A Subalgebras and direct sums

1. May 18, 2018 at 9:10 AM

mnb96

Hi,

consider a (finite dimensional) vector space $V=U\oplus W$, where the subspaces $U$ and $V$ are not necessarily orthogonal, equipped with a bilinear product $*:V\times V \rightarrow V$.

The subspace $U$ is closed under multiplication $*$, thus $U$ is a subalgebra of $V$.

Does this imply that also $W$ is a subalgebra of $V$?

(Note, I can already prove the special case that if U and W are orthogonal, then both U and W are indeed subalgebras).

2. May 18, 2018 at 9:24 AM

Staff: Mentor

Just as a side note: orthogonal doesn't make sense, as long as you don't specify the quadratic form and the field. Vector spaces in general don't automatically allow inner products.

$h=\begin{bmatrix}1&0\\1&-1\end{bmatrix}\; , \;x=\begin{bmatrix}0&1\\0&0\end{bmatrix}\; , \;y=\begin{bmatrix}0&0\\1&0\end{bmatrix}$ with $V=\operatorname{span}_\mathbb{F}\{\,h,x,y\,\}\; , \;U=\mathbb{F}\cdot h\; , \; W=\operatorname{span}_\mathbb{F}\{\,x,y\,\}$. With the multiplication $v*w= v\cdot w - w \cdot v$ we have $h*h=0\in U$ and $x*y=h \notin W$.