Finding Standard Matrix from Basis Let $S=\{v_1, v_2, v_3\}$ be a basis for $\mathbb{R^3}$ where 
$v_1 = (1, 0, 1)$
$v_2 = (-1, 1, -1)$
$v_3 = (-1, 1, 0)$
Let $L: \mathbb{R^3} \mapsto \mathbb{R^2}$ be the linear transformation defined by 
$L(v_1) = (2, 7)$
$L(v_2) = (4, 2)$
$L(v_3) = (5, 4)$
1.) Find $L(e_1), L(e_2)$ and $L(e_3)$
2.) Find the standard matrix of $L$
3.) Find a nontrivial vector $u$ in the kernel of $L$. Express your answer in the standard basis. 
4.) Write your vector $u$ with respect to the basis $S$.
I'm trying to figure these out, but I don't really understand the concept of getting this information from the basis. 
 A: You must first determine what linear combination of the three basis vectors (v1, v2, and v3) is needed to get each of the standard basis vectors e1, e2, and e3.  This is accomplished by solving three separate 3x3 linear systems:
(1)  a1v1 + a2v2 + a3v3 = e1
(2)  b1v1 + b2v2 + b3v3 = e2
(3)  c1v1 + c2v2 + c3v3 = e3
where a=(a1, a2, a3) and b=(b1, b2, b3), and c= (c1, c2, c3) are the scalar coefficients you must determine.
The systems solve easily enough to:
a = (1, 1, -1)   b = (1, 1, 0)  and c = (0, 1, -1)
which means that:
e1 = v1 + v2 - v3
e2 = v1 - v2
e3 = v2 - v3
Now it is straightforward to compute L(e1) = a1(2,7) + a2(4,2) + a3(5,4) = 
 (1)(2,7) + (1)(4,2) + (-1)(5,4)  =  (1, 5)
and L(e2) = b1(2,7) + b2(4,2) + b3(5,4)  =  (6,9)
and L(e3) = c1(2,7) + c2(4,2) + c3(5,4)  =  (-1,-2)
So the standard matrix of L is then [ (1,5), (6,9), (-1,-2) ]
To find a vector u in the kernel of L you must solve for a non-trivial solution to L(u) = 0
Finally you can write u with respect to the basis S, by using the [a,b,c] transformation you solved earlier:
u(S) = (a,b,c)*u
where u(S) is the u vector expressed with respect to the S basis.
A: The matrix of $L$ rel $S$ and the standard basis is $$\begin{pmatrix}2\quad 4\quad 5\\7\quad 2\quad 4\end {pmatrix}$$
The transition matrix is $$T=\begin {pmatrix}1\quad-1\quad-1\\0\quad 1\quad 1\\1\quad-1\quad 0\end {pmatrix}$$.
We need to multiply by $T^{-1}$ on the right:  $$ LT^{-1}=\begin {pmatrix}1\quad 6\quad 1\\5\quad 9\quad 2\end {pmatrix}$$, is what I got, after computing $T^{-1}$.  So, this is the matrix for $L$ rel the standard bases.
$1)$ is now the columns.
$2)$ is done.
For $3)$, we need nonzero $x$ such that $LT^{-1}(x)=0$.
For that we could do Gaussian elimination,  but instead I'll do the cross product of the rows:  $$(1,6,1)×(5,9,2)=\begin {vmatrix}i\quad j\quad k\\1\quad 6\quad 1\\5\quad 9\quad 2\end {vmatrix}=(3,3,-21)$$.
For $4$, we need to multiply by $T^{-1}$:  get $$T^{-1}\begin {pmatrix}3\\3\\-21\end {pmatrix}=\begin {pmatrix}6\\27\\-24\end {pmatrix}$$.
