Answer to Question #100469 in Linear Algebra for Joseph Yep

Question

5. Explain why the set S is not a basis for R3 .

u1= (-1, 3, 2), u2= ( 6,1,1) for R^3.

6. Given matrix A in row cchelon form (R),
find the (a) basis for row space of A and rank (A).
(b) basis for the solution space of A and nullity (A).

1 -2 0 0 3
R= 0 1 3 2 0
0 0 1 1 0
0 0 0 0 0

Answer 1

5. A set $S = (\bold e_1, ..., \bold e_n)$ is a basis in linear space $V$ if for any $\bold v \in V$ exists a unique combination of coefficients $v_i, i = 1..n$, such that $\bold v = \sum_{i=1}^n v_i \bold e_i$. Hence, it suffices to show that some vector in $\R^3$ can not be expressed as linear combination of $\bold u_1, \bold u_2$. Choose, for instance vector $\bold u_3 = \bold u_1 \times \bold u_2 = (1, 13, -19)$ (Geometric intuition: since $\bold u_1$, $\bold u_2$ span a plane in $\R^3$, vector $\bold u_3$, which is cross product of two, will be perpendicular to the plane, hence it is impossible to express it as linear combinations of them). Now, solve $\lambda_1 \bold u_1 + \lambda_2 \bold u_2 = \bold u_3$, which is $\left(\begin{array}{cc|c} -1 & 6 & 1\\ 3 & 1 & 13\\ 2 & 1 & -19 \end{array}\right)$. Reducing the augmented matrix to RREF, obtain $\left(\begin{array}{cc|c} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right)$.

Obviously, this system has no solutions, because of the last row. Hence, S is not a basis in $\R^3$.

6.

a) Rank of the matrix is the number of non-zero rows in RREF of matrix, hence $rk (A) = 3$.

Basis of the row space consists of three non-zero row-vectors of the RREF of the matrix: $\bold v_1 = (1, -2, 0, 0, 3), \bold v_2 = (0,1,3,2,0), \bold v_3 = (0,0,1,1,0)$.

b) Since first three column vectors of RREF of the matrix are pivots, first three columns of the original matrix constitute the basis of solution space of $A$.

According to rank-nullity theorem, $nullity (A) = 5 - rk (A) = 5 - 3 = 2$.