Answer to Question #219489 in Linear Algebra for kavya

Question

find basis and dimension of the subspace W of V spanned by A = [[1 2] [-1 3]], B = [[2 5] [1 -1]], C = [[5 12] [1 1]], D= [[3 4] [-2 5]]

Answer 1

We first consider each matrix as a vector in $\mathbb{R}^4$ to get a $4×4$ matrix.

$$\begin{bmatrix} 1 & 2 & 5 & 3 \\ 2 & 5 & 12 & 4 \\ -1 & 1 & 1 & -2 \\ 3 & -1 & 1 & 5 \end{bmatrix}$$

We now reduce the matrix to row echelon form.

Add $-2$ times the 1st row to the 2nd row to get

$$\begin{bmatrix} 1 & 2 & 5 & 3 \\ 0 & 1 & 2 & -2 \\ -1 & 1 & 1 & -2 \\ 3 & -1 & 1 & 5 \end{bmatrix}$$

Add the 1st row to the 3rd row to get

$$\begin{bmatrix} 1 & 2 & 5 & 3 \\ 0 & 1 & 2 & -2 \\ 0 & 3 & 6 & 1 \\ 3 & -1 & 1 & 5 \end{bmatrix}$$

Add $-3$ times the 1st row to the 4th row to get

$$\begin{bmatrix} 1 & 2 & 5 & 3 \\ 0 & 1 & 2 & -2 \\ 0 & 3 & 6 & 1 \\ 0 & -7 & -14 & -4 \end{bmatrix}$$

Add $-3$ times the 2nd row to the 3rd row to get

$$\begin{bmatrix} 1 & 2 & 5 & 3 \\ 0 & 1 & 2 & -2 \\ 0 & 0 & 0 & 7 \\ 0 & -7 & -14 & -4 \end{bmatrix}$$

Add $7$ times the 2nd row to the 4th row to get

$$\begin{bmatrix} 1 & 2 & 5 & 3 \\ 0 & 1 & 2 & -2 \\ 0 & 0 & 0 & 7 \\ 0 & 0 & 0 & -18 \end{bmatrix}$$

Multiply the 3rd row by $\frac{1}{7}$ to get

$$\begin{bmatrix} 1 & 2 & 5 & 3 \\ 0 & 1 & 2 & -2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & -18 \end{bmatrix}$$

Add $18$ times the 3rd row to the 4th row to get

$$\begin{bmatrix} 1 & 2 & 5 & 3 \\ 0 & 1 & 2 & -2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Add $2$ times the 3rd row to the 2nd row to get

$$\begin{bmatrix} 1 & 2 & 5 & 3 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Add $-3$ times the 3rd row to the 1st row to get

$$\begin{bmatrix} 1 & 2 & 5 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Add $-2$ times the 2nd row to the 1st row to get

$$\begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

The pivots are in the 1st, 2nd and 4th column. Also, the 3rd column is the sum of the 1st column and two times the 2nd column, so it is linearly dependent. So the 1st, 2nd and 4th columns are linearly independent and form a basis for $W$. Hence a basis for $W$ is $\{A,B,D\}$, and the dimension of $W$ is $3$.