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[SOLVED] 072 is Q(theta) a linear transformation from R^2 to itself.

karush

Well-known member
Jan 31, 2012
3,066
if $Q(\theta)$ is

$\left[\begin{array}{rr}
\cos{\theta}&- \sin{\theta}\\
\sin{\theta}&\cos{\theta}
\end{array}\right]$

how is $Q(\theta)$ is a linear transformation from R^2 to itself.

ok I really didn't know a proper answer to this question but presume we would need to look at the unit circle

not sure if this helps

Screenshot 2021-03-13 1.27.12 PM.png
 
Last edited by a moderator:

karush

Well-known member
Jan 31, 2012
3,066
Ok thanks
I usually don't get much replies on these linear algebra posts
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
1,005
Ok thanks
I usually don't get much replies on these linear algebra posts
I'd help out, but it's been since my sophomore year in school (1973) since I've taken a course in Linear Algebra ... except for the very basic stuff, I haven't used it so I've "losed" it.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
Better is to understand what "linear transformation" means! ANY transformation that can be written as a matrix multiplication is linear!

A transformation, L, on a vector space is "linear" if and only if
L(u+ v)= Lu+ Lv, for any vectors u and v, and
L(au)= aLu, for any vector u and scalar, a.

Here if $u= \begin{pmatrix}x \\ y \end{pmatrix}$ and $v= \begin{pmatrix} a \\ b\end{pmatrix}$, $L(u+ v)= \begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{pmatrix}\begin{pmatrix}x+ a \\ y+ b\end{pmatrix}= \begin{pmatrix}(x+ a)cos(\theta)- (y+ b)sin(\theta) \\ (x+ a)sin(\theta)+ (y+ b)cos(\theta)\end{pmatrix}$.

While $Lu+ Lv= \begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}+ \begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix}= \begin{pmatrix} xcos(\theta)- ysin(\theta) \\ xsin(\theta)+ y cos(\theta)\end{pmatrix}+ \begin{pmatrix} acos(\theta)- bsin(\theta) \\ asin(\theta)+ bcos(\theta)\end{pmatrix}= \begin{pmatrix}(x+ a)cos(\theta)- (y+ b)sin(\theta) \\ (x+ a)sin(\theta)+ (y+ b)cos(\theta)\end{pmatrix}$.

And $L(au)= \begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{pmatrix}\begin{pmatri
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072 is Q(theta) is a linear transformation from R^2 to itself.
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karush
Well-known member

Jan 31, 2012 2,838
if Q(θ)Q(θ) is

[cosθsinθ−sinθcosθ][cos⁡θ−sin⁡θsin⁡θcos⁡θ]

how is Q(θ)Q(θ) is a linear transformation from R^2 to itself.

ok I really didn't know a proper answer to this question but presume we would need to look at the unit circle

not sure if this helps


Screenshot 2021-03-13 1.27.12 PM.png


Last edited by a moderator: Today at 12:07 AM
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x}ax \\ ay \end{pmatrix}= \begin{pmatrix}ax coz
 

karush

Well-known member
Jan 31, 2012
3,066
I'd help out, but it's been since my sophomore year in school (1973) since I've taken a course in Linear Algebra ... except for the very basic stuff, I haven't used it so I've "losed" it.
wow.... my senior year was 1970 but my highest level in math was algebra II which today is much more advanced
2021_03_07_16.48.50.jpg