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3.17 Projections

Theorem

Let \(w_{1},...,w_{r}\) be subspaces of \(V\), then \(V=W_{1}\oplus...\oplus W_{r}\;\char"27FA \;\forall v\in V,\exists!v_{i}\in W_{i} ,\forall i,1\leq i\leq r\) such that \(v=v_{1}+...+v_{r}\\\iff \;1)V=W_{1}+W_{2}+...+W_{r}\quad2)\text{ not enough }\forall i,j,1\leq i,j\leq r,W_i\cap W_j=\{0_V\}\)

In order to have uniqueness of the decomposition \(v=v_1+...+v_r\).

Suppose that we also have \(v=v_{1}'+...+v_{r}',v_i'\in W_i,\forall i,1\leq i\leq r\)

Then \(v_1+...+v_r=v_1'+...+v_r'\), then \(v_{1}-v_{1}'=v_{2}'-v_{2}+...+v_{r-1}'-v_{r-1}\)

Thus \(2)\) is not enough, we need to ensure the common part of \(v_1-v_1'\) and \(v_{2}^{\prime}-v_{2}+...+v_{r-1}^{\prime}-v_{r-1}\) should only be \(0\)

That is \(2)\forall i,1\leq i\leq r,W_{i}\cap (W_{1}+...+W_{i-1}+W_{i+1}+...+W_{r})=\{0_{v}\}\)

Example

\(V=\R^2\)

\(W_1=\lang (1,0)\rang=\lang e_1\rang\), \(W_{2}=\langle(0,1)\rangle=\langle e_{2}\rangle\), \(W_{3}=\langle(1,1)\rangle=\langle e_{1}+e_{2}\rangle\)

image​

Clearly, \(\R^2=W_1+W_2+W_3\) and \(W_1\cap W_2=W_1\cap W_3=W_2\cap W_3=\{0\}\)

But \(\R^2\neq W_1\oplus W_2\oplus W_3\) since checking dimension

Here \(W_1\cap(W_2+W_3)=\lang e_1\rang \neq \{0\}\)

Example

\(V = \mathbb{R}^2\), \(S = \langle e_1 + e_2 \rangle\), \(T = \langle e_2 \rangle\) Verify: \(\mathbb{R}^2 = S + T\) and \(S \cap T = \{ 0_{\mathbb{R}^2} \}\), so \(\mathbb{R}^2 = S \oplus T\).

Given \(v \in \mathbb{R}^2\), \(v = (x_1, x_2) = x_1 e_1 + x_2 e_2\) \(= x_1 e_1 + x_2 e_2 - x_1 e_2 + x_1 e_2 = x_1 (e_1 + e_2) + (x_2 - x_1)e_2\) where \(x_1 (e_1 + e_2) \in S\) and \((x_2 - x_1)e_2 \in T\).

We define projection \(P_S : \mathbb{R}^2 \to \mathbb{R}^2\) as \(P_S(v) = P_S(x_1, x_2) = x_1 \cdot (e_1 + e_2)\), then \(\operatorname{\mathrm{Im}}P_{S}=S,~\ker P_{S}=\{(x_{1},x_{2})\in\mathbb{R}^{2}\mid x_{1}=0\}=\langle e_{2}\rangle=T\)

We also have: \(P_T : \mathbb{R}^2 \to \mathbb{R}^2\) defined by \(P_T(v) = P_T(x_1, x_2) = (x_2 - x_1)e_2\), then \(\operatorname{Im} P_T = \langle e_2 \rangle = T\) and \(\ker P_T = \{ (x_1, x_2) \in \mathbb{R}^2 \mid x_2 - x_1 = 0 \} = \langle e_1 + e_2 \rangle = S\)

Lemma

Let \(V\) be a \(k\)-v.s and \(W_1, \dots, W_r\) subspaces of \(V\) such that \(V = W_{1} \oplus W_{2} \oplus \dots \oplus W_{r}\)

Then \(\text{Id}_{V} = P_{W_1}+ P_{W_2}+ \dots + P_{W_r}\)

This lemma is about if we can decompose vector space, then we can decompose Identity map by projector

Example

let us see that \(\text{Id}_{\mathbb{R}^2} = P_S + P_T\)

Given \(v = (x_1, x_2) \in \mathbb{R}^2,\) \(\text{Id}_v(v) = v = (x_1, x_2) = x_1 e_1 + x_2 e_2\)

