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## Main Question or Discussion Point

how would I go about answering the above question I need some pointers on how to start?

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- Thread starter Lauren1234
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- #1

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how would I go about answering the above question I need some pointers on how to start?

- #2

Math_QED

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You have to show that ##(1,0)## gets mapped to ##(1/2, \sqrt{3}/2)## and ##(0,1)## gets mapped to ##(-\sqrt{3}/2, 1/2)##. Showing that it is a linear isomorphism is intuitively clear: linearity is obvious, injectivity is also obvious because two different points get rotated to two different points with the same distance between them, and a rotation in the other direction over the same angle shows that your map is surjective. Of course, if you use some theorems like rank-nullity theorem you get surjectivity from injectivity or vice versa.

- #3

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Fab I’ll give it a go. Is there a specific way I could draw the above also?

You have to show that ##(1,0)## gets mapped to ##(1/2, \sqrt{3}/2)## and ##(0,1)## gets mapped to ##(-\sqrt{3}/2, 1/2)##. Showing that it is a linear isomorphism is intuitively clear: linearity is obvious, injectivity is also obvious because two different points get rotated to two different points with the same distance between them, and a rotation in the other direction over the same angle shows that your map is surjective. Of course, if you use some theorems like rank-nullity theorem you get surjectivity from injectivity or vice versa.

- #4

Math_QED

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Draw? Sure. Draw the unit circle in the ##x-y##-plane. The vector ##(1,0)## is the vector both on the unit circle and on the ##x##-axis. Now, after applying the rotation it ends up at ##\pi/3## on the unit circle (60 degrees). Which vector is that? Basic trigoniometry will help here. Similarly you do the same for ##(0,1)##.Fab I’ll give it a go. Is there a specific way I could draw the above also?

- #5

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Yeah it says to draw the matrix bit as a carefully drawn diagram. So basically I need to draw a circle in the plane right think I’ve got you.Draw? Sure. Draw the unit circle in the ##x-y##-plane. The vector ##(1,0)## is the vector both on the unit circle and on the ##x##-axis. Now, after applying the rotation it ends up at ##\pi/3## on the unit circle (60 degrees). Which vector is that? Basic trigoniometry will help here. Similarly you do the same for ##(0,1)##.

- #6

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ive done this bit but I’m not exa sure what it shows does it tell me the matrix is a linear transformation?

You have to show that ##(1,0)## gets mapped to ##(1/2, \sqrt{3}/2)## and ##(0,1)## gets mapped to ##(-\sqrt{3}/2, 1/2)##. Showing that it is a linear isomorphism is intuitively clear: linearity is obvious, injectivity is also obvious because two different points get rotated to two different points with the same distance between them, and a rotation in the other direction over the same angle shows that your map is surjective. Of course, if you use some theorems like rank-nullity theorem you get surjectivity from injectivity or vice versa.

- #7

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Do you know how to associate a matrix to a linear transformation, relative to some fixed bases?ive done this bit but I’m not exa sure what it shows does it tell me the matrix is a linear transformation?

Suppose we have a linear transformation ##T: V \to W## and ##\{e_1, \dots, e_n\}## a basis for ##V## and ##\{f_1, \dots, f_m\}## a basis for ##W##. Then you calculate ##T(e_1)## and write it in the form ##T(e_1) = \sum_{i=1}^m a_i f_i##. The coefficients ##(a_1, \dots, a_m)## come in the first column of the matrix. Similarly you calculate ##T(e_2)## to get the second column etc. The idea here is that a linear transformation is known completely if we know what the map does to a basis, so we put this information in a matrix.

In your case ##V = W = \mathbb{R}^2## and the basis for both ##V## and ##W## is ##\{(1,0), (0,1)\}##. So, you calculate ##T(1,0) ##. What coefficients do you get when you write this as a linear combination of ##(1,0)## and ##(0,1)##? These will go in the first column.

- #8

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this is what I’ve done so far. Do I but them together and show they’re the same? And that means they’re a linear transformation.Do you know how to associate a matrix to a linear transformation, relative to some fixed bases?

Suppose we have a linear transformation ##T: V \to W## and ##\{e_1, \dots, e_n\}## a basis for ##V## and ##\{f_1, \dots, f_m\}## a basis for ##W##. Then you calculate ##T(e_1)## and write it in the form ##T(e_1) = \sum_{i=1}^m a_i f_i##. The coefficients ##(a_1, \dots, a_m)## come in the first column of the matrix. Similarly you calculate ##T(e_2)## to get the second column etc. The idea here is that a linear transformation is known completely if we know what the map does to a basis, so we put this information in a matrix.

In your case ##V = W = \mathbb{R}^2## and the basis for both ##V## and ##W## is ##\{(1,0), (0,1)\}##. So, you calculate ##T(1,0) ##. What coefficients do you get when you write this as a linear combination of ##(1,0)## and ##(0,1)##? These will go in the first column.

- #9

Math_QED

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Next, write ##T(1,0) = a (1,0) + b(0,1) = (a,b)## with ##a,b \in \mathbb{R}##. The coefficients ##(a,b)## will be the first column of your matrix.

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