Q 2.9. Subspaces. For A ={(| }. Since. are elements of.

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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.