\(P_S(v) = x_1(e_1 + e_2)\) and \(P_T(v) = (x_2 - x_1)e_2\)
\((P_{S}+P_{T})(v)=\) \(P_S(v) + P_T(v) = x_1(e_1 + e_2) + (x_2 - x_1)e_2\) \(= x_1 e_1 + x_2 e_2\) \(= v\)

Proof:

Given \(v \in V\), unique \(v_{i} \in W_{i} \ \forall i, \ 1 \leq i \leq r,v = v_{1} + v_{2} + \dots + v_{r}\)

We know \(\text{Id}_{V}(v)=v,\text{ given }i,\ 1\leq i\leq r,\) \(P_{W_i}(v) = v_{i}\), then \(P_{W_1}(v) + P_{W_2}(v) + \dots + P_{W_r}(v) = v_{1} + v_{2} + \dots + v_{r} = v\)

Thus \(\text{Id}_{V}(v) = \left( P_{W_1}+ \dots + P_{W_r}\right)(v)\)​

Remark

Given \(i \neq j\), \(1 \leq i,j \leq r\), \(P_{W_j} \circ P_{W_i} = 0 \text{ because } \operatorname{Im} P_{W_i} = W_i \subseteq \ker P_{W_j}\)​\(=W_{1} + W_{2} + \dots +\overset{\wedge}{W_{j}} + \dots + W_{r} \quad \text{(NOTATION)}\)

Theorem

If \(V = W_1 \oplus \dots \oplus W_r\), then there exist \(P_1, \dots, P_r: V \to V\) \(k\)-linear maps such that:

  1. \(\forall i, 1 \leq i \leq r, \quad P_i^2 = P_i\).

  2. \(\forall i,j, 1 \leq i,j \leq r, i \neq j, \quad P_i \circ P_j = 0\).

  3. \(\operatorname{Id}_V = P_1 + \dots + P_r\).

  4. \(\operatorname{\mathrm{Im}}P_{i}=W_{i},\forall i,1\leq i\leq r\).

Conversely, if \(P_1, \dots, P_r: V \to V\) are \(k\)-linear maps such that (1), (2), (3), and (4) hold, then:

\(V = \operatorname{Im} P_1 \oplus \operatorname{Im} P_2 \oplus \dots \oplus \operatorname{Im} P_r\)

PROOF:

We just need to prove the converse part.

Suppose we are given linear maps \(P_1, \dots, P_r: V \to V\) satisfying (1), (2), (3), and (4).
Let \(W_i = \operatorname{Im} P_i\), \(\forall i, 1 \leq i \leq r\).

Given \(v \in V\), \(v = \operatorname{Id}_V (v) = (P_1 + \dots + P_r)(v) = P_1(v) + P_2(v) + \dots + P_r(v)\) where \(P_i(v) \in W_i\). by (3) and linear map

So \(V = W_1 + W_2 + \dots + W_r\).

Moreover, the expression of \(v\) as a sum of vectors in \(W_1, W_2, \dots, W_r\) is unique.

Suppose \(0=w_{1}+w_{2}+\dots+w_{r}\quad\text{with }w_{i}\in W_{i},\forall i.\)

Given \(j, 1 \leq j \leq r\), we have \(0=P_{j}(0)=P_{j}(w_{1})+P_{j}(w_{2})+\dots+P_{j}(w_{r})\)

\(\overset{(1)}{=}P_{j}\left(P_{1}\left(w_{1}\right)\right)+\cdots+P_{j}\left(P_{r}\left(w_{r}\right )\right)\overset{(2)}{=}P_{j}\left(P_{j}\left(w_{j}\right)\right)\overset{(1)}{=}P_{j}\left(w_{j}\right)\overset{(1)}{=}w_{j}\)​

So the sum is direct.

Proof of \(P_j(w_j)=w_j\) using (1)

For all \(i\), \(P_i^2 = P_i\). Given that: \(V = W_1 \oplus \dots \oplus W_r\)

Suppose \(v = v_1 + \dots + v_r, \quad \text{with } v_i \in \operatorname{Im} P_i, \forall i.\)

Then, by definition \(P_i(v) = v_i \in \operatorname{Im} P_i.\)

Applying \(P_i\) again: \(P_i(v) = P_i^2(v) = P_i(v_i).\) Since \(v_i \in \operatorname{Im} P_i\), we get \(P_i(v_i) = v_i.\)

Thus, in other words, if \(P_i^2 = P_i\), then \(P_{i} \bigg|_{\operatorname{Im} P_i}= \operatorname{Id}.\)

Example

  1. Prove if \(p:V\to V\) is a projector, then \(Id_{V}-p:V\to V\) is a projector

    Use definition: \(\left(Id_{V}-p\right)^{2}=\left(Id_{V}-p)\circ\left(Id_{V}-p\right)=Id_{V}^{2}+p ^{2}-2pId_{V}\right.=Id_{V}+p-2p=Id_{V}-p\)​

  2. Find a projector \(p: \mathbb{R}^3 \to \mathbb{R}^2\) such that \(\operatorname{\mathrm{Im}}p=\langle(1,2,3)=v_{1}\rangle=S.\)

    We want \(\mathbb{R}^3 = S \oplus T\)

    To construct the projector, we extend the basis \(\{v_1\}\) of \(S\) to a basis of \(\mathbb{R}^3\).

