Answer to Question #235241 in Algebra for Ash

Question

1). Determine whether set S given below is a basis for ℝ 3 . If not, explain why.

S = {(1,0,0),(0,1,0),(0,0,0)}

2). Find the rank of matrix D given below.

D = [0 3 9 0]

[-2 -1 1 -1]

[0 -1 -3 1]

Answer 1

a) A subset $S$ of a vector space $V$ is called a basis if

1. $S$ is linearly independent, and

2. $S$  is a spanning set

Any three linearly independent vectors form a basis of $\R^3.$

Let us check that whether $S$ is a linearly independent set.

Consider the linear combination

$x_1\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+x_2\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}+x_3\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

This is equivalent to the matrix equation

$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ $A=\begin{pmatrix} 1 & 0 & 0 & & 0 \\ 0 & 1 & 0 & & 0 \\ 0 & 0 & 0 & & 0 \\ \end{pmatrix}$

Thus, the general solution is $x_1=0, x_2=0, x_3=t, t\in \R$ is a free variable.

Hence, in particular, there is a nonzero solution.

So $S$ is linearly dependent, and hence $S$ cannot be a basis for $\R^3.$

b)

$D= \begin{bmatrix} 0 & 3 & 9 & 0 \\ -2 & -1 & 1 & -1 \\ 0 & -1 & -3 & 1 \\ \end{bmatrix}$

Swap $R_1$ and $R_2$

$\begin{bmatrix} -2 & -1 & 1 & -1 \\ 0 & 3 & 9 & 0 \\ 0 & -1 & -3 & 1 \\ \end{bmatrix}$

$R_3=R_3+R_2/3$

$\begin{bmatrix} -2 & -1 & 1 & -1 \\ 0 & 3 & 9 & 0 \\ 0 & 0 & 0& 1 \\ \end{bmatrix}$

The rank of a matrix is the number of nonzero rows in the reduced matrix, so the rank is 3.

$rank(D)=3$