ThmDex – An index of mathematical definitions, results, and conjectures.
Columns of a real matrix which span the whole space need not be real-linearly independent
Formulation 0
Let $A \in \mathbb{R}^{2 \times 3}$ be a D4571: Real matrix such that
(i) $a_1, a_2, a_3 \in \mathbb{R}^{2 \times 1}$ are each a D5200: Real column matrix
(ii) \begin{equation} A = \begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix} \end{equation}
Then
(1) \begin{equation} \text{Span}_{\mathbb{R}} \langle a_1, a_2, a_3 \rangle = \mathbb{R}^{2 \times 1} \end{equation}
(2) \begin{equation} 0 \cdot a_1 + 1 \cdot a_2 - 1 \cdot a_3 = \boldsymbol{0} \end{equation}
Proofs
Proof 0
Let $A \in \mathbb{R}^{2 \times 3}$ be a D4571: Real matrix such that
(i) $a_1, a_2, a_3 \in \mathbb{R}^{2 \times 1}$ are each a D5200: Real column matrix
(ii) \begin{equation} A = \begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix} \end{equation}
The second claim is clear. For any $x \in \mathbb{R}^{2 \times 1}$, we can just use the first two columns to write \begin{equation} x = x_1 \cdot a_1 + x_2 \cdot a_2 \end{equation} which establishes the first claim. $\square$