Find a basis of a subspace spanned by matrices $G = \langle \begin{bmatrix}1&1\\1&1\\ \end{bmatrix}, \begin{bmatrix}1&1\\1&0\\ \end{bmatrix}, \begin{bmatrix}2&-3\\1&1\\ \end{bmatrix},\begin{bmatrix}4&-1\\3&2\\ \end{bmatrix}$
First I write all the matrices as vectors:
$M_1 = \begin{bmatrix}1\\1\\1\\1 \end{bmatrix}$
$M_2 = \begin{bmatrix}1\\1\\1\\0 \end{bmatrix}$
$M_3 = \begin{bmatrix}2\\-3\\1\\1 \end{bmatrix}$
$M_4 = \begin{bmatrix}4\\-1\\3\\2 \end{bmatrix}$
Then I found the rref:
$$\begin{bmatrix}1 & 1 & 1 & 1\\1 & 1 & 1 & 0\\2 & -3 & 1 & 1\\4 & -1 & 3 & 2 \end{bmatrix} \rightarrow (...) \rightarrow \begin{bmatrix}1 & 0 & \frac{4}{5} & 0\\0 & 1 & \frac{1}{5} & 0\\0 & 0 & 0 & 1\\0 & 0 & 0 & 0 \end{bmatrix}$$
The basis I get is
$$(\begin{bmatrix}1 &0\\4/5&0 \end{bmatrix},\begin{bmatrix}0 &1\\\frac{1}{5}&0 \end{bmatrix}),\begin{bmatrix}0 &0\\0&1 \end{bmatrix}$$
Is this correct? It doesn't even come close to the solution in my book which is
$$(\begin{bmatrix}1 &1\\1&1 \end{bmatrix},\begin{bmatrix}1 &1\\1&0 \end{bmatrix}),\begin{bmatrix}2 &-3\\1&1 \end{bmatrix}$$
 A: Let's check it out.
$$e_1=\begin{bmatrix}1 & 0 \\ \tfrac{4}{5} & 0 \end{bmatrix} \\ e_2 = \begin{bmatrix}0 & 1 \\ \tfrac{1}{5} & 0 \end{bmatrix} \\ e_3 = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$$
We have:
$$\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix} = e_1+e_2+e_3 \\ \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} = e_1+e_2 \\ \begin{bmatrix}2 & -3 \\ 1 & 1\end{bmatrix} = 2e_1-3e_2+e_3 \\ \begin{bmatrix}4 & -1 \\ 3 & 2\end{bmatrix} = 4e_1-e_2+2e_3$$
You have three matrices, which is definitely the size of the basis given in the book. Each of the four matrices you were given are in the span of the matrices you found. The three matrices you found are linearly independent. This means the two bases span the same subspace.
A: Your solution isn’t wrong per se. There’s an infinite number of them. It looks like you were meant to pick out a subset of the original four matrices that form a basis, which you can do by assembling those same flattened vectors into as columns instead of rows. After row-reduction, you can pick out the original vectors that correspond to pivot columns in the reduced matrix.
