Rotation Matrix 

Awwwwww ❤️

Two legends in One Comment ❤

Beautiful!

Loved your video ... thank you very much for the visualisation

Happy to hear that!

Oh wow that’s neat!

Thanks so much!!!

Thanks so much!!!

Brown and Churchill Complex Variables and Applications

@@drpeyam Thank you!!!😊😊

Believe in the math not in geogebra

You are not hallucinating, the plots of 'clockwise' are correct. In short, think of how shifting 'b units up from y' gives a new y-coord of y + b But shifting y = f(x) up by b units requires 'y - b' or y - b = f(x) Essentially the question you should be asking is 'how can i return back to my normal (x,y) coordinates or x,y, axes from my new (X',Y') which requires an 'inverse operation' inverse of anti-clockwise is clockwise. ------------------------------------------------------------------------------------------------------- Longer explanation: If you think about the fixed coordinate of (X', Y') compared to (x, y) then using the rotation matrix would work But substituting these values back into an equation, say that equation of the ellipse will not give you the anti-clockwise rotation of said graph. To explain the reason for that, I will go back to basics with transformations relating to just horizontal translations. If we take an arbitrary point (x,y) and move it to the right a units to the point (X', Y'), this new point will be (x + a, y) But does this mean the graph f(x) shifted to the right 8 units is f(x + a) = (x+a)^2. The answer to that is no, because essentially our point of view has changed, we want to input X' = x+a to map back to x Rearranging X' = x + a gets you x = X' - a so we should actually be considering f(X' - a) or using x as the variable again f(x - a) when shifting to the right a units,. We can consider a similar argument for horizontal translations to the left, vertical translations both up and down, dilations in both directions or a combination of any of these sorts of transformations. We always have to think of the 'opposite' operation when thinking about transforming graphs (or equations). So if we come back to rotations = A where A is the rotation matrix Similar to the translations example, we actually want to go back to Rearranging gets you A^-1 = = A^-1 where A^-1 is a clockwise rotation [this can be verified by actually taking the inverse] ^This shows why when plotting on geogebra (without the extra manipulation of taking inverses), it rotates it clockwise.

Thank youuuu

You’re welcome!!

Haven’t really done a video on it, but it’s the same idea. The matrix will be of the same form, except with the fixed axis having zeros and ones

Thank you!

Isn’t that obtained by solving Ax = x since the axis doesn’t change when rotated?

@@drpeyam sorry for the dumb question but what is Ax? For more context I'm not a Mathematician or Physicist. I just need help with this problem to complete a personal project. The gyroscope sensor axis x and y axis swaps every time my phone orientation rotates from portrait mode and I'm trying my best to lock them in.

Desmos must be incorrect

actually, you're not wrong. If we consider that same ellipse in the video (x^2)/2 + y^2 = 1 And we consider the transformation = {{cos(pi/4), - sin(pi/4)}, {sin(pi/4), cos(pi/4)}} The new point (X', Y') does indeed simplify to But if we just substitute this into x = X' and y = Y' We get that same equation: (((x-y)/sqrt(2))^2)/2 + ((x+y)/sqrt(2))^2 = 1 However, if we think about this equation for a second, an arbitrary point in this equation is (x,y), not (X', Y') So in other words, we have to now think of the transformation going from (X', Y') back to (x,y) instead, which is clockwise. That is why subbing it straight in actually gives you a clockwise rotation of the graph. This is basically analogous to how f(x+a) is shifting to the left a units, not left. = + The new point to the right a units of (x,y) is (x+a, y) but this does not mean f(x+a) is a shift to the right, because we need to think about 'how we can return back to f(x), or our original (x,y)) so = - = or f(X' - a) changing the dummy variable back to x f(x-a)

Yes haha, you’re overthinking the thumbnail. Also the thumbnail is counterclockwise rotation

@@drpeyam haha Yes I see today I was taught about this .I had watched 3blb essence of linear algebra series and tried to relate everything but messed up with this formula for counterclockwise...I found I had done some mistkes in calculations while trying to make sense lol now its ok..

