Assignment 4
Math 2121: Linear Algebra
RAWAT Sudarshan (21110030)
RAWAT Sudarshan (21110030)
Assignment 4
Problem 1
Let (𝐴 𝐵) be the matrix formed by 𝐴 on top of 𝐵. Show that Col(𝐴 𝐵) =
Col𝐴 +Col𝐵. You may assume 𝐴 has two columns and 𝐵 has three columns.
Then prove rank (𝐴 𝐵) ≤rank𝐴 +rank𝐵.
Solution:
𝑏 𝑏 𝑏
𝑎 𝑎
Let 𝐴 = (𝑎1 𝑎2 ) and 𝐵 = (𝑏1 𝑏2 𝑏3 )
3
4
4
5
6
𝑎 𝑎 𝑏 𝑏 𝑏
𝐵) = ( 1 2 1 2 3 )
𝑎3 𝑎4 𝑏4 𝑏5 𝑏6
⇒ (𝐴
By defintion of column space,
𝑎
𝑎
Col𝐴 = ℝ( 1 ) + ℝ( 2 )
𝑎3
𝑎4
And Col(𝐴
∴ Col(𝐴
𝑎
𝑎
3
4
and
𝑏
𝑏
𝑏
𝑏
Col𝐵 = ℝ( 1 ) + ℝ( 2 ) + ℝ( 3 )
𝑏4
𝑏5
𝑏6
𝑏
𝑏
𝐵) = ℝ(𝑎1 ) + ℝ(𝑎2 ) + ℝ(𝑏1 ) + ℝ(𝑏2 ) + ℝ(𝑏3 ) = Col𝐴 + Col𝐵
𝐵) = Col𝐴 + Col𝐵
4
5
6
◾
Furthermore, by Proposition 2.4.2 we have:
rank(𝐴
𝐵) = dim Col(𝐴
from the first half of the solution we have: dim Col(𝐴
From the result of Exercise 2.39, we have:
𝐵)
𝐵) = dim (Col𝐴 + Col𝐵)
dim (Col𝐴 + Col𝐵) ≤ dim (Col𝐴) + dim (Col𝐵)
Also,
dim (Col𝐴) + dim (Col𝐵) = rank𝐴 + rank𝐵
∴ rank(𝐴
𝐵) ≤ rank𝐴 + rank𝐵
Page 2 of 6
◾ Q.E.D
RAWAT Sudarshan (21110030)
Assignment 4
Problem 2
3.2 Describe the elementary matrices for applying the row operations on a
vector in ℝ4 .
2) Row1 − 𝑐Row4
5) 𝑐Row2
2) Soluton:
In ℝ4 we have:
𝑥1
𝑥 − 𝑐𝑥4
1 0 0 −𝑐
𝑥1
(
)
( 1
)
(
)
(
)
Row1 −𝑐Row4 (
(
)
)
(
)
(
𝑥
𝑥
0
1
0
0
𝑥2 )
(
)
(
)
(
)
(
)
2
2
) →→→→→→→→→→→→→ (
)=(
)(
)
𝑥⃗ = (
(
(
(𝑥3 )
)
( 𝑥3 )
) (
(0 0 1 0 )
)(
(𝑥3 )
)
𝑥
𝑥
0
0
0
1
𝑥
4
4
4
( )
(
) (
)( )
1 0 0 −𝑐
(
)
(
0 1 0 0)
)
(
)
∴ The required elementary matrix is: (
(
(0 0 1 0 )
)
0
0
0
1
(
)
5) Solution:
In ℝ4 we have:
𝑥1
𝑥1
1 0 0 0
𝑥1
(
) 𝑐Row (
) (
)
(
)
(
(
)
(
1 (𝑐𝑥 )
𝑥2 )
0
𝑐
0
0
𝑥2 )
(
)
(
)
(
)
(
)
2
)
)
)
(
)
𝑥⃗ = (
→→→→→→ (
=(
(
)
(
)
(
)
(
𝑥
𝑥
0
0
1
0
𝑥
( 3)
( 3) (
)( 3 )
)
𝑥
𝑥
0
0
0
1
𝑥
4
4
4
( )
( ) (
)( )
1 0 0 0
(
)
(
0 𝑐 0 0)
(
)
)
∴ The required elementary matrix is: (
(
(0 0 1 0)
)
0
0
0
1
(
)
The elementary matrix depends on the row operation being applied, based on aforementioned
solution we have elementary matrix for the row operations Row𝑖 + 𝑐Row𝑗 and 𝑐Row𝑖 . For the
row operation Row𝑖 ↔ Row𝑗 , we just do the same row operation on the Identity matrix to
obtain the elementary matrix.
