Does there exist a nonzero linear transformation $T$ such that ${\alpha}^{T} T \alpha = 0$ for all $\alpha \in \mathbb{R}^n$? Does there exist a nonzero linear transformation $T$ such that
$$\alpha^t T \alpha = 0\text{ for all $\alpha\in\mathbb{R}^n$,}$$
Where ${\alpha}^{t}$ denotes the transpose of the matrix. 
Thanks! =D
 A: Presumably, your vectors are column vectors and $T$ is $n\times n$, so that $\alpha^t T\alpha$ is a $1\times 1$ matrix (i.e., a scalar). 
Multiplying $v^t$ by $w$ amounts to taking the dot product of $v$ with $w$. Since $v\cdot w=0$ if and only if $v$ and $w$ are orthogonal, the question is equivalent to asking if you can find a linear transformation that sends every vector to a vector orthogonal to itself, and does not send all vectors to zero. In two dimensions, this is trivial. If you can do it in two dimensions, you can do it in any larger number (think about it). 
A: Sure. If $n=2$, let $T$ be a $90^\circ$ rotation. A suitable modification of this works for all $n\ge 2$: Simply rotate the first two coordinates $90^\circ$, and map the rest to 0. 
If $n=1$, there is an $r\in{\mathbb R}$ such that $T\alpha=r\alpha$ for all $\alpha$, and $\alpha^T T\alpha$ is simply $r\alpha^2$, so if $n=1$, $T$ must be 0.
A: If $n$ is even then the block matrix  
$$T = \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}$$
works. For odd $n$ the matrix $T$ may be adjusted by simply including $\mathbb{R}^n$ in $\mathbb{R}^{n-1}$ by sending one component to zero. Of course this fails for $n=1$, and as noted in other answers $T$ must be zero for $n=1$.
A: The property holds if and only if $T$ is skew-symmetric (meaning that $T^t=-T$). (The other answers so far are special cases of this.)
Proof: Let $x=\alpha^t T \alpha$. Since $x$ is a $1\times 1$ matrix, it equals its own transpose. Thus $2x=x+x^t=\alpha^t T \alpha+(\alpha^t T \alpha)^t =\alpha^t (T+T^t)\alpha$, which is a quadratic form in $\alpha$ that corresponds to the symmetric matrix $T+T^t$; thus it is zero for all $\alpha$ iff $T+T^t=0$.
