If the row-reduced form of matrix $A$ has a row of zeros, its columns do not span $\mathbb{R}^n$ Can someone explain why it is that, if the row-reduced form of an $n\times m$ matrix $A$ has a row of zeros, the columns of matrix $A$ do not span $\mathbb R^n$?
I'm not seeing the bigger picture here.
 A: Presuming the matrix $\mathbf{A}$ in question is $n \times n$, as a $n \times m$ matrix implies there's $m$ column vectors and if $m<n$, it's clear those columns vectors do not span $\mathbb{R}^n$. I'll leave you to think about $m>n$ after reading my response for the $n \times n$ case:
A row of zeroes in the matrix $\mathbf{A}$ directly implies $\mathbf{A}$ is not invertible, or has a nontrivial solution to the homogenous matrix equation $\mathbf{A\vec{x}}=\mathbf{0}$, or there exists a $\mathbf{\vec{x}}\ne \vec{\mathbf{0}}$ such that $\mathbf{A\vec{x}}=\mathbf{0}$.
And since there's a nontrivial solution, we have that the columns of the matrix $\mathbf{A}=[\mathbf{v_1}\ \mathbf{v_2}\  ...\  \mathbf{v_n}]$ are not linearly independent (since a collection of vectors $\mathbf{v}_1, \mathbf{v}_2, ...\mathbf{v}_n$ are linearly independent iff $\mathbf{v}_1x_1+\mathbf{v}_2x_2+...+\mathbf{v}_nx_n=\vec{\mathbf{0}} \implies x_1=x_2=...=x_n=0$). Since a set of $n$ vectors are not linearly independent, we directly have that the dimension of the span of these $n$ linearly dependent vectors is less than $n$, which is less than the dimension of $\mathbb{R}^n$. Or: $\text{dim}\{\text{span} \{\mathbf{v_1},\mathbf{v_2}, ..., \mathbf{v_n}\}\}<n=\text{dim}\{\mathbb{R}^n\}$. Thus the columns of $\mathbf{A}$ cannot span all of $\mathbb{R}^n$.
For the original poster, muros, did this help?
A: If one assumes that row reduction does not add or remove any zero rows (which would be a reasonable operation if the goal is to solve a linear homogeneous system of equations), then it is even true that the row-reduced form$~R$ of $A$ has a zero row if and only if the columns of $A$ do not span their whole space.
It is pretty obvious that the columns of $R$ itself do not span the whole space if and only if $R$ contains a zero row: in the presence of a zero row, the coordinate corresponding to the zero row will always be zero in a linear combination of the columns, while in the absence of any zero row, every row as a distinct pivot column so that an equation $R\cdot x=b$ always has a solution (and a unique one in which nonzero entries of $x$ occur only at positions corresponding to pivot columns).
So the question becomes to show that row operations do not alter the condition whether the columns span their whole space. This is so because these operations (by design) correspond to simple changes of basis applied in that space. An $n\times m$ matrix $A$ naturally corresponds to a linear map $\Bbb R^m\to\Bbb R^n$, but it helps for the mental picture to think of it as composed of $\Bbb R^m\to V\to\Bbb R^n$ where $V$ is some abstract space of dimension$~n$, and the second arrow expresses vectors of$~V$ in coordinates for some basis $(b_1,\ldots,b_n)$ of$~V$. Now row operations correspond to modifications of the basis (keeping the first arrow unchanged) as follows: interchanging rows $i$ and $k$ means interchanging $b_i$ and $b_k$ in the basis; multiplying row$~i$ by a scalar $\lambda\neq0$ means replacing $b_i$ by $\frac1\lambda b_i$ in the basis; adding a scalar multiple by $\mu$ of row$~i$ to row$~k$ means replacing $b_i$ by $b_i-\mu b_k$ in the basis. (The columns of the matrix can be thought of as holding coordinates in the current basis of specific vectors of $V$, and coordinate conversion is multiplication by the inverse of the matrix whose columns give the coordinates of the new basis vectors in the old basis.) Given that row operation are only changes of basis, the question of whether the column span all of $\Bbb R^n$ is just whether the first arrow in the composition has all of $V$ as its image, and this arrow does not change.
