very simple procedure used to obtain the matrix. All I did was applied it on the basis vectors (which was easy, there are so many zeros), and just joined the vectors together.

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  • $\begingroup$ how do you know that the basis vectors are (1,0) (1,0) and (0,1)(0,1). ? $\endgroup$ – Anonymous Sep 29 '16 at 1:11
  • $\begingroup$ the questions does not give any basis vectors $\endgroup$ – Anonymous Sep 29 '16 at 1:11
  • $\begingroup$ Well, when matrices of linear transformations are computed, they require the presence of a basis to be in context. The first and foremost basis that comes to mind, is the standard basis on $\mathbb{R}^2$, namely $(1,0), (0,1)$. When a basis is not specified, we can pick any basis we like, and in this case I picked the standard basis. However, if you give me a basis, I can return you the transformation in that base. The trick is the same: watch where the basis vectors go. $\endgroup$ – астон вілла олоф мэллбэрг Sep 29 '16 at 1:14
  • $\begingroup$ I would like to add : The matrix of a linear transformation cannot be talked about without the presence (or choice) of a basis. $\endgroup$ – астон вілла олоф мэллбэрг Sep 29 '16 at 1:16
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Hint: Any linear transformation of $\mathbb{R}^2$ is fully determined by where it sends the standard basis vectors of $\mathbb{R}^2$, namely $i=[1,0]$ and $j=[0,1]$. Can you determine what happens to these basis vectors under your linear transformation? Also, if you know what happens to these basis vectors, do you know how to find the matrix corresponding to that linear transformation in that basis?

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