Undergrad Linear Isomorphisms: Understand & Apply

[Lauren1234]

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

 

[member 587159]

Hints:

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.

 

[Lauren1234]

Hints:

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.

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

 

[member 587159]

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

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 $(0,1)$.

 

[Lauren1234]

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 $(0,1)$.

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.

 

[Lauren1234]

Hints:

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.

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

 

[member 587159]

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

 

[Lauren1234]

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.

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.

 

[member 587159]

I told you what to do: Calculate $T(1,0)$ where $T$ is the rotation in your exercise.

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.