For example, in ℝ4 if we Row2 ↔ Row4 , we have the elementary matrix:
1 0 0 0
(
)
(
0 0 0 1)
(
)
(
)
(
(0 0 0 1)
)
0
1
0
0
(
)
Page 3 of 6
◾
RAWAT Sudarshan (21110030)
Assignment 4
Problem 3
3.4 Use the correspondence between the linear transormation and the matrix
to find the matrices for the flips in Exercise 3.1. What is the matrix of the
flip with respect to the line of angle 𝜌?
Solution:
• For the flip with respect to y-axis in ℝ2 , we have:
𝑒⃗1 = (1, 0) is changed to 𝐿𝑦(⃗⃗⃗⃗⃗⃗⃗⃗
𝑒1 ) = (−1, 0), and 𝑒⃗2 = (0, 1) is changed to 𝐿𝑦(⃗⃗⃗⃗⃗⃗⃗⃗
𝑒2 ) = (0, 1).
Therefore, by the one-to-one correspondence between linear transformations 𝐿 : ℝ𝑚 → ℝ𝑛
and 𝑚 × 𝑛 matrices 𝐴, the matrix reponsible for the flipping is:
[𝐿y ] = [𝐿𝑦(⃗⃗⃗⃗⃗⃗⃗⃗
𝑒1 ) , 𝐿𝑦(⃗⃗⃗⃗⃗⃗⃗⃗
𝑒2 ) ] = (
−1 0
)
0 1
Similarly,
• For the flip with respect to diagonal line 𝑦 = 𝑥 in ℝ2 , we have:
𝑒⃗1 = (1, 0) is changed to 𝐿𝑦=𝑥 (𝑒⃗1 ) = (0, 1) and 𝑒⃗2 = (0, 1) is changed to 𝐿𝑦=𝑥 (𝑒⃗2 ) = (1, 0).
Therefore by same reason as above, the matrix responsible for the flipping is:
0 1
[𝐿𝑦=𝑥 ] = [𝐿𝑦=𝑥 (𝑒⃗1 ), 𝐿𝑦=𝑥 (𝑒⃗2 )] = ( )
1 0
Now, consider the flip with respect to the line of angle 𝜌,
Without loss of generality, we can assume that the line always passes through origin based on
the figure given.
Flipping 𝑒⃗1 = (1, 0) about the line is equivalent to rotating 𝑒⃗1 by 2𝜌 about O in anticlockwise
direction
⇒ 𝐿𝜌 (𝑒⃗1 ) = (cos(2𝜌), sin(2𝜌))
Similarly, flipping 𝑒⃗2 = (0, 1) about the line is equivalent to rotating 𝑒⃗2 by 𝜋 − 2𝜌 about O in
clockwise direction
⇒ 𝐿𝜌 (𝑒⃗2 ) = (cos(
= (cos(2𝜌 −
𝜋
𝜋
𝜋
− (𝜋 − 2𝜌)), sin( − (𝜋 − 2𝜌))) [∵ 𝑒2 makes
with +ve x-axis]
2
2
2
𝜋
𝜋
𝜋
𝜋
), sin(2𝜌 − )) = (cos( − 2𝜌), − sin( − 2𝜌)) = (sin(2𝜌), − cos(2𝜌))
2
2
2
2
Finally, by the one-to-one correspondence between linear transformations 𝐿 : ℝ𝑚 → ℝ𝑛 and
𝑚 × 𝑛 matrices 𝐴, the matrix responsible for the flipping is:
cos(2𝜌) sin(2𝜌)
[𝐿𝜌 ] = [𝐿𝜌 (𝑒⃗1 ), 𝐿𝜌 (𝑒⃗2 )] = (
)
sin(2𝜌) − cos(2𝜌)
Page 4 of 6
◾
RAWAT Sudarshan (21110030)
Assignment 4
Problem 4
3.8 Find the matrices of linear transformations.
2) 𝐿(1, 2) = (0, 1), 𝐿(3, 4) = (1, 0).
