Question about linear transformations on the plane:?
That (-6 0 0 -5) is not a projection, is clear due its determinant is not zero. To find the orthogonal projection T1 onto the line -6x1 - 5x2 = 0 I used that T1(x)=x for every x=(x1 x2) on the line and that for every x on line 6x2-5x1=0 you get T1(x)=0, as the second line is orthogonal to the given line and crosses it at the origin. Let T1 be (a b c d) Then by what I said you get the following linear equations for the coefficients of T1: c+5/6 d=0 a+5/6 b=0 a-6/5 b=1 c-6/5 d=-6/5 => T1= (25/61 -30/61 -30/61 36/61)
Linear algebra help with transformations on a plane?
(1) If P is the projection matrix then x → Px. Let P = ( a, b ) where a & b are column vectors Then e1 = ( 1, 0 ) → a and e2 = ( 0, 1 ) → b So to find columns of P we just need to find the vector projections of e1 & e2 A vector in direction of 2x1+x2=0 is ( 1, −2 ). Unit vector = d = ( 1/√5, −2/√5 ) Scalar projection of e1 onto d is ( 1, 0 )•( 1/√5, −2/√5 ) = 1/√5 Vector projection in direction d is (1/√5)( 1/√5, −2/√5 ) = ( 1/5, -2/5 ) In a similar way projection of e2 onto d is (−2/√5)( 1/√5, −2/√5 ) = ( −2/5, 4/5 ) T = | 1/5 -2/5 | …...| -2/5 4/5 | Another way to do this is to use the formula T = dd' If { … } means row vector then T = ( 1/√5, −2/√5 ){ 1/√5, −2/√5 } = as before dd' is just a special case with m=1 of the formula P=A(A'A)‾¹A for projecting onto the space spanned by the columns of the nXm matrix A. The other answers are : (2) For reflection in the line a'x=0 where a is a unit vector the reflection matrix is I−2aa'. This is an orthogonal matrix with determinant = −1. This result is easy to remember and also applies in 3D for reflection in a the plane a'x=0. Note this time the relevant vector a is the normal vector to the line. Normal to line 4x1−3x2 = 0 is a = (4,-3) Normalised, this is a = (1/5)(4,−3) aa' = (1/25)| 16 -12 | …...............| -12 9 | T = I−2aa' = | -7/25 24/25 | ….................| 24/25 7/25 | (3) Repeat the above for a = (6,1) to get T1 = | -35/37 -12/37 | ….....| -12/37 35/37 | Then for a = (4,1) to get T2 = | -15/17 -8/17 | ….....| -8/17 15/17 | Applying T2 and then T1 is same as applying matrix T1.T2 T1.T2 = | 621/629 100/629 | …..........| -100/629 621/629 | The matrix for rotating a vector by θ anticlockwise is | cos(θ) -sin(θ) | | sin(θ) .cos(θ) | Comparing this with above, T1.T2 represents a rotation if there exits θ such that cos(θ)=621/629 and sin(θ)=-100/629. These values are consistent because (621/629)²+(100/629)²=1. With sin(θ)<0 and cos(θ)>0, θ is Q4 and θ = 2π−sin‾¹(100/629) = 6.1235 rads If you intended T2.T1 you'll have to reverse this. Answer then = +0.1597 rads
Linear Algebra: For what value of h is y in the plane spanned by v1 and v2?
take the cross product of v1 and v2 to find the direction perp to the plane they define then take the dot product of y, the value of h that yield y dot [v1 cross v2] =0 is the proper value of h [v1 cross v2] = (-1,8-12) y dot[v1crossv2]=-12+96-12h =0 h=7