Can you explain their leap with the word "thus" in the following problem. 
I'm asking about part (c).  Here's what the solution manual has:

I don't get how they can put < vi , vj > = 1 or 0, THUS Q*Q=I.  Where can I find an explanation for that leap?
 A: Note that the question says "For any orthonormal basis $\beta$ for $V$...".
In an orthonormal basis, firstly you know that every basis vector is orthogonal to every other basis vector, so $\langle v_i, v_j \rangle = 0$ for $i\neq j$. Secondly, you know the basis is normalised, so $\langle v_i, v_j \rangle = 1$ for $i = j$. Therefore:
\begin{align*}
Q^*Q = \begin{pmatrix}-&v_1^*&-\\ -&v_2^*&-\\ &\vdots\ \\- &v_n^*&-\end{pmatrix}\begin{pmatrix}|&|& &|\\ v_1&v_2& \cdots & v_n\\ | & | & & |\end{pmatrix} = \begin{pmatrix} v_1^*v_1 &  v_1^*v_2 & \cdots & v_1^*v_n\\ v_2^*v_1 &  v_2^*v_2 & \cdots & v_2^*v_n\\
\vdots & \vdots & \ddots & \vdots\\
v_n^*v_1 &  v_n^*v_2 & \cdots & v_n^*v_n\\ \end{pmatrix} = \begin{pmatrix} \langle v_1,v_1\rangle &  \langle v_1,v_2 \rangle & \cdots & \langle v_1,v_n\rangle \\ \langle v_2,v_1 \rangle &  \langle v_2,v_2\rangle & \cdots & \langle v_2,v_n\rangle \\
\vdots & \vdots & \ddots & \vdots\\
\langle v_n,v_1\rangle &  \langle v_n,v_2 \rangle & \cdots & \langle v_n,v_n\rangle\\ \end{pmatrix} = \begin{pmatrix} 1 &  0 & \cdots & 0 \\ 0 &  1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots\\
0 &  0 & \cdots & 1 \end{pmatrix} = I
\end{align*}
The equality between the identity matrix and the matrix with inner products in it results from the restrictions on the inner products due to the orthonormality of the vectors, as explained above.
A: One way to think about matrix multiplication is that if $m \times n$ matrix $A$ has rows $a_1^*, a_2^*, \dots, a_m^*$ and $n \times p$ matrix $B$ has columns $b_1, b_2, \dots, b_p$, then $$\newcommand{\<}{\langle}\renewcommand{\>}{\rangle}AB = \begin{bmatrix}
  \<a_1, b_1\> & \<a_1, b_2\> & \dots & \<a_1, b_p\> \\
  \<a_2, b_1\> & \<a_2, b_2\> & \dots & \<a_2, b_p\> \\
 \vdots & \vdots & \ddots & \vdots \\
 \<a_m, b_1\> & \<a_m, b_2\> & \dots & \<a_m, b_p\>
\end{bmatrix}.$$
In your case, the condition that $\<v_i, v_j\> = 1$ if $i=j$ and $0$ otherwise is therefore precisely the condition you need to know that $Q^*Q = I$: it tells you what $Q^*Q$ is, entry by entry.