@@drpeyam and thank you so much ..the one and only math youtuber who replies to the viewers taking their questions seriously...I will be asking you more and thanks for your videos I will definately share them with my friends, they are really intuitive😊

Should be counterclockwise

@@drpeyam I used the same equation that is in the video but the software is showing a clockwise rotation. Can you try it and see what I'm saying?

Same here i dont know why it is clockwise not counter clockwise

@@patrickdaniel2363 You are correct. It actually rotates it clockwise. If you think about the fixed coordinate of (X', Y') compared to (x, y) then using the rotation matrix would work But substituting these values back into an equation, say that equation of the ellipse will not give you the anti-clockwise rotation of said graph. To explain the reason for that, I will go back to basics with transformations relating to just horizontal translations. If we take an arbitrary point (x,y) and move it to the right a units to the point (X', Y'), this new point will be (x + a, y) But does this mean the graph f(x) shifted to the right 8 units is f(x + a) = (x+a)^2. The answer to that is no, because essentially our point of view has changed, we want to input X' = x+a to map back to x Rearranging X' = x + a gets you x = X' - a so we should actually be considering f(X' - a) or using x as the variable again f(x - a) when shifting to the right a units,. We can consider a similar argument for horizontal translations to the left, vertical translations both up and down, dilations in both directions or a combination of any of these sorts of transformations. We always have to think of the 'opposite' operation when thinking about transforming graphs (or equations). So if we come back to rotations = A where A is the rotation matrix Similar to the translations example, we actually want to go back to Rearranging gets you A^-1 = = A^-1 where A^-1 is a clockwise rotation ^This shows why when plotting on geogebra, it rotates it clockwise (if no extra manipulation is done, because we never found out how to 'go back to the original axes/point). Just subbing in = A straight off directly in your equations is equivalent to saying The 'new' point = A^-1 where (X', Y') is the old point But this is clockwise.... which is why your new points are going clockwise

That’s literally the definition of the matrix of a linear transformation, it’s how it’s defined

@@drpeyam Thx !

Sure! It’s then just the matrix followed by a rotation!

you can apply any matrix to another matrix, since a matrix is just a list of its columns (or a list of rows, both views are necessary), so you apply each column of the right matrix to the left matrix to produce new columns in the result

Yeah

yes. It depends on whether you just want to find the new coordinates (X', Y') or whether you want to find the equation of said rotation. The latter requires 'rotating back clockwise' I will copy and paste an explanation I provided to another comment If you think about the fixed coordinate of (X', Y') compared to (x, y) then using the rotation matrix would work But substituting these values back into an equation, say that equation of the ellipse will not give you the anti-clockwise rotation of said graph. To explain the reason for that, I will go back to basics with transformations relating to just horizontal translations. If we take an arbitrary point (x,y) and move it to the right a units to the point (X', Y'), this new point will be (x + a, y) But does this mean the graph f(x) shifted to the right 8 units is f(x + a) = (x+a)^2. The answer to that is no, because essentially our point of view has changed, we want to input X' = x+a to map back to x Rearranging X' = x + a gets you x = X' - a so we should actually be considering f(X' - a) or using x as the variable again f(x - a) when shifting to the right a units,. We can consider a similar argument for horizontal translations to the left, vertical translations both up and down, dilations in both directions or a combination of any of these sorts of transformations. We always have to think of the 'opposite' operation when thinking about transforming graphs (or equations). So if we come back to rotations = A where A is the rotation matrix Similar to the translations example, we actually want to go back to Rearranging gets you A^-1 = = A^-1 where A^-1 is a clockwise rotation ^This shows why when plotting on geogebra, it rotates it clockwise.

Yeah! I think there’s a neat 3b1b video on that!

Not length, but coordinate

Михаил Ужов the y coordinate is still positive

@@akivas2034 Yes but the x coordinate isn't. You're literally going backwards from the origin, and by definition that results in a negative coordinate. Remember, rotations preserve only the lengths, not the coordinates themselves.

Hilbert Black now I get it, thank you!

Check out linear algebra, you’d be surprised 😉

What’s that?

Steinitz Replacement Theorem in Linear Algebra en.wikipedia.org/wiki/Steinitz_exchange_lemma ☝🏻