$2\times 2$ matrix such $A$ such that $Ax.x=0$ for all $x$ in $\mathbb{R}^2$ I need to find a $2\times2$ matrix $A$ where $Ax. x=0$ for all $x$ in $\mathbb{R}^2$ (. is dot product).
I tried using a general matrix 
\begin{align}
\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
\end{align}
and a general vector $(v_1, v_2)$ but then I multiply out and I'm lost when it comes to finding actual values for $a,b,c,d$. I'm assuming this has something to do with orthogonality and maybe I'm missing a trick somewhere?
 A: I think you began going in a correct way.
Probably, you reached to the equation:
$$ax^2 + (b+c)xy +dy^2 = 0$$
after having multiplied out.
You are not supposed to solve this equation.
You should give values to a,b,c,d that satisfies the equation for all x and y.
For example: a = d = 0 and b=-c (any c)
$$\left( \begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array} \right)$$
$$\left( \begin{array}{rr} 0 & -\pi \\ \pi & 0 \end{array} \right)$$
$$\left( \begin{array}{rr} 0 & 1889 \\ -1889 & 0 \end{array} \right)$$
A: You probably want to exclude the zero matrix.
How about the matrix representing rotation by 90 degrees? It takes $(1,0)$ to $(0,1)$ and $(0,1)$ to $(-1,0)$. Hence it looks like
$$
\begin{align}
\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}
\end{align}.
$$
You could also consider the appropriate equation directly and get $$av_1^2 + (b+c)v_1v_2+dv_2^2=0,$$ from which you can easily guess the non-trivial solution above.
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
$Ax\cdot x = 0\quad\imp\quad A\+ x\cdot x = 0\imp\quad \pars{A + A\+} x\cdot x
= 0$
$$
A + A\+
=
\pars{%
\begin{array}{cc}
a & b
\\
c & d
\end{array}}
+
\pars{%
\begin{array}{cc}
a & c
\\
b & d
\end{array}}
=
\pars{%
\begin{array}{cc}
2a & b + c
\\
c + b & 2d
\end{array}}
$$
which has real and null eigenvalues $\lambda_{\pm}$ which satisfy:
$$
\pars{\lambda - 2a}\pars{\lambda - 2d} - \pars{b + c}^{2} = 0
\quad\imp\quad
\lambda^{2} - 2\pars{a + d}\lambda + 4ad - \pars{b + c}^{2} = 0 
$$
such that $\quad 0 = \lambda_{+}\lambda_{-} = 4ad - \pars{b + c}^{2}\quad$
and $\quad 0 = \lambda_{+} + \lambda_{-} = 2\pars{a + d}$. Those conditions lead to
$$
d = -a\qquad\mbox{and}\qquad 4a^{2} + \pars{b + c}^{2} = 0
$$ 
The solutions are given by $a = d = 0$ and $c = -b$:
$${\large%
A
=
\pars{%
\begin{array}{cc}
0 & b
\\
-b & 0
\end{array}}}
$$
A: The condition $(Ax)\cdot x = 0$ means that for every $x$, $Ax$ is either zero or is perpendicular to $x$. That means $A$ has to be a (scaled) 90-degree rotation, by which I mean, there is some constant $C$ so that
$$A = C\bigg( \mbox{$\pm$90 degree rotation}. \bigg)$$
