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Recitation Notes 1 Administrative Matters Spring 16, 21-241: Matrices and Linear Transformations

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Recitation Notes 1 Administrative Matters Spring 16, 21-241: Matrices and Linear Transformations
Recitation Notes
Spring 16, 21-241: Matrices and Linear Transformations
February 2, 2016
Abstract
[TOPICS COVERED] Forms of linear equations. Linear transformations.
1
Administrative Matters
• Notes on homework: Make a clear distinction between the additive inverse, −v, and the scalar
product, (−1)v, even though they are equal.
2
Definitions
1. Normal vector
2. Linear transformation
3
Problems
1. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 1.3.7-8. Write the equation
of the plane passing through P with normal vector n in
• normal form and
• general form
for the following:


3
(a) P = (0, 1, 0), n =  2 
1
 
2

(b) P = (3, 0, −2), n = 5 
0
Solution.
1
(a) Any point (x, y, z) on the line must satisfy the equation
     
3
x
0
 2  ·  y  −  1  = 0
1
z
0
Thus

      
3
x
3
0
 2 · y = 2 · 1 
1
z
1
0
Thus the normal form is
   
3
x
 2  ·  y  = (3)(0) + (2)(1) + (1)(0) = 2,
1
z
and the general form is
3x + 2y + 1z = 2 =⇒ 3x + 2y + z = 2.
(b) Any point (x, y, z) on the line must satisfy the equation

