..READ THIS
FIRST..Problems
about Linear transformations
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FIRST
If a problem is solved. It is not 'the' answer.No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.
Problems about Linear transformations
Level 1 problems
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Is the following transformation linear t : R x R --> R x R : (x,y) --> (2x - y, 0)For all vectors (x,y) , (x',y') and all real numbers r, s we have t(r(x,y) + s(x',y')) = t(rx+sx',ry+sy') =(2rx + 2sx' -ry -sy',0) Otherwise r.t(x,y) + s.t(x',y') = r.(2x - y, 0) + s.(2x' - y', 0) = (2rx + 2sx' -ry -sy',0) Hence t(r(x,y) + s(x',y')) = r.t(x,y) + s.t(x',y') and t is a linear transformation. -
Find the matrix of the following linear transformations with respect to a natural basis. t1: R x R --> R x R : (x,y) --> (2x - y, 0) t2: R x R --> R x R : (x,y) --> (2x - y, x) t3: R x R x R --> R x R x R: (x,y,z) --> (2x - y, 0, y +z) t4: R x R x R --> R x R x R: (x,y,z) --> (0, 0,y)The basis of R x R is ( (1,0) , (0,1) ) t1(1,0) = (2,0) and has coordinates (2,0) t1(0,1) = (-1,0) and has coordinates (-1,0) The matrix of t1 is then [2 -1] [0 0] t2(1,0) = (2,1) and has coordinates (2,1) t2(0,1) = (-1,0) and has coordinates (-1,0) The matrix of t2 is then [2 -1] [1 0] The basis of R x R x R is ( (1,0,0), (0,1,0), (0,0,1) ) t3(1,0,0) = (2,0,0) and has coordinates (2,0,0) t3(0,1,0) = (-1,0,1) and has coordinates (-1,0,1) t3(0,0,1) = (0,0,1) and has coordinates (0,0,1) The matrix of t3 is then [2 -1 0] [0 0 0] [0 1 1] t4(1,0,0) = (0,0,0) and has coordinates (0,0,0) t4(0,1,0) = (0,0,1) and has coordinates (0,0,1) t4(0,0,1) = (0,0,0) and has coordinates (0,0,0) The matrix of t4 is then [0 0 0] [0 0 0] [0 1 0] -
Find the image of the vector (-2,4) with respect to the following linear transformations.
Do this first without the matrix, and next with the matrix of t.t1 : R x R --> R x R : (x,y) --> (2x - y, 0) t2 : R x R --> R x R : (x,y) --> (2x - y, x)t1(-2,4)=(-8,0) With the matrix of t1 [2 -1][-2] [-8] [0 0][4] = [ 0] t2(-2,4)=(-8,-2) With the matrix of t2 [2 -1][-2] [-8] [1 0][4] = [-2] -
Find the eigenvalues and the characteristic vectors of t1 : R x R --> R x R : (x,y) --> (2x - y, 0) t2 : R x R --> R x R : (x,y) --> (2x - y, x) t3 : R x R x R --> R x R x R: (x,y,z) --> (2x - y, 0, y +z) t4 : R x R x R --> R x R x R: (x,y,z) --> (0, 0,y)First, we calculate the eigenvalues r of t1 The characteristic equation is |2-r -1| | | = 0 <=> -r(2-r) = 0 |0 0-r| The eigenvalues are r = 0 and r = 2 For r=0 the characteristic vectors are the non trivial solutions of the system 2x - y = 0 0x +0y = 0 The characteristic vectors are all the multiples of (1,2). For r=2 the characteristic vectors are the non trivial solutions of the system 2x - y = 2x 0x +0y = 2y This system is equivalent with 0x + y = 0 The characteristic vectors are all the multiples of (1,0). You find the eigenvalues and characteristic vectors of t2 in the same way. Next, we calculate the eigenvalues r of t3 The characteristic equation is |2-r -1 0 | |0 -r 0 | = 0 <=> (r-1).r.(2-r) = 0 |0 1 1-r| The eigenvalues are r = 0 , r = 1 and r = 2 For r=0 the characteristic vectors are the non trivial solutions of the system 2x - y + 0z = 0 0x +0y + 0z = 0 0x + y + z = 0 This system is equivalent with 2x - y = 0 y + z = 0 The characteristic vectors are all the multiples of (1,2,-2). For r=1 the characteristic vectors are the non trivial solutions of the system 2x - y + 0z = x 0x +0y + 0z = y 0x + y + z = z This system is equivalent with x - y = 0 y = 0 The characteristic vectors are all the multiples of (0,0,1). For r=2 the characteristic vectors are the non trivial solutions of the system 2x - y + 0z = 2x 0x +0y + 0z = 2y 0x + y + z = 2z This system is equivalent with y=0 z=0 The characteristic vectors are all the multiples of (1,0,0). You find the eigenvalues and characteristic vectors of t4 in the same way.
Level 2 problems
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The rotation, with angle u radians (u not 0), about a fixed point o is a linear transformation of the vector space of all vectors in a plane.
