Math 3130
Spring 2016 Abrams
Practice Exam for Exam 2 - SOLUTIONS
Show all your work. No credit will be given for answers which are not accompanied by supporting computations. Use the back of the sheet if you need more space. A non-graphing calculator with no QWERTY keyboard
is allowed, but definitely not needed. Circle answers when appropriate. Good luck.
1. (6 pt total) Linear Transformations.
(a) Define precisely: If T : V → W is a function from the vector space V to the vector space W ,
the T is called a linear transformation in case:
and
(i)
T (u + v) = T (u) + T (v) for all u, v in V
(ii)
T (ku) = kT (u) for all u in V and k in R.
(b) Use the definition you gave in part (a) to show that the function T : R3 → R3 given by the formula
T ((x, y, z)) = (x + y, y, z + 2x) is a linear transformation.
For property (i): Pick u = (x, y, z) and v = (x0 , y 0 , z 0 ) in R3 . Then
T (u + v) = T ((x, y, z) + (x0 , y 0 , z 0 ))
= T ((x + x0 , y + y 0 , z + z 0 ))
(definition of addition in R3 )
= ((x + x0 ) + (y + y 0 ), (y + y 0 ), (z + z 0 ) + 2(x + x0 ))
(definition of T )
0
0
0
0
0
= (x + y + x + y , y + y , z + 2x + z + 2x )
(arithmetic)
= (x + y, y, z + 2x) + (x0 + y 0 , y 0 , z 0 + 2x0 )
(definition of addition in R3 )
= T ((x, y, z)) + T ((x0 , y 0 , z 0 ))
= T (u) + T (v)
For property (ii): Pick u = (x, y, z) in R3 and k in R Then
T (ku) = T (k(x, y, z))
= T ((kx, ky, kz))
(definition of scalar multiplication in R3 )
= (kx + ky, ky, kz + 2kx)
(definition of T )
= (k(x + y), ky, k(z + 2x))
(arithmetic)
= k(x + y, y, z + 2x)
(definition of scalar multiplication in R3 )
= kT (u)
Since properties (i) and (ii) are both true, we conclude that T is a linear transformation
2. (1 pt each) Circle the functions T which are linear transformations.
ALL OF THESE FUNCTIONS ARE LINEAR TRANSFORMATIONS. (See class notes and homework.)
(i) Let A be an m × n matrix. Define T : Rn → Rm by T (v) =Av for each column vector v in Rn .
(ii) T : R2 → R2 where T rotates each vector in R2 π/4 radians clockwise around the origin.
(iii) T : P3 → P2 where T (p(x)) = p0 (x).
(iv) T : P2 → P2 where T (a0 + a1 x + a2 x2 ) = a0 + a1 (x + 3) + a2 (x + 3)2 .
1
3. (3 pt) If B = {1, x, x2 } is the standard basis for P2 , and T : P2 → P2 is a linear transformation with the
property that T (1) = x, T (x) = 3x2 + 1, and T (x2 ) = 1 + 2x, compute T (4 − x + 2x2 ).
T (4 − x + 2x2 ) = T (4 · 1 − x + 2x2 ) = 4T (1) − T (x) + 2T (x2 ) (since T is a linear transformation)
= 4(x) − (3x2 + 1) + 2(1 + 2x)
(using the given information)
= 1 + 8x − 3x2
(arithmetic).
4. (3 pt total) (i) Let V be an n-dimensional vector space with basis B = {u1 , u2 , ..., un } and let W be
an m-dimensional vector space with basis B 0 = {w1 , w2 , ..., wm }. Let T : V → W be any linear transformation.
Let A denote the matrix
A = [T (u1 )]B 0 ... [T (u2 )]B 0 ... · · · ... [T (un )]B 0
(ii) (Do just ONE of these.)
Then for every vector v in V,
A[v]B = [T (v)]B 0
OR
T (v) =
GB 0 (A[v]B )
(iii) Draw a diagram (containing four vector spaces, and four arrows) which presents the information
given in parts (i) and (ii). Make sure to clearly include and label all appropriate sets, functions, vector spaces,
arrows, etc ...
[ ]B
V
↓
Rn
T
→
→
W
↑ GB 0
Rm
mult by
A
5. (8 pt total) Define T : P3 → P3 by setting T (p(x)) = 4p0 (x) − 2p(x) for every p(x) ∈ P3 . (So, for example,
T (x2 ) = 8x − 2x2 .). It is not hard to show that T is a linear transformation. (You need not show it here, but
you may assume it’s true.)
(a) Let B = {1, x, x2 , x3 } be the standard basis for P3 . Find the matrix of T with respect to B. Call
this matrix A.


