Let B= {P, Q, R), where P = (1 1 0), Q = (0 1 1), and R = (1 0 1). Show that B s

Question

Let B= {P, Q, R), where P = (1 1 0), Q = (0 1 1), and R = (1 0 1). Show that B spans R^3. Let U = -2P. Show that B' = {U, Q, R} also spans R^3. Show that B" = {P - Q + 2R, Q, R} also spans R^3. The nonzero rows of a reduced matrix are linearly independent and the columns with the leading 1's are linearly independent. Let S be a set of m points of R^n and let L_1 = Span(S). Let M be the m-by-n matrix obtained by stacking the points of S in a column and let M' be the row reduce matrix obtained from M (also called Gaussian elimination). Then the nonzero rows of M' form a basis for L_1 and the set S' of columns of M corresponding to the columns of M with leading 1's form a basis for L_2 = Span(S'). The linear space L_1 = Span(S) is called the row space of the matrix M and L_2 = Span(S') is called the column space of M. Find a basis for the subspace of R^5 spanned by the points (0 5 1 2 3), (1 0 1 6 4), (0 0 1 2), and (1 5 2 8).

Explanation / Answer

Let B= {P, Q, R}. We shall reduce B to its RREF by the following row operations:

1. Add -1 times the 1st row to the 2nd row

2. Add -1 times the 2nd row to the 3rd row

3. Multiply the 3rd row by ½

4. Add 1 times the 3rd row to the 2nd row

5. Add -1 times the 3rd row to the 1st row

The RREF of B is I3. This implies that the columns of B are linearly independent and span R3.

Let B’ = {U , Q,R}. We shall reduce B’ to its RREF by the following row operations

1. Multiply the 1st row by -1/2

2. Add 2 times the 1st row to the 2nd row

3. Add -1 times the 2nd row to the 3rd row

4. Multiply the 3rd row by ½

5. Add 1 times the 3rd row to the 2nd row

6. Add 1/2 times the 3rd row to the 1st row

The RREF of B’ is I3. This implies that the columns of B’ are linearly independent and span R3.

Let B” = { P-Q+2R, Q,R}. We shall reduce B” to its RREF by the following row operations:

1. Multiply the 1st row by 1/3

2. Add -1 times the 1st row to the 3rd row

3. Add -1 times the 2nd row to the 3rd row

4. Multiply the 3rd row by 3/2

5. Add -1/3 times the 3rd row to the 1st row

The RREF of B’’ is I3. This implies that the columns of B’’ are linearly independent and span R3.

Problem 2.5.6

The last row of the last two vectors has got truncated. Without this information, the question cannot be answered. However, the method can be explained. We can form a matrix with the given vectors as columns. Then we reduce M to its RREF. It can be easily determined from the RREF of M whether all or some of the columns of M are linearly independent. Further, if we can create the vector(1,1,1,1,1)T by a linear combination of the linearly independent columns of the RREF of M, then the corresponding columns of M span R5.

Please post the question regarding the theorem 2.5.5 again.

1. Multiply the 1st row by 1/3

2. Add -1 times the 1st row to the 3rd row

3. Add -1 times the 2nd row to the 3rd row

4. Multiply the 3rd row by 3/2

5. Add -1/3 times the 3rd row to the 1st row

The RREF of B’’ is I3. This implies that the columns of B’’ are linearly independent and span R3.

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