Irreducible representation of $\mathfrak{so}_n(\mathbb{R})$ in $\mathbb{R}^n$ Let $\mathfrak{so}_n(\mathbb{R})$ be the Lie algebra of real skew symmetric matrices. The natural basis representation of this algebra is given by, 
$$
\Delta_{i,j} = E_{i,j}-E_{j,i},\quad 1\leq i<j\leq n,
$$
where $E_{i,j}\in\mathfrak{gl}_n(\mathbb{R})$ is the matrix whose elements are $0$ except for the $(i,j)$th which is $1$. This basis descibe the infinitesimal rotations.
Questions: Is there other irreducible matrix representations of $\mathfrak{so}_n(\mathbb{R})$ in $\mathfrak{gl}_n(\mathbb{R})$ ? If yes, how to identify the canonical representation above among all representations ? 
Thank you. 
 A: I'll consider all representations $\mathfrak{so}_n(\mathbf{R})\to\mathfrak{gl}_n(\mathbf{R})$, not only irreducible ones.
Of course there's no uniqueness (for $n\ge 2$) since you can conjugate the standard representation by a fixed matrix in $\mathrm{GL}_n(\mathbf{R})$. Also, you can consider the identity representation (which for $n\ge 2$ is not the standard one and is not irreducible). Nevertheless, for $n\notin\{2,4,8\}$, this are the only representations.
To show that this is the case, first assume $n\ge 3$. Let $f:\mathfrak{so}_n(\mathbf{R})\to\mathfrak{gl}_n(\mathbf{R})$ be a nontrivial $n$-dimensional representation. Then it induces a nontrivial continuous representation $f':\mathrm{Spin}(n)\to\mathrm{GL}_n(\mathbf{R})$. Since $n\ge 3$, $\mathrm{Spin}(n)$, the two-fold (simply) connected covering of $\mathrm{SO}(n)$ (with kernel $Y_n$, is compact, so after conjugation we can suppose that the image of $f'$ preserves a the standard scalar product. Hence, by connectedness, $f'$ maps into $\mathrm{SO}(n)$. Now exclude $n=4$: this ensures that the only closed normal subgroups of $\mathrm{Spin}(n)$ are the whole group and the subgroups of the center (which has order $2$ if $n$ is odd, $4$ if $n$ is even). Since $f'$ is not trivial, we see that $f':\mathrm{Spin}(n)\to\mathrm{SO}(n)$ is a two-fold covering, say with kernel $Y'$ of order $2$.
I claim, assuming in addition $n\neq 8$, that $Y'=Y_n$. If $n$ is not divisible by $4$, the center of $\mathrm{Spin}(n)$ is cyclic and hence $Y'=Y_n$. If $n$ is divisible by $4$ but not equal to $8$, the automorphism group of $\mathrm{Spin}(n)$ preserves $Y_n$: indeed, it is induced by the action of $\mathrm{O}(n)$. So if $Y'\neq Y_n$ is another central subgroup of order $2$, then $\mathrm{Spin}(n)/Y'$ (which is known as "half-spin group") is not isomorphic to $\mathrm{SO}(n)$. Hence $f'$ induces an automorphism of $\mathrm{SO}(n)$. Hence, after conjugation by some element of $\mathrm{O}(n)$, we have the identity map.

About exceptions $n=2,4,8$:
$n=2$: any one-parameter subgroup yields a representation of $\mathrm{so}(2)$, and most such representation do not preserve a scalar product. Several such representations are irreducible.
$n=4$: we have nontrivial representations factoring through a surjective homomorphism $\mathfrak{so}(4)\to\mathfrak{so}(3)$. The latter has 2 kinds of nontrivial 4-dimensional representations: those non-irreducible ones (1+3 decomposition), and the irreducible ones, whose image can be identified to $\mathbf{SU}(2)$ acting on $\mathbf{C}^2\simeq \mathbf{R}^4$.
$n=8$: "triality": consider the composite representation $\mathrm{Spin}(8)\to\mathrm{SO}(8)\to\mathrm{GL}_8(\mathbf{R})$, and precompose by an automorphism of $\mathrm{Spin}(8)$ that does not preserve $Y_8=\mathrm{Ker}(\mathrm{Spin}(8)\to\mathrm{SO}(8))$. Of course these are (absolutely) irreducible, since the image is $\mathrm{SO}(8)$.
