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$V$ is finite-dimensional over $\Bbb{C}$ and the form $\langle \cdot , \cdot \rangle$ is Hermitian. $U$ is a subspace of $V$.

Show that $V = U \oplus U^\perp$

I've been able to show that $U \cap U^\perp = \{0\}$. I don't know how to approach the problem showing that every vector $v\in V$ can be written as $v = u + u'$, where $u, u'$ are in $U$ and $U^\perp$ respectively.

Community
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mathsishard
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  • I have been wondering if there was a direct proof which does not take an orthonormal basis to construct projections. However, such proof may not exist since it is not true in infinite dimensional case (link). – MathDrifter Mar 29 '22 at 18:08

2 Answers2

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Let $\{e_1,\ldots,e_k\}$ be an orthonormal basis of the subspace $U$. For each $v\in V$, let$$P(v)=\sum_{j=1}^k\langle v,e_j\rangle e_j.$$Then$$(\forall v\in V):v=\overbrace{P(v)}^{\phantom{U}\in U}+\overbrace{\bigl(v-P(v)\bigr)}^{\phantom{U^\perp}\in U^\perp}.$$The fact that $v-P(v)\in U^\perp$ can be justified as follows: if $j\in\{1,2,\ldots,k\}$, then\begin{align}\bigl\langle v-P(v),e_j\bigr\rangle&=\left\langle v-\sum_{l=1}^k\langle v,e_l\rangle e_l,e_j\right\rangle\\&=\langle v,e_j\rangle-\langle v,e_j\rangle\\&=0.\end{align}Since $\{e_1,\ldots,e_k\}$ is a basis of $U$, this proves that $v-P(v)\in U^\perp$.

José Carlos Santos
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  • it is only true if $U$ is closed linear subspace ?(w.r.t topology induced by inner product).bcz somewhere i see that this direct implies there exist projection $T$ ,whose range is $U$. and one can prove range of projection is closed – Meet Patel Dec 10 '22 at 14:41
  • @MeetPatel Every vector subspace $U$ of a finite-dimensional vector space $V$ is a closed subset of $V$. – José Carlos Santos Dec 10 '22 at 14:47
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Choose any orthonormal basis $E=\{e_1 ,\ldots, e_k\}$ in $U$. Continue $E$ to a basis $E^+ = \{e_1 ,\ldots, e_k;\; v_{k+1},\ldots, v_n\} $ in $V$. Apply Gram-Schmidt to $E^+$. The first $k$ vectors in $E^+$ already are orthogonormal, and so Gram-Schmidt alters noting in them. The remaining $n-k$ vectors may change, and we get the orthonormal basis $E^{++} = \{e_1 ,\ldots, e_k;\; e_{k+1},\ldots, e_n\} $ in $V$. The vectors $e_{k+1},\ldots, e_n$ are orthogonal to $e_1 ,\ldots, e_k$, and so they all are in $U^\perp$. Present any $v\in V$ as the linear combination $v = (a_1 e_1 +\cdots+ a_k e_k)\;+\; (a_{k+1}e_{k+1}+\cdots + a_n e_n) = u+u'$. Clearly, $u \in U$ and $u' \in U^\perp$.

V. Mikaelian
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