Is there a scalar product s.t. the following list is orthogonal? Let $A_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}\quad A_2=\begin{pmatrix}1&1\\0&0\end{pmatrix},\quad A_3=\begin{pmatrix}1&1\\1&0\end{pmatrix},\quad A_4=\begin{pmatrix}1&1\\1&1\end{pmatrix}$.
Is there a scalar product s.t. $\|A_k\|=k$ for $k=1,2,3,4$ and $A_i\perp A_j$ for $i\neq j$ ? 

With $\langle A, B \rangle = \mbox{Tr}(A^\top B)$ we have that $\|A_i\|=i$, but unfortunately it's not orthogonal. So, how can we conclude?
 A: Let $E_1:=\begin{bmatrix}1&0\\0&0\end{bmatrix}$, $E_1:=\begin{bmatrix}0&1\\0&0\end{bmatrix}$, $E_3:=\begin{bmatrix}0&0\\1&0\end{bmatrix}$, and $E_4:=\begin{bmatrix}0&0\\0&1\end{bmatrix}$.  Then, the matrix of transformation sending $(E_1,E_2,E_3,E_4)$ to $(A_1,A_2,A_3,A_4)$ is
$$T:=\begin{bmatrix}1&1&1&1\\0&1&1&1\\0&0&1&1\\0&0&0&1\end{bmatrix}\,.$$
The required bilinear form in the basis $(A_1,A_2,A_3,A_4)$ is given by the matrix
$$B:=\begin{bmatrix}1&0&0&0\\0&4&0&0\\0&0&9&0\\0&0&0&16\end{bmatrix}\,.$$
Therefore, in the basis $(E_1,E_2,E_3,E_4)$, the bilinear form is given by the matrix
$$\left(T^{-1}\right)^\top\,B\,T^{-1}=\begin{bmatrix}1&-1&0&0\\-1&5&-4&0\\0&-4&13&-9\\0&0&-9&25\end{bmatrix}\,.$$
In other words, this bilinear form is given by
$$\left\langle \begin{bmatrix}a&b\\c&d\end{bmatrix},\begin{bmatrix}x&y\\z&w\end{bmatrix}\right\rangle=ax-ay-bx+5by-4bz-4cy+13cz-9cw-9dz+25dw$$
for $a,b,c,d,x,y,z,w\in\mathbb{R}$.
If the base field is $\mathbb{C}$, then the bilinear form (or rather, the sesquilinear form) is given by
$$\left\langle \begin{bmatrix}a&b\\c&d\end{bmatrix},\begin{bmatrix}x&y\\z&w\end{bmatrix}\right\rangle=a\bar x-a\bar y-b\bar x+5b\bar y-4b\bar z-4c\bar y+13c\bar z-9c\bar w-9d\bar z+25d\bar w$$
for $a,b,c,d,x,y,z,w\in\mathbb{C}$.
A: Lemma. Let $V$ be a real vector space and $v_1,...,v_n$ a basis. Then there exists an inner product on $V$ that makes $(v_i)_i$ an orthonormal basis.
Proof. Define $\langle v_j,v_k\rangle = \delta_{jk}$ and extend it linearly.

For your Problem: Check that the $A_j$ form a basis of $\mathbb{R}^{2\times 2}$. Then also $(A_j/j)_j$ is a basis and the inner product from the lemma does the job.
