Question

Q. Find the dimension of the subspace of R4 that is span of the vectors
(█(1¦(-1)@0@1)), (█(2¦1@1@1)),(█(0¦0@0@0)),(█(1¦1@-2@-5))

Answer 1

We denote the subspace $V$.of $R^4$ that is span of the vectors (1,-1,0,1), (2,1,1,1), (0,0,0,0), (1,1,-2,-5).

The dimension of $V$ is 3.

The dimension of the subspace R4 is the number of vectors in any basis for the space to be spanned.

 Dimension is the number of vectors in any basis for the space to be spanned.

                      (2.) Rank of a matrix is the dimension of the column space.

 

Known that dimension is the maximum number of linearly independent vectors in a subspace.

Consider  4×4  matrix and row reduce it:

$$A= \begin{pmatrix} 1 & -1 & 0 & 1 \\ 2 & 1& 1& 1 \\ 0 & 0& 0& 0 \\ 1 & 1 & -2 & -5 \end{pmatrix}$$

The dimension of the subspace $V$ is equal to the matrix $rank(A)$.

From the 2nd row subtract the 1st line, multiplied by 2; from the 4th row subtract the 1st row:

$$\begin{pmatrix} 1 & -1 & 0 & 1 \\ 0 & 3& 1& -1 \\ 0 & 0& 0& 0 \\ 0 & 2 & -2 & -6 \end{pmatrix}$$

2nd row divide by 3:

$$\begin{pmatrix} 1 & -1 & 0 & 1 \\ 0 & 1& \frac{1}{3} & -\frac{1}{3} \\ 0 & 0& 0& 0 \\ 0 & 2 & -2 & - 6 \end{pmatrix}$$

Fom the 4th line subtract 2nd row, multiplied by 2:

$$\begin{pmatrix} 1 & -1 & 0 & 1 \\ 0 & 1& \frac{1}{3} & -\frac{1}{3} \\ 0 & 0& 0& 0 \\ 0 & 0 & -\frac{8}{3} & - \frac{16}{3} \end{pmatrix}$$

Сhange the 3rd row and the 4th row in places:

$$\begin{pmatrix} 1 & -1 & 0 & 1 \\ 0 & 1& \frac{1}{3} & -\frac{1}{3} \\ 0 & 0 & -\frac{8}{3} & -\frac{16}{3} \\ 0 & 0& 0& 0 \end{pmatrix}$$

Since non-zero lines are 3, then Rank(A) = 3.

The dimension of $V$ is 3.