The function T : R
2→R3
is defined by:
x
(
4 x2
)
x
Show that T islinear.
A proof of linearity begins by assigning v1 and v2 to be two arbitrary vectorsin the domain of
T , and α to be a scalar to prove the con
...
The function T : R
2→R3
is defined by:
x
(
4 x2
)
x
Show that T islinear.
A proof of linearity begins by assigning v1 and v2 to be two arbitrary vectorsin the domain of
T , and α to be a scalar to prove the conditions:
i. T ( v1+ v2 )=T (v1 )+T (v2
) and
ii. T ( α v1 )=αT (v1
) for all scalars α (usually α∈ R ) and vectors
v1
,v2∈T .
Let v , v ∈ R
2
. Then, v =
(
x1
)
and v =
(
y1
)
. Therefore,
x + y
(
4 ( x2+ y2
) 4 x2+4 y2
Now, to prove linearity we must compare the values of T ( v1+v2 ) with T ( v1 )+T(v2
) .
x y
4 x2 4 y2 4 x2+4 y2
T ( v1)+T ( v2)=T (
1
)
+T ( 1
)
=
(
4 x
+5 x
)
+(
4 y +5 y )
=(
4 x +4 y +5 x +5 y )
.
Note that the first condition T ( v1+v2 )=T (v1 )+T (v2
) , is satisfied.
Now, for α ∈ R
T ( α v1)=T (
α x1
)
=
(
4 α x2
4α x +5α x .
α x2
1 2
x1
To satisfy the second condition we compare the values of T ( α v1 ) and αT (v1) .
x
(
4 x2
) (
4α x2
1+4 y +5 x 2+5 y 2
.
x 1
1+5 α x2
.
x1
T (
1
)
∈R
2
.
2
1
T ( v1+ v2 )=T (
+ y
+ y 1
2 1
αT (v1)=αT (
) (
1 2 3
( )
( )
x3 α x3
Therefore, T (α v1 )=αT (v1
) . Hence, by proving these two conditions are true, we have shown
T is a linear transformation.
Question 3
Show that
S={x∈R
3
: 6 x +4 x +x =0 }
is a subspace of R
3
.
To verify that S is a subspace we need to prove:
I. Closure under vector addition: v1 , v2∈ S then v1+ v2∈ S .
II. Closure under scalar multiplication: v1 ∈ S and α a scalar, then α v1∈ S .
III. The existence of a zero vector 0∈S .
The remaining seven vector space axioms are inherited from these three.
First to prove closure under addition, we let v1 and v2 to be two arbitrary vectors,so
v1
,v2∈S .
x1 y1
Then, v1=
(
x2)
and v2=
(
y2 )
.This gives us:
x3 y3
6 x1+4 x2+x3 ¿ 0
6 y1+4 y2+ y3 ¿ 0
x1+ y1
Now, v1+ v2= x2+ y2 =6 ( x1+ y1)+4 ( x2+ y2)+( x3+ y3) .
x3+ y3
Simplifying this we get, (6 x1+ 4 x2+ x3 )+(6 y1+ 4 y2+ 4 y3 )=0+ 0=0. Therefore, S is closed
under vector addition.
Proving closure under scalar multiplication requires usto let α∈ R . Then:
x1 α x1
α v1=α (
x2)
=
(
α x2)
.
This becomes, 6 (α x1 )+4 (α x2 )+α x3=α (6 x1+4 x2+x3 )=α (0)=0. Therefore, S is closed
under scalar multiplication.
0
To prove the existence of a zero vector, such that 0∈S , we consider 0= 0 . Since:
0
6 (0)+ 4 (0)+0=0
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