# First column zero matrix: Studying the basis

I have the following matrix $\mathbf{U}$ which is in echelon form. The strange to me is that I havent met a matrix with first column zero.

$$\mathbf{U} = \begin{bmatrix} 0 & 5 & 4 & 3 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

According to theory columns with pivots define a basis. Also according to theory, a matrix can have more that one basis. So, according to these facts, we may say that the basis are:

$$B_1= \begin{Bmatrix}\begin{bmatrix} 5 \\ 0 \\ 0 \\0 \\ \end{bmatrix}, \begin{bmatrix} 4 \\ 2 \\ 0 \\ 0 \\ \end{bmatrix}\end{Bmatrix} ,~ B_2= \begin{Bmatrix}\begin{bmatrix} 5 \\ 0 \\ 0 \\0 \\ \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix}\end{Bmatrix} \text{and}~B_3= \begin{Bmatrix}\begin{bmatrix} 4 \\ 2 \\ 0 \\0 \\ \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix}\end{Bmatrix}$$

Taking one of these bases ($B_1$ or $B_2$ or $B_3$) we may represent the non basis column as a combination of the basis columns. I think this is by definition of the column space basis. However, the last hold only when taking $B_3$ as column space basis.

Can you please correct possible flaws on my logic?

Thanks!!

Any of $B_1$, $B_2$ or $B_3$ is a basis for the column space of $\mathbf{U}$. It is pretty clear that the last column is a linear combination of either $B_2$ or $B_3$, with the coefficients 0 and 1. There is a little arithmetic to do to find the coefficients to express the last column in terms of $B_1$, but you can probably figure out that $1/5$ and $1/2$ will work. And you can express the first column in terms of any of these bases with the coefficients 0 and 0.
The three sets $B_1,B_2,B_3$ serve as a basis for a unique two dimensional subspace of $\Bbb R^4$ because each set has two linearly independent vectors.