    
3
x
2
 5  ·  y  −  0  = 0
−2
z
0
Thus

     
3
2
x
2
 5 · y = 5 · 0 
−2
0
z
0

Thus the normal form is
   
x
2
 5  ·  y  = (2)(3) + (5)(0) + (0)(−2) = 6,
z
0
and the general form is
2x + 5y + 0z = 6 =⇒ 2x + 5y = 6.
Page 2
2. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 6.4.8-11. Determine whether
T is a linear transformation.
(a) T : P2 → P2 defined by T (a + bx + cx2 ) = (a + 1) + (b + 1)x + (c + 1)x2 .
(b) T : P2 → P2 defined by T (a + bx + cx2 ) = a + b(x + 1) + b(x + 1)2 .
(c) T : F → F defined by T (f ) = f (x2 ).
(d) T : F → F defined by T (f ) = (f (x))2 .
Solution.
(a) No. The zero vector is 0 + 0x + 0x2 . However, T (0 + 0x + 0x2 ) = 1 + x + x2 is not
the zero vector.
(b) Yes. There are two things to check.
Step 1: Let k be a constant. Then
T (k(a + bx + cx2 )) = T (ka + kbx + kcx2 )
= ka + kb(x + 1) + kb(x + 1)2
= k(a + b(x + 1) + b(x + 1)2 )
= kT (a + bx + cx2 )
Step 2:
T ((a1 + b1 x + c1 x2 ) + (a2 + b2 x + c2 x2 ))
= T ((a1 + a2 ) + (b1 + b2 )x + (c1 + c2 )x2 )
= (a1 + a2 ) + (b1 + b2 )(x + 1) + (b1 + b2 )(x + 1)2
= (a1 + b1 (x + 1) + b1 (x + 1)2 ) + (a2 + b2 (x + 1) + b2 (x + 1)2 )
= T (a1 + b1 x + c1 x2 ) + T (a2 + b2 x + c2 x2 )
Therefore T is a linear transformation.
(c) Yes. There are two things to check.
Step 1: Let k be a constant. Then
T (kf (x)) = T ((kf )(x)) = (kf )(x2 ) = kf (x2 ) = kT (f (x)).
Step 2: Let f and g be functions. Then
T (f (x) + g(x)) = T ((f + g)(x)) = (f + g)(x2 ) = f (x2 ) + g(x2 ) = T (f (x)) + T (g(x)).
(d) No. Note that T (x) = x2 , so 2T (x) = 2x2 . On the other hand, T (2x) = (2x)2 = 4x2 .
Clearly they are not the same function, so T (2x) 6= 2 T (x).
Page 3
3. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 6.4.15. Let T : R2 → P2 be
a linear transformation of which
1
3
T
= 1 − 2x and T
= x + 2x2 .
1
−1
−7
a
Find T
and T
.
9
b
Solution.
Method 1: Note: This method is best suited when you are trying to find
the transform
−7
−7
of a specific vector. For example, the transform of
, i.e. T
.
9
9
We know that T (cx + dy) = cT (x) + dT (y). Hence we can obtain the answer by solving
−7
1
3
=c
+d
.
9
1
−1
We have
1 3 −7 −−−−−−−−→ 1 3 −7
r2 7→r2 −r1
1 −1 9
0 −4 16
−−−−−−−−→ 1 3 −7
r2 7→−r2 /4
0 1 −4
1 0 5
−−
−
−
−
−
−
−
→
r1 7→r1 −3r2
0 1 −4
Thus
T
−7
9
3
1
−4
=T 5
−1
1
3
1
−4 T
=5T
−1
1
= 5(1 − 2x) − 4(x + 2x2 )
= 5 − 19x − 8x2
Method 2: Note: This method allows us to find the transform of any given vector.
We haven’t yet learnt what a basis
guess that every
in R2 can
is, but one
could
vector
1
0
1
0
be expressed as a combination of
and
. Thus if we find T
and T
,
0
1
0
1
then using the identity T (cx + dy) = cT (x) + dT (y), we can find any
a
1
0
1
0
T
=T a
+b
= aT
+b
b
0
1
0
1
We solve
c1
1
1
+ d1
3
−1
=
1
0
and
Page 4
c2
1
1
+ d2
3
−1
=
0
1
simultaneously.
1 0
1 3 1 0 −−−−−−−−→ 1 3
r2 7→r2 −r1
1 −1 0 1
0 −4 −1 1
−−−−−−−−→ 1 3 1 0
r2 7→−r2 /4
0 1 14 − 14
3
1 0 14
−−
−
−
−
−
−
−
→
4
r1 7→r1 −3r2
0 1 14 − 14
Thus we have
1
1 1
3
1
+
=
0
4 1
4 −1
and
0
1
3
=
4
1
1
1
−
4
3
−1
Thus
T
T
1
0
0
1
1
T
4
3
= T
4
=
1
1
1
T
4
3
−1
1
1
1 1
1
= (1 − 2x) + (x + 2x2 ) = − x + x2
4
4
4 4
2
1
1
1
− T
4
3
−1
3
1
1 7
1
= (1 − 2x) − (x + 2x2 ) = − x − x2
4
4
2 4
2
0
1
+
Finally
T
a
b
=aT
1
0
+b T
=a
1 1
1
− x + x2
4 4
2
+b
1 7
1
− x − x2
2 4
2
4. Analyze the proof for Theorem 6.14(a). Can you give a reason for each step? Suppose we
were lazy to prove that 0x = 0. How else can we prove the theorem?
Solution. The proof is as follows:
T (0) = T (0v) = 0T (v) = 0
The proof relies on the property 0x = 0, which we have yet to justify. Let us prove it.
0x = (0 + 0)x
=⇒ 0x = 0x + 0x
(distributivity)
=⇒ (−0x) + 0x = (−0x) + 0x + 0x
=⇒ 0 = 0 + 0x
(axiom on inverse)
=⇒ 0 = 0x
(axiom on zero vector)
We can still prove that T (0) = 0 without using this. In fact, the proof is similar.
T (0) = T (0 + 0) = T (0) + T (0)
(def of lin. trans.)
=⇒ (−T (0)) + T (0) = (−T (0)) + T (0) + T (0)
=⇒ 0 = 0 + T (0)
(axiom on inverse)
=⇒ 0 = T (0)
(axiom on zero vector)
Page 5
4
Additional Notes
1. The dot product is the inner product applied to vector spaces in Rn . It is the function
· : Rn × Rn → R
defined by
(x1 , . . . , xn ) · (y1 , . . . , yn ) := x1 y1 + · · · xn yn .
It satisfies the following properties of inner products
• x·y =y·x
• x · (y + z) = x · y + x · z
• c(x · y) = c(x · y)
• x · x ≥ 0, with equality iff x = 0
2. Let T : U → V be a linear transformation from vector space U to vector space V . We have seen
that T (0) = 0. Note the difference in the two 0. More carefully, we can write T (0U ) = 0V ,
signifying that the linear transformation of the zero vector in U is the zero vector in V .
5
Exercises
1. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 6.4.12.
2. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 6.4.14.
3. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 6.4.17.
4. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 6.4.18.
Page 6
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