−2
 0 

T (1) = 4 · 0 − 2 · 1 = −2 = −2 · 1 + 0x + 0x2 + 0x3 , so [T (1)]B = 
 0  . Similarly, the computations
0
2
3
2
2
T (x) = 4 − 2x = 4 · 1 + −2x + 0x + 0x , T (x ) = 8x = 2x = 0 · 1 + 8x + −2x2 + 0x3 , and T (x3 ) = 12x2 − 2x3 =

−2 4
0
0
 0 −2 8
0 
.
gives that A = 
 0
0 −2 12 
0
0
0 −2

0 · 1 + 0x + 12x2 + −2x3
(b) Let q(x) = 2 − x2 + x3 . Find the coordinate vector [q(x)]B .


2
 0 

q(x) = 2 · 1 + 0x − 1x2 + 1x3 , so [q(x)]B = 
 −1  .
1
2
(c) Compute the matrix product A[q(x)]B .


 
−2 4
0
0
2
 0 −2 8

 
0 

 0  = 
 0
0 −2 12   −1  
0
0
0 −2
1

−4
−8 

14 
−2
(d) Compute T (q(x)).
T (q(x)) = T (2−x2 +x3 ) = 4(2−x2 +x3 )0 −2(2−x2 +x3 ) = 4(−2x+3x2 )−4+2x2 −2x3 = −4−8x+14x2 −2x3
(e) Find the coordinate vector [T (q(x))]B .
TWO WAYS. The Linear Transformation Theorem says this is just A[q(x)]B , which from part


