# Find a $4\times4$ matrix $B$ with $\operatorname{Col}(B)=W$. What is the rank of $B$?

Consider the subspace $W = \{(x_1,x_2,x_3,x_4) \in R^4 \mid x_1 - 2x_2+x_3-x_4=0\}$

Find a $4\times4$ matrix $B$ with $\operatorname{Col}(B)=W$. What is the rank of $B$?

I have the solution but I don't understand it quite well. I need an answer with explanation.

• Note that $4\times4$ rather than 4$x$4 is standard usage. Also, I included \operatorname{} and \mid in my edits here. Jan 5, 2018 at 18:29
• Being specific about what was in the solution that you don't understand might result in a better answer. Jan 5, 2018 at 18:30
• What have you tried? To start, can you find a basis of $W$ ? If yes, taking the basis vectors and the standard ordered basis vectors of $\Bbb R^4$ together to form a $4\times 7$ matrix and applying Gauss-Jordan elimination does the trick. Jan 5, 2018 at 18:31

Assuming that $\operatorname{Col}(B)$ is the vector space spanned by the columns of $B$, a possible solution is:$$\begin{bmatrix}1&0&0&1\\0&1&0&1\\0&0&1&1\\1&-2&1&0\end{bmatrix}.$$As you can see, each column belongs to $W$. Furthermor, the first three columns are linearly independent and therefore $\dim\operatorname{Col}(B)\geqslant3$. But, since all columns belong to $W$, $$\dim\operatorname{Col}(B)\leqslant\dim W=3.$$
The rank of $B$ can only be $3$, since the span of the columns of $B$ is $W$ and $\dim W=3$.