Introduction
The problem asks us to express the basis vectors
e
1
=
(
1
,
0
)
e
1
=
(
1
,
0
)
e_(1)=(1,0) e_1 = (1, 0) e 1 = ( 1 , 0 ) and
e
2
=
(
0
,
1
)
e
2
=
(
0
,
1
)
e_(2)=(0,1) e_2 = (0, 1) e 2 = ( 0 , 1 ) as linear combinations of the vectors
α
1
=
(
2
,
−
1
)
α
1
=
(
2
,
−
1
)
alpha_(1)=(2,-1) \alpha_1 = (2, -1) α 1 = ( 2 , − 1 ) and
α
2
=
(
1
,
3
)
α
2
=
(
1
,
3
)
alpha_(2)=(1,3) \alpha_2 = (1, 3) α 2 = ( 1 , 3 ) . In other words, we want to find constants
c
1
c
1
c_(1) c_1 c 1 and
c
2
c
2
c_(2) c_2 c 2 such that:
e
1
=
c
1
α
1
+
c
2
α
2
e
1
=
c
1
α
1
+
c
2
α
2
e_(1)=c_(1)alpha_(1)+c_(2)alpha_(2) e_1 = c_1 \alpha_1 + c_2 \alpha_2 e 1 = c 1 α 1 + c 2 α 2
e
2
=
d
1
α
1
+
d
2
α
2
e
2
=
d
1
α
1
+
d
2
α
2
e_(2)=d_(1)alpha_(1)+d_(2)alpha_(2) e_2 = d_1 \alpha_1 + d_2 \alpha_2 e 2 = d 1 α 1 + d 2 α 2 Work/Calculations
Step 1: Setting up the Equations for
e
1
e
1
e_(1) e_1 e 1
The equation for
e
1
e
1
e_(1) e_1 e 1 can be written as:
(
1
,
0
)
=
c
1
(
2
,
−
1
)
+
c
2
(
1
,
3
)
(
1
,
0
)
=
c
1
(
2
,
−
1
)
+
c
2
(
1
,
3
)
(1,0)=c_(1)(2,-1)+c_(2)(1,3) (1, 0) = c_1 (2, -1) + c_2 (1, 3) ( 1 , 0 ) = c 1 ( 2 , − 1 ) + c 2 ( 1 , 3 ) Breaking it down into components, we get:
1
=
2
c
1
+
c
2
1
=
2
c
1
+
c
2
1=2c_(1)+c_(2) 1 = 2c_1 + c_2 1 = 2 c 1 + c 2
0
=
−
c
1
+
3
c
2
0
=
−
c
1
+
3
c
2
0=-c_(1)+3c_(2) 0 = -c_1 + 3c_2 0 = − c 1 + 3 c 2 Step 2: Solving for
c
1
c
1
c_(1) c_1 c 1
c
2
c
2
c_(2) c_2 c 2
After Calculating, we get:
c
1
=
3
7
,
c
2
=
1
7
c
1
=
3
7
,
c
2
=
1
7
c_(1)=(3)/(7),quadc_(2)=(1)/(7) c_1 = \frac{3}{7}, \quad c_2 = \frac{1}{7} c 1 = 3 7 , c 2 = 1 7 Step 3: Setting up the Equations for
e
2
e
2
e_(2) e_2 e 2
The equation for
e
2
e
2
e_(2) e_2 e 2 can be written as:
(
0
,
1
)
=
d
1
(
2
,
−
1
)
+
d
2
(
1
,
3
)
(
0
,
1
)
=
d
1
(
2
,
−
1
)
+
d
2
(
1
,
3
)
(0,1)=d_(1)(2,-1)+d_(2)(1,3) (0, 1) = d_1 (2, -1) + d_2 (1, 3) ( 0 , 1 ) = d 1 ( 2 , − 1 ) + d 2 ( 1 , 3 ) Breaking it down into components, we get:
0
=
2
d
1
+
d
2
0
=
2
d
1
+
d
2
0=2d_(1)+d_(2) 0 = 2d_1 + d_2 0 = 2 d 1 + d 2
1
=
−
d
1
+
3
d
2
1
=
−
d
1
+
3
d
2
1=-d_(1)+3d_(2) 1 = -d_1 + 3d_2 1 = − d 1 + 3 d 2 Step 4: Solving for
d
1
d
1
d_(1) d_1 d 1
d
2
d
2
d_(2) d_2 d 2
After Calculating, we get:
d
1
=
−
1
7
,
d
2
=
2
7
d
1
=
−
1
7
,
d
2
=
2
7
d_(1)=-(1)/(7),quadd_(2)=(2)/(7) d_1 = -\frac{1}{7}, \quad d_2 = \frac{2}{7} d 1 = − 1 7 , d 2 = 2 7 Conclusion
The basis vector
e
1
=
(
1
,
0
)
e
1
=
(
1
,
0
)
e_(1)=(1,0) e_1 = (1, 0) e 1 = ( 1 , 0 ) can be expressed as a linear combination of
α
1
α
1
alpha_(1) \alpha_1 α 1 and
α
2
α
2
alpha_(2) \alpha_2 α 2 as follows:
e
1
=
3
7
α
1
+
1
7
α
2
e
1
=
3
7
α
1
+
1
7
α
2
e_(1)=(3)/(7)alpha_(1)+(1)/(7)alpha_(2) e_1 = \frac{3}{7} \alpha_1 + \frac{1}{7} \alpha_2 e 1 = 3 7 α 1 + 1 7 α 2 Similarly, the basis vector
e
2
=
(
0
,
1
)
e
2
=
(
0
,
1
)
e_(2)=(0,1) e_2 = (0, 1) e 2 = ( 0 , 1 ) can be expressed as:
e
2
=
−
1
7
α
1
+
2
7
α
2
e
2
=
−
1
7
α
1
+
2
7
α
2
e_(2)=-(1)/(7)alpha_(1)+(2)/(7)alpha_(2) e_2 = -\frac{1}{7} \alpha_1 + \frac{2}{7} \alpha_2 e 2 = − 1 7 α 1 + 2 7 α 2 Thus, both
e
1
e
1
e_(1) e_1 e 1 and
e
2
e
2
e_(2) e_2 e 2 can be represented as linear combinations of
α
1
α
1
alpha_(1) \alpha_1 α 1 and
α
2
α
2
alpha_(2) \alpha_2 α 2 .