Projection proofs

Question

Let $V$ be a finite-dimensional inner-product space. Let $U$ be a subspace of $V$ and $(e_1, . . . , e_m)$ an orthonormal basis of $U.$ The orthogonal projection of $V$ onto $U$ is the operator $P_U \in \mathcal L(V)$ defined by $$P_U v = \langle v,e_1 \rangle e_1 +\dots+\langle v,e_m\rangle e_m$$ (a) Use the above definition to prove that $U^{\perp} \subseteq nullP_u$ and $nullP_u \subseteq U^{\perp}$ thus conclude $nullP_u =U^{\perp}$

What I have: $u \in P_uv$ by definition. $(e_1,...e_m)$ is a basis, so $P_uv$ is linearly independent. This implies that if $P_uv=0$ (the $nullP_u$) then $\langle v,e_1 \rangle=\langle v,e_2 \rangle= \space... \space=\langle v,e_m\rangle =0$ But as this is an inner product space, this implies that $v=e_j$ by definiteness. $U^{\perp}=\{v \in V:\langle v,u \rangle=0 \space \forall u\in U \}$ since $u \in P_uv$, $U^{\perp}=\{v \in V:\langle v,\langle v,e_1 \rangle e_1 +\dots+\langle v,e_m\rangle e_m \rangle=0 \space \forall u\in U \}$ I'm stuck on where to go after this though.

b)Use the above definition to prove that $P_u^2 = P_u$

What I have so far

$$P_U(P_Uv) = \langle \langle v,e_1 \rangle e_1 +\dots+\langle v,e_m\rangle e_m,e_1 \rangle e_1 +\dots+\langle \langle v,e_1 \rangle e_1 +\dots+\langle v,e_m\rangle e_m,e_m\rangle e_m = (\langle v,e_1 \rangle e_1 +\dots+\langle v,e_m\rangle e_m)(\langle 1,e_1\rangle e_1 +\dots + \langle 1,e_m\rangle e_m = (\langle v,e_1 \rangle e_1 +\dots+\langle v,e_m\rangle e_m)(\langle e_1, e_1 \rangle +\dots + \langle e_m, e_m\rangle)$$

Answer 1

Here we have $\;U\;$ is a subspace of $\;V\;$ and $\;E\;=\;\{e_{1},e_{2},...e_{m} \}\;$ is an orthonormal basis for $\;U\;.\;$

(a).$\;$Here, $\;\;P_{U}:\;V \to\;U\;\;\text{and}\;\;P_{U}(v)\;=\;\displaystyle\sum_{i=1}^{i=m}\;\langle v,e_{i} \rangle e_{i}\;\;$

For any $\;v \in U^{\perp}\;$ we have $\;\langle v,e_{i} \rangle =0\;\;\forall\;i=1,2,..m\;.\;$Hence $\;\;P_{U}(v)\;=\;\displaystyle\sum_{i=1}^{i=m}\;\langle v,e_{i} \rangle e_{i}\;=0\;\;$ Hence $\;v \in N\big(P_{U}\big)\;.\;$

Conversely, $\;$for each $\;v \in N\big(P_{U}\big)\;$ we have$\;\;P_{U}(v)\;=\;0\;\;$and hence $\;\displaystyle\sum_{i=1}^{i=m}\;\langle v,e_{i} \rangle e_{i}\;=0\;$ which implies $\;\langle v,e_{i} \rangle =0\;\;\forall\;i=1,2,..m\;,\;$by linear independence of $\;E\;.\;$ Hence $\;v \in U^{\perp}\;.\;$

Thus $\; N\big(P_{U}\big)\;=\; U^{\perp}\;.\;$

(b). Note that by orthonormality of $\;E\;$ we get $\;\langle e_{i},e_{j} \rangle \;=\;\delta_{ij}\;=\;\begin{cases}1\;\;\;\text{if}\;\;\;i=j\\0\;\;\;\text{if}\;\;\;i\ne j \end{cases}\;$

For each $\;v \in V\;\;$we have $\;\;P^{2}_{U}(v)\;=\;P_{U}\;\left(\displaystyle\sum_{i=1}^{i=m}\;\langle v,e_{i} \rangle e_{i}\right)\;=\;\displaystyle\sum_{i=1}^{i=m}\;\langle v,e_{i} \rangle \; P_{U} \big(e_{i}\big)\;$

$\;\;=\;\;\displaystyle \sum_{i=1}^{i=m}\;\langle v,e_{i} \rangle \; \;\displaystyle\sum_{j=1}^{j=m}\;\langle e_{i},e_{j} \rangle e_{j}\;=\;\;\displaystyle \sum_{i=1}^{i=m}\;\langle v,e_{i} \rangle \;\displaystyle\sum_{j=1}^{j=m}\;\;\delta_{ij} e_{j}\; \;=\;\displaystyle\sum_{i=1}^{i=m}\;\langle v,e_{i} \rangle e_{i}\;=\;P_{U}\big(v\big)\;.\;$

Hence $\;\;P^{2}_{U}\;\equiv \;P_{U}\;$

Answer 2

I cannot make sense of most of your arguments:

• You write "$u\in P_uv$; this makes no sense as $P_Uv$ is a vector.

• You say "$P_Uv$ is linearly independent"; linear independence is a property of sets of vectors. You could say that a single vector is linear independent, but that's exactly zero new information about the vector: every nonzero vector is linearly independent.

• Your notation with inner products has a mix of numbers and vectors that again makes no sense. For example, what would "$\langle 1,e_1\rangle$" be?

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Let $v\in U^\perp$. Then $\langle v,e_j\rangle=0$ for all $j$ (since $e_j\in U$), and so $PUv=0e_1+\cdots0e_m=0$. So $U^\perp\subset \text{null}\,P_U$. Conversely, if $P_Uv=0$, then $$\langle v,e_1\rangle\,e_1+\cdots+\langle v,e_m\rangle\,e_m=0.$$ As $e_1,\ldots,e_m$ are linearly independent, we get that $\langle v,e_1\rangle=\cdots=\langle v,e_m\rangle=0$. Then $v\in U^\perp$ and so $\text{null}\,P_U\subset U^\perp$.

For the second part, since $P_U$ is linear $$ P_U(P_Uv)P_U(\langle v,e_1\rangle\,e_1+\cdots+\langle v,e_m\rangle\,e_m) =\langle v,e_1\rangle\,P_Ue_1+\cdots+\langle v,e_m\rangle\,P_Ue_m =\langle v,e_1\rangle\,e_1+\cdots+\langle v,e_m\rangle\,e_m=PUv. $$ As this works for all $v$, we get $P_U^2=P_U$.