Description

Dane Johnson
Exercise 1
Consider the linear transformation T : P3 → P2 defined by T(p) = p0. For P3 fix the basis {1 + x,1 − x,x + x2,x2 + x3} and for P2 use the canonical basis for P2. We want to find the matrix representation of T with respect to these bases.
A linear transformation is characterized by its effect on the set of basis vectors we have chosen, so we first perform these calculations.
T(1 + x) = 1(1) + 0(x) + 0(x2)
T(1 − x) = −1(1) + 0(x) + 0(x2)
T(x + x2)1(1) + 2(x) + 0(x2)
T(x2 + x3) = 0(1) + 2(x) + 3(x2) .
Since we are using the canonical basis for the codomain of this transformation, these calculations already show the coefficients of the resulting vectors with respect to the basis we have fixed for P2. So we take the coefficients above for use as columns in the matrix representation of the linear transformation. The matrix representation is then,
1 −1 1 0
0 0 2 2
0 0 0 3
To check our work, let p(x) = c1(1+x)+c2(1−x)+c3(x+x2)+c4(x2+ x3) ∈ P3. We expect T(p) = (c1 −c2 +c3)+(2c3 +2c4)x+3c4x2. Using the matrix,
.
This is what we would expect for the transformation to look like in matrix representation.
Exercise 2
Let us find the dimension of S = span{v1,v2,v3} ⊆ R3, where v1 = (1,0,1), v2 = (1,1,0), v3 = (1,−1,2) .
Some trial and error gives the equation 2v1 − 1v1 = v3 so that 2v1 − 1v1 − 1v3 = 0, where 0 = (0,0,0) ∈ R3. Since we have found a nontrivial solution to the equation α1v1 + α2v2 + α3v3 = 0, the vectors are dependent and since v3 ∈ span{v1,v2}, span{v1,v2} = span{v1,v2,v3}. Note further that v2 ∈/ span{v1} or vice versa, so we conclude that the dimension of span{v1,v2,v3} is equal to the dimension of span{v1,v2}, which is 2, and that {v1,v2} is basis for S.
To expand this basis to a basis for R3, try v4 = (0,0,1). To see if span{v1,v2,v4} = R3, let (x1,x2,x3) ∈ R3. Using matrix algebra we have
1 1 0 x1 1 0 0 x1 − x2 
0 1 0 x2 ∼ 0 1 0 x2  .
1 0 1 x3 0 0 1 x2 + x3 − x1
So taking α1 = x1 − x2,α2 = x2,α3 = x2 + x3 − x1, this implies that (x1,x2,x3) = α1v1 + α2v2 + α3v4. Therefore {v1,v2,v4} spans R3.
Exercise 3
Let T : R3 → R3 be the linear transformation given by the orthogonal projection of the vector (x1,x2,x3) ∈ R3 onto the plane x2 = 0.
(1) First we find the matrix representation of T. Since we seek to keep the first and third components of each vector unaffected by the transformation, but ensure that the second component becomes 0 under the transformation, we apply this requirement to each basis vector. Take the canonical basis for both the domain and codomain of T. Then we have T((1,0,0)) = (1,0,0),T((0,1,0)) = (0,0,0),T((0,0,1)) = (0,0,1) and the matrix representation
1 0 0 1 0 0x1 x1
0 0 0 so that 0 0 0x2 =  0  as desired.
0 0 1 0 0 1 x3 x3
.
(3)Since range(T) = {(x1,0,×2) | x1,x2 ∈ R}, and we have been using the canonical basis for R3 so far in this exercise, let us propose
{(1,0,0),(0,0,1)}
as a basis for the range of T. Clearly, then by taking α1 = x1 and α3 = x3 we have for any (x1,0,×3) ∈ range(T) that α1(1,0,0)+α3(0,0,1) = (x1,0,×3).
Exercise 4
Let B = {(1,0,1),(0,2,2),(3,0,1)}, where the coordinates of these vectors are with respect to the canonical basis for R3. First consider the linear transformation T : R3 → R3 (wrt the canonical basis in both sets) such that
T((1,0,0)) = (1,0,1), T((0,1,0)) = (0,2,2), T((0,0,1)) = (3,0,1) .
The matrix representation of this transformation is then
1
0

1 0
2
2 .
To check this, let (a,b,c) ∈ R3. Then, using ei for canonical basis shorthand,
T((a,b,c)) = T(ae1 + be2 + ce3) = aT(e1) + bT(e2) + cT(e3)
= a(1,0,1) + b(0,2,2) + c(3,0,1) = (a + 3c,2b,a + 2b + c).
Next using matrices,
.
So the two calculations agree with one another.
In order to find the matrix representation of the transformation from B to the the canonical basis, we would need to have a matrix that inverts the linear transformations effect on a vector. But this suggests that we should just take the matrix

in order to represent the linear transformation from B back to the canonical basis.
Exercise 5
Consider the linear transformation given by a clockwise rotation of π/4 in the plane spanned by e1,e2 along the e3 axis. Call this linear transformation T : R3 → R3. Applying the transformation to the canonical basis vectors (and using radians instead of degrees to achieve the same results) we have the mappings
.
For now we are using the canonical basis, so to represent this transformation using a matrix, we use the resulting vectors as the columns of the matrix that represents the transformation. This gives:

Note that these vectors are with respect to the canonical basis, so we need to compute their coordinates with respect to B. Recall that we have a matrix from exercise 4 that will assist in the process:

The matrix representation of the transformation with respect to the basis B is then given by:

Note that we could also have found this matrix in one step by multiplying: