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Chapter 1: Problem 11

Let \(u\) and \(v\) be distinct vectors of a vector space \(\mathrm{V}\). Show that if \(\\{u, v\\}\) is a basis for \(\mathrm{V}\) and \(a\) and \(b\) are nonzero scalars, then both \(\\{u+v, a u\\}\) and \(\\{a u, b v\\}\) are also bases for \(\mathrm{V}\).

Short Answer

Expert verified
In conclusion, both \(\{u+v, au\}\) and \(\{au, bv\}\) are bases for the vector space V. We showed this by verifying that each set of vectors is linearly independent and spans the entire vector space V.

Step by step solution

01

Check linear independence of {u+v, au}

Suppose we have a linear combination of u+v and au equal to the zero vector: \(c_1(u + v) + c_2(au) = 0\) Let's isolate the scalars c1 and c2: \((c_1 + ac_2)u + c_1v = 0\) Since {u, v} is a basis, it is linearly independent. So, the only solution for the above equation is: \(c_1 + ac_2 = 0\) \(c_1 = 0\) Since a is nonzero (by problem statement), the solution is unique and vectors u+v and au are linearly independent. ###Step 2: Spanning the Vector Space by {u+v, au}###
02

Show that {u+v, au} spans V

We need to show that any vector w 鈭� V can be written as a linear combination of u+v and au. Since {u, v} is a basis for V, w can be written as a linear combination of u and v: \(w = x_1u + x_2v\) Now, we need to find the coefficients, say y1 and y2, such that: \(w = y_1(u+v) + y_2(au)\) Solving for y1 and y2, \(w = (y_1+ay_2)u + y_1v\) Comparing the coefficients, we have \(y_1+ay_2 = x_1\) \(y_1 = x_2\) Since a and x2 鈮� 0, there exists a unique solution for y1 and y2. Hence, {u+v, au} spans the entire vector space V. Since {u+v, au} is both linearly independent and spans V, it is a basis for V. ##Part 2: {au, bv}## ###Step 1: Linear Independence of {au, bv}###
03

Check linear independence of {au, bv}

Suppose we have a linear combination of au and bv equal to the zero vector: \(c_1(au) + c_2(bv) = 0\) \((ac_1)u + (bc_2)v = 0\) Since {u, v} is a basis, it is linearly independent, meaning the only solution for the above equation is: \(ac_1 = 0\) \(bc_2 = 0\) Since a and b are nonzero (by problem statement), the only possible solution is c1 = c2 = 0. Hence, vectors au and bv are linearly independent. ###Step 2: Spanning the Vector Space by {au, bv}###
04

Show that {au, bv} spans V

We need to show that any vector w 鈭� V can be written as a linear combination of au and bv. Since {u, v} is a basis for V, w can be written as a linear combination of u and v: \(w = x_1u + x_2v\) Now, we need to find the coefficients, say z1 and z2, such that: \(w = z_1(au) + z_2(bv)\) Solving for z1 and z2, \(w = (az_1)u + (bz_2)v\) Comparing the coefficients, we have \(az_1 = x_1\) \(bz_2 = x_2\) Since a and b are nonzero, there exists a unique solution for z1 and z2. Hence, {au, bv} spans the entire vector space V. Since {au, bv} is both linearly independent and spans V, it is a basis for V. In conclusion, both {u+v, au} and {au, bv} are bases for the vector space V.

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