−4
 −8 

(c) is 
 14  . OR we can compute this same coordinate matrix directly from part (d) to get the same answer!
−2
6. (12 pt total) Let T : R2 → R2 be the linear transformation defined by T ((x, y)) = (x + 7y, 3x − 3y).
(a) Find the matrix [T ]B of T with respect to the standard basis B = {e1 , e2 } of R2 . Call this matrix A.
T (e1 ) = T ((1, 0)) = (1 + 7 · 0, 3 · 1 − 3 · 0) = (1, 3)
T (e2 ) = T ((0, 1)) = (0 + 7 · 1, 3 · 0 − 3 · 1) = (7, −3)
Since we are using standard basis, these computations give [T (e1 )]B =
1 7
A=
3 −3
(b) Let C be the basis C = {(1, 1), (0, 1)} of R2 .
this matrix M .
1
3
and [T (e2 )]B =
7
−3
. Thus
Find the matrix [T ]C of T with respect to C.
Call
T ((1, 1)) = (1 + 7 · 1, 3 · 1 − 3 · 1) = (8,
0). We
solve (8, 0) = k1 (1, 1) + k2 (0, 1) for k1 and k2 ; this easily
8
gives k1 = 8 and k2 = −8. Thus [T ((1, 1))]C =
. Similarly, T ((0, 1)) = (x + 7 · 1, 3 · 0 − 3 · 1) = (7, −3).
−8
We
solve (7, −3) = k1 (1,1) + k2 (0, 1)for k1 and k2 ; this easily gives k1 = 7 and k2 = −10. Thus [T ((0, 1))]C =
7
8
7
. Thus M =
.
−10
−8 −10
(c) Let P denote the transition matrix from C to B. Find P.
1 0 .
P = [(1, 1)]B .. [(0, 1)]B =
since B is standard basis.
1 1
(d) Let Q denote the transition matrix from B to C.
Find Q.
There are two ways to do this. One is to use the theorem
P −1 where P is given in the
that says Q = previous part of this question. The other is to do it directly, Q = [e1 ]C ... [e2 ]C . Using this second way,
1
we write (1, 0) = k1 (1, 1) + k2 (0, 1) and solve for k1 , k2 . This gives k1 = 1, k2 = −1, so that [e1 ]C =
.
−1
0
1 0
Similarly (and even more easily), we get that [e2 ]C =
. So Q =
.
1
−1 1
3
(e) Give an equation which relates the four matrices A, M, P, and Q. (Your answer will contain these
four letters.)
The theorem that connects these (Theorem 8.5.2) says that M = QAP . You can check that the multiplication
of the three matrices on the right really does give M .
7. (3 pt) Let T : R2 → R2 be the linear transformation ’rotate through 45 degrees counterclockwise’. Find
the matrix of T with respect to the standard basis of R2 .
√
√
We compute T (e1 ) and T (e2 ) by simply doing the geometric rotation. So we get T (e1 ) = ( 22 ,√22 )√(it’s the
vector of length 1 in the direction 45 degrees counterclockwise from the x-axis), while T (e2 ) = (− 22 , 22 ) (it’s
!
√
the vector of length 1 in the direction 45 degrees counterclockwise from the y-axis). Thus [T (e1 )]B =
√
√ !
√ !
2
2
−
−√ 22
√2
√2
.
So
the
matrix
of
T
in
the
standard
basis
is
while [T (e2 )]B =
2
2
2
2
2
2
√2
2
2
2
8. (2 pt) Let T : R3 → R3 be the projection of R3 into the xz-plane. (So T ((x, y, z)) = (x, 0, z).) Find a
basis of the kernel of T .
Since T ((0, 1, 0)) = (0, 0, 0), we have that (0, 1, 0) is in the kernel of T . But since T applied to the other
two standard basis vectors is nonzero, the kernel is one dimensional. So {(0, 1, 0)} is a basis of the kernel.
9. (1 pt each) True - False
(a)
TRUE Suppose V and W are finite dimensional vector spaces. By appropriately using
coordinate vectors, it is possible to view any linear transformation T : V → W as a matrix multiplication. (This
is the point of the diagram given above.)
(b)
TRUE Suppose T : V → W is a linear transformation, and suppose B = {b1 , b2 , ..., bn } is a
basis of V . If you are told the values of T (b1 ), T (b2 ), ..., T (bn ), then you have enough information to compute
T (v) for every v in V . (Use properties of linear transformations and the fact that a basis spans the vector space.)
(c)
TRUE It is possible to have an n × n matrix A with the property that for every invertible
n × n matrix P we have A = P −1 AP. (For instance, the identity matrix A = In has this property. BUT NOT
ALL matrices have this property (i.e. there are some matrices which do NOT have this property.))
(d)
TRUE If V is a vector space and k is any scalar, then the function T : V → V given by defining
T (v) = kv for every v ∈ V is a linear transformation. (Easily shown.)
(e)
FALSE If T : P∞ → P∞ is the linear transformation given by T (f (x)) = f 0 (x) (i.e., T is
the ’derivative’ linear transformation on polynomials), then the kernel of T is {0}. (The kernel consists of all
constant polynomials.)
(f)
TRUE If T : R∞ → R∞ is the linear transformation given by T ((a0 , a1 , a2 , . . . )) = (0, a0 , a1 , a2 , . . . )
(i.e., T is the “right shift” linear transformation), then the kernel of T is the zero vector of R∞ . (The zero vector
of R∞ is clearly the only vector which can produce the zero vector of R∞ as an output.)
(g)
TRUE If M, A, and P are n × n matrices for which M = P −1 AP , then det(M ) = det(A).
(Shown in class, or just use properties of determinants.)
4