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Known: b⃗is projected onto the vector a⃗to form a projection p⃗.p⃗=xa⃗,where x is a expansion coefficient and x is a scalar. e⃗=b⃗−p⃗,where e is a error.证明:1.将向量投影到向量a⃗∵a⃗⊥e⃗∴a⃗T(b⃗−p⃗)=0⇒a⃗T(b⃗−xa⃗)=0⇒a⃗Tb⃗−xa⃗Ta⃗=0⃗∴xa⃗Ta⃗=a⃗Tb⃗∴x=a⃗Tb⃗a⃗Ta⃗(x is a scalar and a⃗Ta⃗is a number.)∵p=xa⃗=a⃗x∴p=a⃗x=a⃗a⃗Tb⃗a⃗Ta⃗⇒p=a⃗a⃗Ta⃗Ta⃗⏟projectionb⃗⇒projectionp=p⃗×b⃗证明：2.推广:将向量投影到平面A⃗∵p⃗=x1^a1+x2^a2=[a1a2][x1x2]=A⃗X^;b⃗−A⃗X^⊥planeA;∴[a1Ta2T](b−A⃗x^)=0⃗⇒A⃗T(b⃗−A⃗X^)=0⃗⇒A⃗TA⃗X^=A⃗Tb⃗(e⃗=b⃗−A⃗X^,e⃗位于A转置的零空间)∵e⊥C(A)(Column space of A)∴如果A为一维向量且定义为a,则ATA为一个数。∴A⃗TA⃗X^=A⃗Tb⃗⇒X^=(A⃗TA⃗)−1A⃗Tb⃗⇒P⃗=A⃗X^=A⃗(A⃗TA⃗)−1AT⏟Projectionmatrixb\begin{aligned} \text{Known: } \ \ &\vec{b} \text{ is projected onto the vector } \vec{a} \text{ to form a projection } \vec{p}. \\ & \vec{p}=x \vec{a},\text{where x is a expansion coefficient and x is a scalar. } \\ & \vec{e}=\vec{b}-\vec{p},\text{where e is a error.} \\ &\text{证明:1.将向量投影到向量} \vec{a} \\ \because \ \ & \vec{a} \perp \vec{e} \\ \therefore \ \ & \vec{a}^T(\vec{b}-\vec{p})=0 \Rightarrow \vec{a}^T(\vec{b}-x\vec{a})=0 \Rightarrow \vec{a}^T\vec{b}-x\vec{a}^T\vec{a}=\vec{0} \\ \therefore \ \ & x\vec{a}^T\vec{a}=\vec{a}^T\vec{b} \\ \therefore \ \ & x=\frac{\vec{a}^T\vec{b}}{\vec{a}^T\vec{a}}(\text{x is a scalar and} \ \ \vec{a}^T\vec{a}\text{ is a number.}) \\ \because \ \ & p=x\vec{a}=\vec{a}x \\ \therefore \ \ & p=\vec{a}x= \vec{a}\frac{\vec{a}^T\vec{b}}{\vec{a}^T\vec{a}} \Rightarrow p=\underbrace{\boxed{\frac{\vec{a}\vec{a}^T}{\vec{a}^T\vec{a}}}}_{projection}\vec{b} \Rightarrow projection \ \ p = \vec{p} \times \vec{b} \\ &\text{证明：2.推广:将向量投影到平面} \vec{A} \\ \because \ \ & \vec{p}=\hat{x_1}a_1+\hat{x_2}a_2 = \left[\begin{array} {c} a_1 \ \ a_2 \end{array}\right] \left[\begin{array} {c} x_1 \\ x_2 \end{array}\right ]={\vec{A}}\hat{X};\\ \ \ & \boxed{ \vec{b}-{\vec{A}}\hat{X} } \perp \boxed{plane \ \ A} ; \\ \therefore \ \ & \left[\begin{array} {c} a_1^T \\ a_2^T \end{array}\right](b-\vec{A}\hat{x})=\vec{0} \\ \ \ & \Rightarrow \vec{A}^T(\vec{b}-\vec{A}\hat{X})=\vec{0}\Rightarrow \vec{A}^T\vec{A}\hat{X}=\vec{A}^T\vec{b}(\vec{e}=\vec{b}-\vec{A}\hat{X},\vec{e}位于A转置的零空间)\\ \because \ \ & e \perp C(A) (\text{Column space of A}) \\ \therefore \ \ & \text{如果A为一维向量且定义为a,则} A^TA \text{为一个数。} \\ \therefore \ \ & \vec{A}^T\vec{A}\hat{X}=\vec{A}^T\vec{b} \Rightarrow \hat{X}=(\vec{A}^T\vec{A})^{-1}\vec{A}^T\vec{b} \Rightarrow \vec{P}=\vec{A}\hat{X}=\underbrace{\boxed{\vec{A}(\vec{A}^T\vec{A})^{-1}A^T}}_{Projection \ \ matrix}b \end{aligned} Known: ∵ ∴ ∴ ∴ ∵ ∴ ∵ ∴ ∵ ∴ ∴ b is projected onto the vector a to form a projection p.p=xa,where x is a expansion coefficient and x is a scalar. e=b−p,where e is a error.证明:1.将向量投影到向量aa⊥eaT(b−p)=0⇒aT(b−xa)=0⇒aTb−xaTa=0xaTa=aTbx=aTaaTb(x is a scalar and aTa is a number.)p=xa=axp=ax=aaTaaTb⇒p=projectionaTaaaTb⇒projection p=p×b证明：2.推广:将向量投影到平面Ap=x1^a1+x2^a2=[a1 a2][x1x2]=AX^;b−AX^⊥plane A;[a1Ta2T](b−Ax^)=0⇒AT(b−AX^)=0⇒ATAX^=ATb(e=b−AX^,e位于A转置的零空间)e⊥C(A)(Column space of A)如果A为一维向量且定义为a,则ATA为一个数。ATAX^=ATb⇒X^=(ATA)−1ATb⇒P=AX^=Projection matrixA(ATA)−1ATb
key 1 : A一定是方阵，所以不存在A−1A^{-1}A−1。
key 2 : 投影矩阵是对称矩阵。⇒\Rightarrow⇒ PT=P,P2=PP^T=P,P^2=PPT=P,P2=P