UPSC 2015 Maths Optional Paper 1 Q4b — Step-by-Step Solution

12 marks · Section A

Bases and dimension; coordinates in a basis · Linear Algebra · asked 7× in 13 yrs · Read the full method →

Question

Find the dimension of the subspace of $\mathbb R^4$ spanned by the set $\{(1,0,0,0),(0,1,0,0),(1,2,0,1),(0,0,0,1)\}$ . Hence find its basis.

Technique

Row-reduce the matrix of spanning vectors; count non-zero rows for dimension; pivot positions in the original rows give a basis.

Solution

Strategy. Form the matrix with the vectors as rows, row-reduce, count pivots. Pivot rows yield a basis (for the row space).

Step 1 — Row matrix

M=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 1 & 2 & 0 & 1\\ 0 & 0 & 0 & 1\end{pmatrix}.

Step 2 — Row reduce

$R_3\to R_3-R_1$ :

\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 2 & 0 & 1\\ 0 & 0 & 0 & 1\end{pmatrix}.

$R_3\to R_3-2R_2$ :

\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1\end{pmatrix}.

$R_4\to R_4-R_3$ :

\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{pmatrix}.

Three non-zero rows. Dimension = 3.

Step 3 — A basis

The first three vectors after reduction are pivot rows (pivots in columns 1, 2, 4). Equivalently from the original set, three linearly independent vectors form a basis. The simplest choice:

Answer

\boxed{\;\{(1,0,0,0),\,(0,1,0,0),\,(0,0,0,1)\},\quad\dim=3.\;}