Prove vectors $v_1, v_2, v_3$ span $\mathbb{R}^3$ given that they are linearly independent Given the vectors $v_1, v_2, v_3 \in \mathbb{R}^3$ are linearly independent, I would like to show that $\mathrm{span}(v_1,v_2,v_3) = \mathbb{R}^3$.
Here is what I have tried: Given that $v_1, v_2$ and $v_3$ are linearly independent, then I know that there is some linear combination of the three vectors that would give me $(0,0,0)$. Now I just want to use that to show that there is some linear combination that will give me any $(a,b,c) \in \mathbb{R}^3$. Am I thinking about this proof the right way? Any insights will be appreciated. Thank you.
 A: A basis for $\mathbb{R}^n$ has exactly $n$ vectors. Since you have $3$ vectors that are linearly independent in $\mathbb{R}^3$, it follows that these vectors span. 
A: Since the three vector are linearly independent, the subspace they span has dimension $3$. But then this subspace must be the whole space $\mathbb R^3$.
A: $m_1(x_1,y_1,z_1) + m_2(x_2,y_2,z_2) + m_3 (x_3,y_3,z_3) = (a,b,c)$  this is what we want to show.
but the left had side is equal to $(m_1x_1+m_2x_2+m_3x_3, m_1y_1+m_2y_2+m_3y_3, m_1z_1+ m_2z_2+m_3z_3)$
So we need to show that $m_1x_1+m_2x_2+m_3x_3 = a$, $m_1y_1+m_2y_2+m_3y_3 = b$ and $m_1z_1+ m_2z_2+m_3z_3 =c$.
but all the $x_i$ s and $y_i$'s are constants so we basically have three equations with 3 unknowns ie equations in the form of for example $6m_1 + 5m_2$ and $7m_3 = 7$ and they are all linearly independent so there exists a solution.
A: Let $v_1, v_2, v_3$ be linearly independent. Construct the $3 \times 3$ matrix $M = (v_1 | v_2 | v_3)$ by appending the vectors side-by-side. If you know the Rank-Nullity Theorem, then you know that this implies that the rows of $M$ are linearly independent. If you don't know this theorem yet, then the following argument shows this. Suppose the rows of $M$ are linearly dependent but the columns are linearly independent. If we consider the determinant as a function of the rows of a matrix, then $\mathrm{det}(M) = 0$ since the rows are linearly dependent. On the other hand, $\mathrm{det}(M^T) \neq 0$ since the columns are linearly independent. But it is very easy to show that $\mathrm{det}(M) = \mathrm{det}(M^T)$. This gives a contradiction so the rows can't be linearly independent.
Now, take any vector $u = (a,b,c) \in \mathbb{R}^3$. We want to show that there exists some vector $v = (x,y,z)$ so that $Mv = u$. But now we may consider the rows of the matrix equation $Mv=u$ to be linear equations. As there are three linearly independent equations in three variables, there is a unique solution (the easiest way to see this is to row-reduce $M$, since $\mathrm{det}(M) \neq 0$ the reduced form must be the identity).
A: There's not much to prove.  $n$ (to be a little more general)  linearly independent vectors in an $n$-dimensional space always form a basis, and hence span.
Plus, your thinking is off a bit, I would say.  In some sense, it's never any problem getting the zero vector in the span of a set of vectors: the trivial linear combination will work.  But, you are correct that the challenge is to show that any triple $(a,b,c)$ at all is in the span.
A: Define the matrix $M$ by $M=(v_1,v_2,v_3)$. Then the function $f:\mathbb{R}^3\to\mathbb{R}^3$ defined by $f(\mathbf{x})=M\mathbf{x}$ is a linear transform and, since columns of $M$ are independent we have that $f(\mathbf{x})=0$ iff $\mathbf{x}=0$. Then $\ker f=\{0\}$ and by rank nullity thoerem
$$\dim \text{im} f=3-\dim \ker f=3$$
which implies $\text{im} f=\mathbb{R}^3$.