Find the matrix of a rotation with respect to an orthonormal basis (e1,e2) in the planeWrite this matrix for u = pi, and calculate the eigenvalues and the characteristic vectors.
The coordinates of the image of e1 are (cos(u),sin(u)) The coordinates of the image of e2 are (cos(u+pi/2),sin(u+pi/2)) or ( -sin(u) , cos(u) ) The matrix of the rotation is [cos(u) -sin(u)] [sin(u) cos(u)] The characteristic equation is [cos(u)-r -sin(u)] [ ] = 0 [sin(u) cos(u)-r] <=> (cos(u) - r)2 + sin2 u = 0 <=> r2 -2.cos(u).r + 1 = 0 The discriminant is 4.cos2 (u) -4 is negative for all u not 0. So, there are no real eigenvalues and no real characteristic vectors. -
Let A = matrix of a linear transformation t.
Prove that t has an eigenvalue 0 if and only if A is singular.
If t has an eigenvalue 0, there is a characteristic vector v such that
t(v) = 0.v <=> t(v) = 0 .Let (x,y,z) be the coordinates of v. (x,y,z) is not (0,0,0).
[x] [0] A.[y] =[0] has a solution different from (0,0,0) [z] [0] From the theory about homogeneous systems, we know that this is equivalent with det(A) = 0 <=> A is singular. -
The linear transformation t has, with respect to an orthonormal basis (e,u) , the matrix [m m] [1 2]a) Calculate m such that (1,1) are the coordinates of a characteristic vector v.
b) Calculate the coordinates of a characteristic vector w, lineair independent of v and such that w is a unit vector.
c) What is the matrix of t if we take v and w as a new basis in the vector space.
d) Calculate the coordinates of e and u with respect to this new basis.
a)(1,1) are the coordinates of a characteristic vector v
<=> [m m][1] = r. [1] <=> 2 m = r [1 2][1] [1] 3 = r <=> m = 3/2 and r = 3b)The eigenvalues are the solutions of|3/2 -r 3/2| = 0 <=> ... <=> r = 3 or r = 1/2 | 1 2- r|Each characteristic vector corresponding with r = 1/2 is a solution of the system/ 3/2 x + 3/2 y = 1/2 x | <=> ... <=> 2x+3y = 0 \ x + 2 y = 1/2 yThe characteristic vectors have coordinates (3t,-2t).
The magnitude has to be = 1._____________ | 2 2 ____ \| 9 t + 4 t = 1 <=> t = 1 /V 13 ____ ____ A unit vector w has coordinates (3/ V 13 , -2/ V 13 )c)[ 3, 0 ] [ 0, 1/2 ]d) We havev = e + u and ____ ___ w = 3/ V 13 e -2/ V 13 u Solving this for w end u, we find ____ e = 2/5 v + V 13 /5 w ____ u = 3/5 v - V 13 /5 w The coordinates of e and u with respect to this new basis are ____ e =(2/5 , V 13 /5 ) ____ u =(3/5 ,- V 13 /5) -
The vector v = (1,-1) is a characteristic vector of the linear transformation with matrix A = [4 3] [7 8] a) What is the corresponding eigenvalue? b) Calculate 1998 A v[4 3][ 1] = [ 1] [7 8][-1] [-1] So the eigenvalue is 1 and A transforms v in v 1998 A v = v -
Calculate all eigenvalues of the linear transformation t with matrix [1 u v] [0 1 u] [0 0 1]Here, u and v are constant, non zero real numbers. Give for each eigenvalue the dimension of the associated vector space.
The characteristic equation is(1-r)3 = 0 ; so the only eigenvalue r = 1 The coordinates of the characteristic vectors are the solutions of the system 0x + uy + vz = 0 0x + 0y + uz = 0 0x + 0y + 0z = 0 The solutions are all real multiples of (1,0,0) . The dimension of the associated vector space is 1.
Level 3 problems
Let A = matrix of a linear transformation t and C is a regular matrix. Prove that A and B = C-1 .A.C have the same eigenvalues.
The eigenvalues of B are the roots of the characteristic equation det(B - rI) = 0 (I is the identity matrix) <=> det(C-1 .A.C- rI) = 0 <=> det(C-1.A.C- r.C-1.I.C) = 0 <=> det(C-1.(A - rI).C) = 0 <=> det(C-1).det(A - rI).det(C) = 0 det(C) and det(C-1) are not 0 since C is a regular matrix. <=> det(A - rI) = 0 And the roots of this equation are the eigenvalues of A.-
A of a linear transformation t is a 2x2 matrix with two different eigenvalues.
Show that the characteristic vectors corresponding with different eigenvalues are linear independent.
Prove that, if you choose two linear independent characteristic vectors, as a new basis, the matrix of t is a diagonal matrix.
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The product of all eigenvalues of a matrix A is not 0.
Show that A is a regular matrix.Suppose that A is a singular matrix, then det(A) = 0 and the characteristic equation det(A - r I)=0 has the solution r=0. Thus, there is at least one eigenvalue = 0 and then the product of all eigenvalues of a matrix A is 0.0
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