3) 𝐿(1, 2) = (3, 4), 𝐿(3, 4) = (1, 2).
8) 𝐿(1, 0, 0) = (1, 2), 𝐿(0, 0, 1) = (3, 4), 𝐿(0, 1, 0) = (5, 6).
2) Solution:
Since linear transformations are preserved by column operations, we can perform column
operations on the following matrix to obtain the matrix responsible for the linear
transformation:
1 0 )
1 3
1 0
1 0
(
(
)
(
)
(
)
(
(
)
(
)
(
)
0 1 )
(
)
2 4)
2 −2)
0 −2)
(
(
(
(
(
)
(
)
(
)
→
→
→
1)
(
)
(
)
(
)
(
)
1
−
0
1
0
1
1
1
(
(
)
(
)
(
)
2)
(
)
−2 32
(1 0)
(1 −3)
(−2 −3)
(
)
1
We can read the required matrix to be: [𝐿] = (−2
− 12
3
2
)
3) Solution:
Using the similar reasoning as above, we can obtain the required matrix by performing
column operations on:
1 3
1 0
1
0
1 0
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
2
4
2
−2
0
−2
0 1)
(
)
(
)
(
)
(
)
(
)→(
)→(
)→(
)
(
(
(
(
(3 1)
)
(3 −8 )
)
(−5 −8 )
)
(−5 4)
)
4
2
4
−10
−6
−10
−6
5
(
)
(
)
(
)
(
)
−5 4
)
−6 5
∴ [𝐿] = (
8) Solution:
We can observe that given transformations are transformations of standard basis vectors of
ℝ3 , and the transformation is effectively going from ℝ3 → ℝ2 , So we expect:
[𝐿] = [𝐿(𝑒1 ), 𝐿(𝑒2 ), 𝐿(𝑒3 )] = (
Page 5 of 6
1 5 3
)
2 6 4
RAWAT Sudarshan (21110030)
Assignment 4
Problem 5
3.9. Can you find linear transformations satisfying the given equalitites?
3) 𝐿(1, 2) = (3, 4), 𝐿(3, 4) = (1, 2), 𝐿(1, 1) = (0, 0)
Solution:
Start by noting that:
1 3
1 1
1
( )= ( )− ( )
1
2 4
2 2
1
1
1 1
1 3
1
3
1
1
−1
0
⇒ 𝐿( ) = 𝐿( ) − 𝐿( ) ⇒ 𝐿( ) = ( ) − ( ) = ( ) ≠ ( )
1
4
2
1
1
0
2
2
2 2
2 4
Hence, there’s no such linear tranformation 𝐿 satisfying the transformations:
𝐿(1, 2) = (3, 4), 𝐿(3, 4) = (1, 2), 𝐿(1, 1) = (0, 0)
◾
Problem 6
3.13. Find the linear transformation that takes the vectors
(0, 1, 2), (1, 2, 0), (2, 0, 1) to the next one, in circular way.
Solution:
Let 𝐿 be the required linear transformation, then according to the requirements, we need:
𝐿(0, 1, 2) = (1, 2, 0), 𝐿(1, 2, 0) = (2, 0, 1),
and 𝐿(2, 0, 1) = (0, 1, 2)
Based on this information, we can construct the following matrix and use the fact that
column operations preseve the linear transformations to obtain the [𝐿]:
0 1 2)
1 0 2)
1 0 0)
1 0 0)
1 0 0)
(
(
(
(
(
(
)
(
)
(
)
(
)
(
1
2
0
2
1
0
2
1
−4
0
1
0
0 1 0)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
(
(
(
(
2 0 1)
0 2 1)
0 2 1)
−4 2 1)
0 0 1)
(
)
(
)
(
)
(
)
(
)
→
→
→
→
(
(
(
(
(
1 2 0)
2 1 0)
2 1 −4)
0 1 0)
0 1 0)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2
0
1
0
2
1
0
2
1
−4
2
1
0
0
1
(
)
(
)
(
)
(
)
(
)
0 1 2
1 0 2
1 0 0
1 0 0
1 0 0
(
)
(
)
(
)
(
)
(
)
0 1 0
(
)
∴ [𝐿] = (
(0 0 1 )
)
(1 0 0 )
Page 6 of 6
◾