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A Subalgebras and direct sums

  1. May 18, 2018 at 9:10 AM #1
    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. jcsd
  3. May 18, 2018 at 9:24 AM #2

    fresh_42

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    2017 Award

    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.

    The answer to your question is no. Example:
    ##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##.
     
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