    Let \(v_{2}=(0,1,0),v_{3}=(0,0,1).\)Then, \(\{v_1, v_2, v_3\}\) forms a basis for \(\mathbb{R}^3\)

    Let \(T=\lang v_2,v_3\rang\) and assume \((x_1, x_2, x_3) = \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3.\)

    Then \(p(x_{1}, x_{2}, x_{3}) = \alpha_{1} v_{1}\)

Let \(P:V\to V\) be a projector and \(\ker P\) and \(ImP\) be subspaces of \(V\)

Prove: \(\ker P\cap ImP=\{0_{V}\}\)

Suppose \(v\in\ker P\cap ImP\)

Since \(v\in ImP\), then \(\exists w\in V:v=p(w)\), then \(P(v)=P^2(w)=P(w)=v\)

And \(v\in\ker P\), then \(P(v)=v=0\)

Another way:

Given \(v\in V\), then \(v=\underbrace{v-P(v)}_{\in \ker P}+\underbrace{P(v)}_{\in ImP}\), \(v-P(v)\in \ker P\) is because \(P(v-P(v))=P(v)-PP(v)=0\)

Thus \(V=\ker p\oplus Im p\)

Suppose \(\dim_k V < \infty\). Let \(B_1 = \{ v_1, \dots, v_s \}\) be a basis of \(\ker p\). Let \(B_2 = \{ v_{s+1}, \dots, v_n \}\) be a basis of \(\operatorname{Im} p\).

Then \(B = B_1 \cup B_2\) forms a basis for \(V\) and \(p(v_i) = 0, \quad \forall 1 \leq i \leq s\), \(p(v_i) = v_i, \quad \forall s+1 \leq i \leq n\)

image​

Remark

\(p: V \to V\) projector is diagonalizable with eigenvalues \(0\) and \(1\).

\(V_0 = \ker p\), \(V_1 = \text{Im } p\), \(V = \ker p \oplus \text{Im } p\)

\(\ker p\) is \(p\)-invariant (\(p\) is linear), \(\text{Im } p\) is \(p\)-invariant

Proposition

Let \(V = W_1 \oplus \dots \oplus W_r\), and let \(p_i: V \to V\) (\(1 \leq i \leq r\)) be the projections to \(W_1, \dots, W_r\) respectively (as in the previous theorem).

Let \(f: V \to V\) be a \(k\)-linear map. Then, for all \(i\), \(1 \leq i \leq r\):

\(W_{i} \text{ is }f\text{-invariant}\iff f \circ p_{i} = p_{i} \circ f\)

Proof:

\(\Leftarrow\)) Fix \(i\), \(1 \leq i \leq r\). Suppose \(f \circ p_i = p_i \circ f\).

Let \(v \in W_i\), we want to prove that \(f(v) \in W_i\).

\(v=p_{i}(v)\implies f(v)=f(p_{i}(v))=f\circ p_{i}(v)=p_{i}(f(v))\in\operatorname{\mathrm{Im}} p_{i}=W_{i}\) (by hypothesis)

\(\downarrow\)

\(\operatorname{Im} p_i = W_i\)

So \(W_i\) is \(f\)-invariant.

\(\Rightarrow\)) Since \(V = W_1 \oplus \dots \oplus W_r\), then \(v=w_1+...+w_r\) where \(w_i\in W_i\) and this is unique

Since \(W_i\) is \(f\)-invariant, then \(f\circ p_{i}(v)=f\circ p_{i}(w_{1}+\dots+w_{r})=f(w_{i})=w_{i}^{\prime}\in W_i\)

\(p_{i}\circ f(v)=p_{i}\circ f(w_{1}+\dots+w_{r})=p_{i}(w_{1}^{\prime}+\dots+w_{r} ^{\prime})=w_{i}^{\prime}\in W_{i}\)

Thus \(f\circ p_{i}=p_{i}\circ f\)

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