Prove that if rref of matrix $\bf (v_1\space v_2 \space \cdots v_m)$ has row of zeroes, vectors $\bf v_1,\cdots v_m$ do not span $\mathbb R^{n}$

Question

Proposition

Consider vectors $\mathbf{v}_1,\mathbf{v}_2,\cdots \mathbf{v}_m$ in $\mathbb R^n$. Define matrix $\mathbf A := \begin{pmatrix} \mathbf{v}_{1} & \cdots & \mathbf{v}_m \end{pmatrix}$. If reduced row echelon form of matrix $\bf A$ has row of zeroes, then vectors $\mathbf{v}_1,\mathbf{v}_2,\cdots \mathbf{v}_m$ do not span $\mathbb R^{n}$

My attempt:

Let $\mathbf{R} := rref(\mathbf A)$ and suppose $\bf R$ has at least one row of zeros. Consider arbitrary n-dimensional vector $\bf b$ such that $\mathbf{b}_{n} ≠ 0$. Since n-th row of $\bf R$ is $\bf O$, linear system $\bf Rx = b$ is inconsistent. Since $\bf R$ is row equivalent to $\bf A$, we have $\bf E_1E_2 \cdots E_kA = R$, where $\bf E$ represents elementary matrix. It follows that linear system $\bf Ax = \bigr(E_1E_2\cdots E_k)^{-1}b$ is inconsistent, which means that the vector $\bf \bigr(E_1E_2\cdots E_k)^{-1}b$ cannot be spanned by $\mathbf{v}_1,\mathbf{v}_2,\cdots \mathbf{v}_m$. $\blacksquare$

Is it correct?

Answer 1

As stated in the problem, the last row consists of all zeros.

Let $w \in \mathbb{R}^n$ and $w_n=1$.

Assume $w= \sum_{j=1}^m \alpha_j v_j$.

All $v_{m,n}=0, \Rightarrow w_n=0$. Contradiction.