University Mathematics Assignment: Subspace and Linear Dependency

This assignment delves into fundamental concepts of linear algebra, specifically focusing on subspace and linear dependency. The solution begins by examining whether given sets of vectors satisfy the properties of a subspace, assessing closure under scalar addition and multiplication. It then proceeds to determine linear dependency using matrix operations and determinant calculations, checking if the determinant of a matrix formed by the vectors is non-zero. Finally, it investigates linear mappings, assessing whether a function's derivative can be linearly mapped. The assignment provides a detailed analysis of these core concepts, offering valuable insights into vector spaces, linear independence, and matrix transformations. The assignment is a great resource for students to understand linear algebra.

Q 2.9
Subspaces
a. For A ={(λ , λ+ μ3 , λ−μ3 ,| λ , μ ϵ R}
Since λ∧μ are elements of R any combination of the two will be subspace of R
b. B= {( λ2 ,−λ2 , 0 ¿∨λ ϵ R }
Zero is a substance of R and also the rest of variables are.
c. Is a subspace of R because it is closed under both scalar addition and multiplication
d. They are not subspace since they are not closed under the scaler addition or
multiplication
Q 2.10
a. ,
x1=
2
−1
3
x2=
1
1
−2
x3=
3
−3
8
for linear dependency the three vectors are combined into a single matrix
A=
[ 2 1 3
−1 1 −3
3 −2 8 ]
=2((8 x 1)-(-2 x-3)) – 1((-1x 8) – (-3 x 3)) +3 ((-1 x -2)-(3 x -3))
= 4 -1 + 11
= 14
Other than row reduction to determine linear dependency, the following condition should
be met
A x det ¿ 0

=2((8 x 1)-(-2 x-3)) – 1((-1x 8) – (-3 x 3)) +3 ((-1 x -2)-(3 x -3))
= 4 -1 + 11
= 14
A x 14 = 0
Therefore, there is not linear dependency
b. ,
B=
[ 1 1 1
2 1 0
1 0 0
0 1 1
0 1 1 ]
The above is reduced to c ¿
1 0 0
0 1 0
0 0 1
0 0 0
0 0 0
For linear dependency {c} x{ λ} = 0
1 0 0
0 1 0
0 0 1
0 0 0
0 0 0
x
λ1
λ2
λ3
λ4
λ5
≠ 0
Therefore they are not linearly dependent
Q 2.11
a. Combining the three
Y= x1+ x2 + x3

Y=
1 1 2
1 2 −1
1 3 1
b. ,
Since Φ :C1 → C0, there is no way a derivative of Φ ( f ) to be linearly mapped to f `. Tis also
especially now that C0 is a continuous function
c. Cos (x) is a constant. For linear mapping, there has to be a scalar multiple. This
means that there linear mapping. The scalar multiple in this scenario is 1.
d. Finding the pivot column
The matrix can be rewritten into
1 2 3
0 4 4
0 0 0
The pivot matrix becomes
x pivot =
1 0 3
0 1 0
0 0 0
e. Taking θ=0
x → [ 1 0
0 1 ] x
This is identity matrix. Linear mapping is possible since multiplying by it will
have the same vector.