TITLE: Basis (linear algebra)
QUESTION [7 upvotes]: Let's have vectors $v_1,v_2,v_3,v_4\in \mathbb R^4$. Prove that these vectors form basis (1), determine if they are orthogonal and/or orthonormal (2). Then make a transition matrix $T_{\epsilon\alpha}$  from basis $\alpha$ to standard basis and use it to get coordinates of vector $w=(2,3,5,1)_\alpha$ in standard basis (3).
$$v_1=(1,1-1,1),v_2=(1,-1-1,-1),v_3=(0,1,0,-1),v_4=(1,0,1,0) $$
(1) Prove that these vectors form a basis.

I did following:
  $$\left( \begin{array}{cccc}
 1 & 1 & -1 &1 \\
1 & -1 & -1 &-1 \\
0 & 1 & 0 &-1 \\
1 & 0 & 1 & 0  
\end{array} \right)=A$$
  $$\left| \begin{array}{cccc}
 1 & 1 & -1 &1 \\
1 & -1 & -1 &-1 \\
0 & 1 & 0 &-1 \\
1 & 0 & 1 & 0  
\end{array} \right|=-8 \ $$
  $|A| \neq 0 \implies$ Vectors $v_1,v_2,v_3,v_4$ are linearly independent and form basis.

(2) Determine if they are orthogonal and/or orthonormal. 

I did following:
  $$v_1\cdot v_2=(1-1+1-1)=0$$
  $$v_1\cdot v_3=(0+1+0-1)=0$$
  $$v_1\cdot v_4=(1+0-1+0)=0$$
  $$v_2\cdot v_3=(0-1+0+1)=0$$
  $$v_3\cdot v_4=(0+0+0+0)=0$$
  $\implies$ Basis is orthogonal.
  $$|v_1|=\sqrt {1+1+1+1}=\sqrt4=2$$
  $$|v_2|=\sqrt {1+1+1+1}=\sqrt4=2$$
  $$|v_3|=\sqrt {0+1+0+1}=\sqrt2$$
  $$|v_4|=\sqrt {1+0+1+0}=\sqrt2$$
  $\implies$ Basis is not orthonormal.

(3) Make a transition matrix $T_{\epsilon\alpha}$  from basis $\alpha$ to standard basis and use it to get coordinates of vector $w=(2,3,5,1)_\alpha$ in standard basis. 

$$T_{\epsilon\alpha}=A^{-1}=\left( \begin{array}{cccc}
\frac14 & \frac14 & 0 & \frac12 \\
\frac14 & -\frac14 & \frac12 &0 \\
-\frac14 & -\frac14 & 0 & \frac12 \\
\frac14 & -\frac14 & -\frac12 & 0  
\end{array} \right)$$
$$u=A^{-1} w$$ Where $u$ is in standard basis?

Is (1) and (2) correct and can you help me out with (3) ?

REPLY [2 votes]: It's important to remember that a vector $w$ written in terms of basis $\alpha = \{v_1, v_2, v_3, v_4\}$ has the form:
$$( a, b, c, d)_{\alpha} = 
av_1 + bv_2 + cv_3 + dv_4$$
It's also important to remember that when your vectors $v_i$ are written in terms of coordinates, that these are coordinates with respect to the standard basis.  For example,
$$( 1,  0,  0, 0)_{\alpha} = 
v_1 = 
(1,  1, -1, 1)_{\epsilon}$$
Therefore, the matrix $T_{\epsilon \alpha}$ should have the property that:
$$T_{\epsilon \alpha} (a,b,c,d) = a( 1, 1, -1, 1)+
b( 1,  -1, -1,-1) + c( 0,  1, 0,  -1)+
d( 1,  0,  1, 0)$$
Thus, $T_{\epsilon \alpha} = A$, the matrix you've written above, whose rows are the standard-basis representations of the vectors $v_i$ in the given order.