Question #105263
Let V be a subspace of R^4 and S = {u1,u2,u3} be a basis for V. Suppose v1, v2, v3 are vectors in V such that (v1)S = (1,−2,0), (v2)S =(2,−7,4), and (v3)S =(−3,8,−1).
Show that {v1, v2, v3} is also a basis for V.
1
Expert's answer
2020-03-13T12:20:14-0400

{u1,u2,u3} is basis of V.

c1u1+c2u2+c3u3=0c_1u_1+c_2u_2+c_3u_3=0

v1=u12u2v_1=u_1-2u_2

v2=2u17u2+4u3v_2=2u_1-7u_2+4u_3

v3=3u1+8u2u3v_3=-3u_1+8u_2-u_3

We can solve these three equations to get v1,v2,v3 in terms of u1,u2,u3.

u1=(23v1+2v2+8v3)/3u_1=(23v_1+2v_2+8v_3)/3

u2=(10v1+v2+4v3)/3u_2=(10v_1+v_2+4v_3)/3

u3=(11v1+2v2+5v3)/3u_3=(11v_1+2v_2+5v_3)/3

Putting these in c1u1+c2u2+c3u3=0c_1u_1+c_2u_2+c_3u_3=0

c1(23v1+2v2+8v3)/3+c2(10v1+v2+4v3)/3+c3(11v1+2v2+5v3)/3=0c_1(23v_1+2v_2+8v_3)/3+c_2(10v_1+v_2+4v_3)/3+c_3(11v_1+2v_2+5v_3)/3=0

v1(23c1+10c2+11c3)/3+v2(2c1+c2+2c3)/3+v3(8c1+4c2+5c3)/3=0v_1(23c_1+10c_2+11c_3)/3+v_2(2c_1+c_2+2c_3)/3+v_3(8c_1+4c_2+5c_3)/3=0

d1v1+d2v2+d3v3=0d_1v_1+d_2v_2+d_3v_3=0

d1=(23c1+10c2+11c3)/3d_1=(23c_1+10c_2+11c_3)/3

d2=(2c1+c2+2c3)/3d_2=(2c_1+c_2+2c_3)/3

d3=(8c1+4c2+5c3)/3d_3=(8c_1+4c_2+5c_3)/3

c1=c2=c3=0c_1=c_2=c_3=0

So, d1=d2=d3=0d_1=d_2=d_3=0

Thus, {v1,v2,v3} spans and is linearly independent.

{v1 , v2 , v3 } is also a basis of V.


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Comments

Assignment Expert
16.03.20, 00:56

There are different methods to solve a linear system of equations with respect to variables u1, u2, u3. For example, Gaussian elimination (reduction), elimination of variables, Cramer's rule, the inverse matrix method. Some hints about these methods can be found at https://web.mit.edu/16.unified/www/FALL/signalssystems/linear_algebra.pdf, https://www.mathcentre.ac.uk/resources/Engineering%20maths%20first%20aid%20kit/latexsource%20and%20diagrams/5_6.pdf .

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15.03.20, 14:45

How do u get the u1​=(23v1​+2v2​+8v3​)/3 u2=(10v1+v2+4v3)/3 u3=(11v1+2v2+5v3)/3 Don't quite really understand this part. Could you kindly elaborate on it?

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