Matrix spanning R$^3$ Why doesn't $(1,-1,1),(1,1,-1),(1,1,1)$ span $R^3$
Let there be any vector $X$ and then we apply the basic thing that i.e for any three scalars $c_1,c_2$ and $c_3$ and $X_1=(1,-1,1)$ $X_2=(1,1,-1)$ , $X_3=(1,1,1)$ and  $X=(x,y,z)$
To span in  $R^3$
we must have 
$$c1X_1 + c_2X_2+C_3X_3=X$$
Making the augmented matrix with this we get
$$ \left[
    \begin{array}{ccc|c}
      1&1&1&x\\
      -1&1&1&y\\
      1&-1&1&z
    \end{array}
\right] $$
Reducing to the upper triangular form, it gives me 
$$ \left[
    \begin{array}{ccc|c}
      1&1&1&x\\
      0&2&2&y+x\\
      0&0&2&z+y
    \end{array}
\right] $$
This shows that I am getting unique solution for $c_1,c_2$ and $c_3$  according to the values of $X=(x,y,z)$ but still this doesn't span $R^3$. Why?
 A: Clearly $(1,-1,1)$ and $(1,1,-1)$ are linearly independent, and for any linear combination the two, the sum of the second and third coordinate is $0$, which means $(1,1,1)$ cannot be such a linear combination. Therefore all three are linearly independent. Since we have three linearly independent vectors in $\Bbb R^3$, they must span all of $\Bbb R^3$.
There must be a mistake somewhere. Either you've taken the wrong there vectors, or there's a mistake your answer key.
A: Theorem: If $V$ is a finite dim vector space and T is a linear transformation $T:V\rightarrow V$ then $dimV=dim ImT+dim KerT$ 
In your case: $V=\mathbb R^3$ , and $T$ is the matrix you gave.
So it is enough to check $dimKerT$ to see if $T$ spans all $V$.
we write:
a(1,1,1)+b(1,-1,1)+c(1,1,-1)=(0,0,0)
(a+b+c,a-b+c,a+b-c)=(0,0,0)
Now you have only 2 options:
If there is only one solution a=b=c=0 then the matrix spans V
If there is at least one more solution then $dimKerT\geq 1$ hence $dimImT\leq 2$ then the matrix doesn't span V.
