Question

Find a Basis for the indicated subspace of R4:The set of all vectors ofthe form (a,b,c,d) for which a + 2b= c + 3d = 0.

Answer 1

A basis for the indicated subspace of R^4 is {(1, 0, -2, 0), (0, 1, 0, -3)}.

Step-by-step explanation:

Express the given conditions as equations:

a + 2b = 0

c + 3d = 0

Solve for one variable in terms of another in each equation:

From equation 1: a = -2b

From equation 2: c = -3d

Substitute these expressions back into the original equations:

(-2b) + 2b = 0 --> -2b = 0 (redundant)

-3d + 3d = 0 (redundant)

Identify two linearly independent vectors:

We see that b and d can be any arbitrary values, while a and c are determined based on their relationship with b and d.

Choose two arbitrary values for b and d (e.g., b = 1 and d = -1).

Substitute these values to get two specific vectors satisfying the original conditions:

(1, 0, -2, 0) and (0, 1, 0, -3)

Verify linear independence:

Check if any linear combination of these two vectors results in the zero vector:

k(1, 0, -2, 0) + l(0, 1, 0, -3) = (0, 0, 0, 0) only when k = l = 0.

Since no non-zero values of k and l make the sum equal to the zero vector, the two vectors are linearly independent.

Therefore, {(1, 0, -2, 0), (0, 1, 0, -3)} forms a basis for the subspace of R^4 described by the given conditions.

Answer 2

The basis for the subspace of R4 where a + 2b = c + 3d = 0 is given by the vectors {(1, 0, -1, -1/3), (0, 1, -2, -2/3)}.

Step-by-step explanation:

To find a basis for the subspace of R4 defined by the set of all vectors (a,b,c,d) for which a + 2b = c + 3d = 0, we need to solve for two of the variables in terms of the others because we have two equations with four unknowns. Let's express the variables c and d in terms of a and b using the given relationships:

• c = -a - 2b
• d = -(a + 2b)/3

Now, any vector in the subspace can be written in terms of a and b:

(a, b, -a - 2b, -(a + 2b)/3)

This can be rewritten as a combination of a and b:

a(1, 0, -1, -1/3) + b(0, 1, -2, -2/3)

Therefore, the set of vectors {(1, 0, -1, -1/3), (0, 1, -2, -2/3)} forms a basis for the subspace since these vectors are linearly independent and span the subspace defined by a + 2b = c + 3d = 0.