Is a 20-Degree Rotation in the XY-Plane a Linear Transformation?

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Homework Statement

Is the function which rotates the xy-plane by 20 degrees is a linear transformation?

From R² -> R²

Homework Equations

x` = xcos$$\theta$$ + ysin$$\theta$$
y` = -xsin$$\theta$$ + ycos$$\theta$$

Where $$\theta$$ = 20 degrees (or $$\pi$$/9 )

The Attempt at a Solution

Apparently the solution to this is true:

Ok so the two properties must hold

T(u + v) = T(u) + T(v)
and T(cu) = cT(u)
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Let u = (u₁, u₂) and v =(v₁, v₂)


T(u) + T(v) = u₁cos$$\theta$$ + u₂sin$$\theta$$, - v₁sin$$\theta$$ + v₂cos$$\theta$$

However this isn't equal to T(u+v) = T((u₁ + v₁), (u₂ + v₂)) which when expanded will give me more cos$$\theta$$ terms than in T(u) + T(v).

I figure I am just not computing this properly, any help would be great, thanks.

 

[anathema]

Yes, the function that rotates the xy-plane by 20 degrees is a linear transformation from R2 -> R2. This can be shown by checking the two properties that you mentioned: T(u+v) = T(u) + T(v) and T(cu) = cT(u). Let's start with the first property:

T(u+v) = T((u1 + v1), (u2 + v2)) = (u1 + v1)cos\theta + (u2 + v2)sin\theta, -(u1 + v1)sin\theta + (u2 + v2)cos\theta

Now, let's expand T(u) + T(v):

T(u) + T(v) = (u1cos\theta + u2sin\theta) + (v1cos\theta + v2sin\theta), -(u1sin\theta + u2cos\theta) + -(v1sin\theta + v2cos\theta)

As you can see, these two expressions are equal, so the first property holds. Now let's check the second property:

T(cu) = T(cu1, cu2) = (cu1)cos\theta + (cu2)sin\theta, -(cu1)sin\theta + (cu2)cos\theta

cT(u) = c(u1cos\theta + u2sin\theta) = (cu1)cos\theta + (cu2)sin\theta, -(cu1)sin\theta + (cu2)cos\theta

Again, these two expressions are equal, so the second property holds as well. Therefore, we can conclude that the function that rotates the xy-plane by 20 degrees is a linear transformation from R2 -> R2.

I hope this helps clarify things for you. Keep up the good work in your studies!

 

[tomcue]

I would like to point out that the function described in the homework statement is indeed a linear transformation. This can be seen by considering the two properties mentioned: T(u + v) = T(u) + T(v) and T(cu) = cT(u).

Firstly, let's consider T(u + v). Using the equations given, we have:

T(u + v) = (u1 + v1)cos\theta + (u2 + v2)sin\theta, -(u1 + v1)sin\theta + (u2 + v2)cos\theta

Expanding this, we get:

T(u + v) = u1cos\theta + v1cos\theta + u2sin\theta + v2sin\theta, -u1sin\theta - v1sin\theta + u2cos\theta + v2cos\theta

Now, using the properties of cosine and sine, we can rewrite this as:

T(u + v) = (u1cos\theta - u1sin\theta) + (v1cos\theta - v1sin\theta) + (u2sin\theta + u2cos\theta) + (v2sin\theta + v2cos\theta)

Which can be simplified to:

T(u + v) = u1(cos\theta - sin\theta) + v1(cos\theta - sin\theta) + u2(sin\theta + cos\theta) + v2(sin\theta + cos\theta)

This is equal to T(u) + T(v), as required by the first property.

Next, let's consider T(cu). Using the equations given, we have:

T(cu) = cu1cos\theta + cu2sin\theta, -cu1sin\theta + cu2cos\theta

Using the properties of multiplication, we can rewrite this as:

T(cu) = c(u1cos\theta + u2sin\theta), -c(u1sin\theta - u2cos\theta)

Which is equal to cT(u), as required by the second property.

Therefore, we can conclude that the function described in the homework statement is indeed a linear transformation from R2 -> R2.