# Finding the basis and dimension of a subspace of the vector space of 2 by 2 matrices

I am trying to find the dimension and basis for the subspace spanned by: $$\begin{bmatrix} 1&-5\\ -4&2 \end{bmatrix}, \begin{bmatrix} 1&1\\ -1&5 \end{bmatrix}, \begin{bmatrix} 2&-4\\ -5&7 \end{bmatrix}, \begin{bmatrix} 1&-7\\ -5&1 \end{bmatrix}$$ in the vector space $M_{2,2}$. I don't really care about the answer, I am just hoping for an efficient algorithm for solving problems like this for matrices.

I am not sure how to account for interdependence within the matrices. My instinct as of now is to find the maximum restriction imposed by the matrices. It is clear that the $1$ in position $a_{1,1}$ in each matrix will allow me to get any number in that position, so one vector in the basis will be: $$\begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}$$ But depending on which of the matrices I scale, I have restrictions on the other entries. So I don't think I can include that matrix in my basis.

It just occurred as I was writing this that I could maybe just think about these as $4$ by $1$ vectors and proceed as usual. Is there any danger in doing so?

• No danger. Just wanted to suggest so. – Berci Jul 12 '16 at 22:51
• ah ok, welp, hopefully the question will help people who have the same question in the future. – qbert Jul 12 '16 at 22:52
• $M_{2,2}(\mathbb{F})$ is isomorphic to $\mathbb{F}^4$, where $\mathbb{F}$ is the underlying field of the vector space. You can just think of these matrices as 4-tuples, as you suggested. – Christian Jul 12 '16 at 22:52
• Write these matrices as row vectors, and use row -reduction to find the rank and a basis. – Bernard Jul 12 '16 at 23:00

Inputs

$$\alpha = \left( \begin{array}{rr} 1 & -5 \\ -4 & 2 \end{array} \right), \qquad \beta = \left( \begin{array}{rr} 1 & 1 \\ -1 & 5 \end{array} \right), \qquad \gamma = \left( \begin{array}{rr} 2 & -4 \\ -5 & 7 \end{array} \right), \qquad \delta = \left( \begin{array}{rr} 1 & -7 \\ -5 & 1 \end{array} \right)$$

Find the basis for these matrices.

Matrix of row vectors

As noted by @Bernard, compose a matrix of row vectors. Flatten the matrices in this manner $$\left( \begin{array}{rr} 1 & -5 \\ -4 & 2 \end{array} \right) \quad \Rightarrow \quad \left( \begin{array}{crrc} 1 & -5 & -4 & 2 \end{array} \right)$$ Compose the matrix $$\mathbf{A} = \left( \begin{array}{crrr} 1 & -5 & -4 & 2 \\\hline 1 & 1 & -1 & 5 \\\hline 2 & -4 & -5 & 7 \\\hline 1 & -7 & -5 & 1 \end{array} \right)$$

Row reduction

Column 1

$$\left( \begin{array}{rccc} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ -2 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ \end{array} \right) % \left( \begin{array}{crrc} 1 & -5 & -4 & 2 \\ 1 & 1 & -1 & 5 \\ 2 & -4 & -5 & 7 \\ 1 & -7 & -5 & 1 \\ \end{array} \right) % = % \left( \begin{array}{crrr} \boxed{1} & -5 & -4 & 2 \\ 0 & 6 & 3 & 3 \\ 0 & 6 & 3 & 3 \\ 0 & -2 & -1 & -1 \\ \end{array} \right) %$$

Column 2

$$\left( \begin{array}{rccc} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ -2 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ \end{array} \right) % \left( \begin{array}{crrr} \boxed{1} & -5 & -4 & 2 \\ 0 & 6 & 3 & 3 \\ 0 & 6 & 3 & 3 \\ 0 & -2 & -1 & -1 \\ \end{array} \right) % = % \left( \begin{array}{cccc} \boxed{1} & 0 & -\frac{3}{2} & \frac{9}{2} \\ 0 & \boxed{1} & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) %$$

The fundamental rows are marked by the unit pivots.

Solution

The basis is $$\mathcal{B} = \left\{ \alpha, \, \beta \right\} = \left\{ \left( \begin{array}{rr} 1 & -5 \\ -4 & 2 \end{array} \right), \ \left( \begin{array}{rr} 1 & 1 \\ -1 & 5 \end{array} \right) \right\}$$