How can an inner product be defined through a proof? I'm having a really hard time with question b of the image below, (Find any non-trivial A and B such that they are orthogonal) and question c, the proof. 
I know that non trivial means a nonzero solution, but how do I interpret the subscript? And does part c utilize Cauchy Schwarz? It looks almost like the pythagorean theorem but I'm not sure if I'm way off. 
Any help or guidance you can provide would be super helpful, thanks! 
 A: For b) we need only find $A$ and $B$ non-zero so that the inner product is $0$, i.e.,
$$ \langle A, B \rangle = \left\langle \begin{bmatrix} a_{11} & a_{12}\\ 0 & a_{22} \end{bmatrix},
\begin{bmatrix} b_{11} & b_{12}\\ 0 & b_{22} \end{bmatrix}\right\rangle 
= a_{11}b_{11} + a_{12}b_{12} + a_{22}b_{22} = 0$$
So one easy example would be taking $a_{11}=a_{12}=b_{22}=0$. This would give a sum of zeros, even if $A$ or $B$ were non-trivial. For instance,
$$ A= \begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}, \quad
B= \begin{bmatrix} 3 & -17\\ 0 & 0 \end{bmatrix},\qquad \text{then $\langle A,B\rangle = 0$}$$
As for c), you are correct in recognizing the Cauchy-Schwartz inequality. As we have an inner product, it must satisfy
$$\langle A, B\rangle^2 \leq \langle A, A\rangle \langle B,B\rangle$$
where
$$\langle A, B\rangle^2 = (a_{11}b_{11}+a_{12}b_{12}+a_{22}b_{22})^2$$
$$\langle A, A\rangle \langle B,B\rangle
= (a_{11}a_{11}+a_{12}a_{12}+a_{22}a_{22})(b_{11}b_{11}+b_{12}b_{12}+b_{22}b_{22})
= (a_{11}^2+a_{12}^2+a_{22}^2)(b_{11}^2+b_{12}^2+b_{22}^2)$$
A: You can map each of this matrix to a vector of size 3. In each matrix there is a a zero entry. Ignore that zero and take the other three entries as components of a vector )that is the x,y,z co-ordinates)..
After this process it is clear that this question is not really about matrices, it is the same question as for $\mathbf{R}^3$.
If you have written computer programs the above process can be viewed this way. In a program we use, 2d-arrays, or 3d-arrays etc. A 2d-array can be scanned row by row and mapped to a 1-d array (that is what we have done).
