Proving that the sum of elements of two bases is a basis I am given a (not necessarily orthonormal) basis of a certain finite vector space $\{e_i\}_{1\leq i\leq n}$.
Now, after the usual Gram-Schmidt orthonormalization procedure, I end up with an orthonormal basis $\{u_i\}_{1\leq i\leq n}$.
I need to prove that, for each $t\in [0, 1]$  the convex combinations $\{te_i+(1-t)u_i\}_{1\leq i\leq n}$ constitute a basis of the same vector space.
I tried two things to no avail:
First, I used the definition of linear independence: I formed the sum:
$$\sum_i \left(\alpha_i te_i+\alpha_i (1-t)u_i\right)=0$$
$$t\sum_i \alpha_i (e_i-u_i)+\sum_i \alpha_iu_i=0$$
which doesn't help me. I tried plugging in the explicit recursive formula for the Gram-Schmidt orthonormalization procedure but  it ends up as a mess.
I also tried to do this with the help of matrices. The arbitrary basis can be put in matrix form(vectors as columns) as an invertible matrix $A$. After the orthonormalization I get an orthogonal matrix $U$. The combination I need to prove is an invertible matrix is $tA+(1-t)U$. For $t\in \{0,1\}$ this is obvious, the case when $t$ is neither is problematic. If I assume this is not invertible, then there must exist a non zero eigenvector $v$ with zero eigenvalue:  $(tA+(1-t)U)v=0$ or $Av=\frac{t-1}{t}Uv$. I can't find a contradiction yet.
Any help would be appreciated. Thanks
 A: You don't need the fact that the $u_i$ are orthonormal, just the fact that $u_i$ lies in the span of $e_1,e_2,\ldots,e_i$ for every$~i$, and its expression involves $e_i$ with a positive coefficient. In other words the matrix $A$ which expresses the vectors $u_i$ in coordinates on the basis $[e_1,\ldots,e_n]$ is upper triangular with positive diagonal coefficients. Then $tI+(1-t)A$ is easily seen to have the same property for all $t\in[0,1]$, and this is the matrix expressing your convex combination in coordinates on the basis $[e_1,\ldots,e_n]$. Since this implies the matrix is invertible, you've got a basis of the vector space.
A: Since $\{u_n\}$ is derived from Gram-Schmidt procedure, we have
$${e_i}\cdot {u_j}=\left\{ \begin{align}
  & 0,\quad j>i; \\ 
 & p_{ij}>0,\quad j=i. \\ 
\end{align} \right.$$
For any non-trival linear combination $\sum_i \left(\alpha_i te_i+\alpha_i (1-t)u_i\right)=0$, suppose $j$ is the last subscript such that $\alpha_j \ne 0$. Then
$$0=u_j\cdot \sum_i{\left( \alpha _ite_i+\alpha _i\left( 1-t \right)u_i \right)}=p_{ij} \alpha_jt+\alpha _j\left( 1-t \right)=\alpha _j \left( p_{ij}t + \left( 1-t \right) \right).$$
Hence, $p_{ij}t+(1-t)=0$, which is impossible since $p_{ij}>0$ and $t \in [0,1]$.
