Find a nonzero matrix $A$ in $M_{2\times 2}(R)$ satisfying $v\cdot Av=0$ for every $v\in R^2$ Ok, so I have this problem: Find a nonzero matrix $A$ in $M_{2\times 2}(R)$ satisfying $v\cdot Av=0$ for every $v \in R^2$.
So if I say that $v=\begin{bmatrix}x\\y\end{bmatrix}$ and that $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$  so that $A*v=\begin{bmatrix}ax+by\\cx+dy\end{bmatrix}$.  So I'm pretty sure that Im being asked to find $a,b,c,d$ such that $v \cdot Av=x(ax+by)+y(cx+dy)=0$
Now, since it has to be a nonzero matrix would it make sense to make the values opposites of each other so that they cancel out?? If not, can someone help in what to do?
 A: Since your result has to hold for every $v \in \mathbb{R}^2$. So try the following three vectors:
$$v_1=\begin{bmatrix}1\\0\end{bmatrix}, \quad v_2=\begin{bmatrix}0\\1\end{bmatrix}, \quad v_3=\begin{bmatrix}1\\1\end{bmatrix}.$$
You will get sufficient information about the matrix $A$.
For example
$$v_1 \cdot (Av_1)=v_1 \cdot \left(\begin{bmatrix}a &b\\c&d\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\right)=v_1 \cdot \begin{bmatrix}a\\c\end{bmatrix}=a(1)+c(0)=a.$$
A: If you look at the requirement $v \cdot Av = 0$, $\forall v \in \Bbb R^2$ carefully you will soon realize that it forces the matrix $A$ to be skew-symmetric, that is, $A^T = -A$.  This follows from a standard argument, for we may write
$(v_1 + v_2) \cdot A(v_1 + v_2) = 0 \tag{1}$
for any $v_1, v_2 \in \Bbb R^2$.  Then
$v_1 \cdot Av_1 + v_1 \cdot Av_2 + v_2 \cdot Av_1 + v_2 \cdot A v_2 = 0, \tag{2}$ 
and since $v_i \cdot Av_i = 0$ for $i = 1, 2$ we have
$v_1 \cdot Av_2 + v_2 \cdot Av_1 = 0. \tag{3}$
(3) yields
$v_1 \cdot Av_2 = - v_2 \cdot Av_1 = -A^Tv_2 \cdot v_1 = - v_1 \cdot A^Tv_2, \tag{4}$
showing that $A^T = -A$.  It is easy to see then that $A^T = -A$ implies $a = -a$, $d = -d$, $c = -b$, so we must have $a = d = 0$ but $b$ may be chosen freely.  If one is not acquainted with the formula $v_1 \cdot B v_2 = B^Tv_1 \cdot v_2$ used in (4), one may simply choose
$v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \tag{5}$
and
$v_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \tag{6}$
and explicitly evaluate (3) together with $v_i \cdot Av_i = 0$, $i = 1, 2$, to obtain the same result.  It is easy to see that with
$A = \begin{bmatrix} 0 & b \\ -b & 0 \end{bmatrix} \tag{7}$
that $v \cdot Av = 0$ holds for all $v \in \Bbb R^2$.
Note: The relation $A^T = -A$ follows from $v \cdot Av = $ for $A \in M_{n \times n}(\Bbb R)$, we don't need $n = 2$ to obtain this result.  End of Note.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
