Basis vectors question Suppose the columns of a $5\times 5$ matrix $A$ are a basis for $\mathbb R^5$
If $b$ is in $\mathbb R^5$ then $Ax = b$ is solvable because the basis vectors span $\mathbb R^5$
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Why is this the case? I don't get how the basis vectors spanning $\mathbb R^5$ says anything about it being solvable.
 A: It is solvable by the definition of basis. Multiplying A with x represents a linear combination of these basis vectors. Since $b \in \mathbb{R}^5$ it is possible to linearly combine it from the columns of A. And the matrix vector  product $Ax$ is exactly a linear combination of the columns of A.
A: Write the matrix $A$ as 5 columns vectors $A = (A_1 \vert A_2 \vert A_3 \vert A_4 \vert A_5)$ and the vector $x = (x_1, \dots, x_5)^T$.
The product $Ax$ is equal to the vector
$$x_1 A_1 + x_2 A_2+x_3 A_3+x_4 A_4+x_5 A_5.$$
Saying that the columns of $A$ span $\mathbb R^5$ means that any vector can be written as a linear combination of the columns $A_1, \dots, A_5$. This should be the case in particular of the vector $b$. This is exactly saying that $Ax=b$ has a solution.
A: Set $A=(C_1, \cdots, C_5)$ where $C_i$ is the $i$-th column of $A.$ Now, $Ax=b$ is solvable if and only if there exists $x_1,\cdots, x_5\in \mathbb{R}$ such that $$x_1C_1+\cdots+x_5C_5=b.$$ Given $b\in \mathbb{R}^5$ the existence of $x_1,\cdots,x_5\in \mathbb{R}$ is guaranteed because $\{C_1, \cdots, C_5\}$ is a basis of $\mathbb{R}^5.$
