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## 3.V. Projection

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**3.V. Projection**3.V.1. Orthogonal Projection Into a Line 3.V.2. Gram-Schmidt Orthogonalization 3.V.3. Projection Into a Subspace 3.V.1. & 2. deal only with inner product spaces. 3.V.3 is applicable to any direct sum space.**3.V.1. Orthogonal Projection Into a Line**Definition 1.1: Orthogonal Projection The orthogonal projectionofvinto the line spanned by a nonzerosis the vector. Example 1.3: Orthogonal projection of the vector ( 2 3 )T into the line y = 2x.**Example 1.4: Orthogonal projection of a general vector in**R3 into the y-axis Example 1.5: Project = Discard orthogonal components A railroad car left on an east-west track without its brake is pushed by a wind blowing toward the northeast at fifteen miles per hour; what speed will the car reach?**Example 1.6: Nearest Point**A submarine is tracking a ship moving along the line y = 3x + 2. Torpedo range is one-half mile. Can the sub stay where it is, at the origin on the chart, or must it move to reach a place where the ship will pass within range? Ship’s path is parallel to vector Point p of closest approach is (Out of range)**Exercises 3.V.1.**1. Consider the function mapping a plane to itself that takes a vector to its projection into the line y = x. (a) Produce a matrix that describes the function’s action. (b) Show also that this map can be obtained by first rotating everything in the plane π/4 radians clockwise, then projecting into the x-axis, and then rotating π/4 radians counterclockwise.**3.V.2. Gram-Schmidt Orthogonalization**Given a vector s, any vector v in an inner product space can be decomposed as where Definition 2.1: Mutually Orthogonal Vectors Vectors v1, …, vk Rnare mutually orthogonalif vi· vj = 0 i j Theorem 2.2: A set of mutually orthogonal non-zero vectors is linearly independent. Proof: → → cj = 0 j Corollary 2.3: A set of k mutually orthogonal nonzero vectors in V kis a basis for the space. Definition 2.5: Orthogonal Basis An orthogonal basisfor a vector space is a basis of mutually orthogonal vectors.**Example 2.6: Turn**into an orthogonal basis for R3.**Theorem 2.7: Gram-Schmidt Orthogonalization**If β1 , …, βk is a basis for a subspace of Rnthen the κjs calculated from the following scheme is an orthogonal basis for the same subspace. Proof: For m 2 , Let β1 , …, βm be mutually orthogonal, and Then QED**If each κj is furthermore normalized, the basis is called**orthonormal. The Gram-Schmidt scheme simplifies to:**Exercises 3.V.2.**1. Perform the Gram-Schmidt process on this basis for R3, 2. Show that the columns of an nn matrix form an orthonormal set if and only if the inverse of the matrix is its transpose. Produce such a matrix.**3.V.3. Projection Into a Subspace**Definition 3.1: For any direct sum V = M N and any v V such that v= m + n with m M and n N The projection ofvinto M along Nis defined as projM, N(v) = m • Reminder: • M & N need not be orthogonal. • There need not even be an inner product defined.**Example 3.2: The space M22 of 22 matrices is the**direct sum of Task: Find projM , N (A), where Solution: Let the bases for M & N be → is a basis for M22 . ∴**Example 3.3: Both subscripts on projM , N (v) are**significant. Consider with basis & and subspaces It’s straightforward to verify Task: Find projM , N (v) and projM , L (v) where Solution: For →**For**→ Note: BML is orthogonal but BMN is not.**Definition 3.4: Orthogonal Complement**The orthogonal complement of a subspace M of Rnis M= { vRn | vis perpendicular to all vectors in M } ( read “M perp” ). The orthogonal projection projM(v) of a vector is its projection into M along M. Example 3.5: In R3, find the orthogonal complement of the plane Solution: Natural basis for P is ( parameter = z) →**Lemma 3.7:**Let M be a subspace of Rn. Then Mis also a subspace and Rn = M M . Hence, v Rn, v projM (v) is perpendicular to all vectors in M. Proof: Construct bases using G-S orthogonalization. Theorem 3.8: Let vbe a vector in Rnand let M be a subspace of Rn with basis β1 , …, βk . If Ais the matrix whose columns are the β’s then projM (v) = c1β1 + …+ ckβk where the coefficients ciare the entries of the vector (ATA) ATv. That is, projM (v) = A (ATA)1ATv. Proof: where c is a column vector → → → By lemma 3.7,**Interpretation of Theorem 3.8:**If B = β1 , …, βk is an orthonormal basis, then ATA = I. In which case, projM (v) = A (ATA)1ATv = A ATv. with In particular, if B = Ek , then A = AT = I. In case B is not orthonormal, the task is to find C s.t. B = AC and BTB = I. → → Hence**Example 3.9:**To orthogonally project into subspace From we get →**Exercises 3.V.3.**1. Project v into M along N, where 2. Find M for**3. Define a projection to be a linear transformation t :**V → V with the property that repeating the projection does nothing more than does the projection alone: ( t t )(v) = t (v) for all v V. (a) Show that for any such t there is a basis B= β1 , …, βn for V such that where r is the rank of t. (b) Conclude that every projection has a block partial-identity representation: