Math Problem Statement

What is the matrix for projW : R

3 → R

3

, where

W = Span

1

0

−1

 ,

1

1

1

Solution

To find the projection matrix for projW:R3R3\text{proj}_W: \mathbb{R}^3 \to \mathbb{R}^3 onto the subspace WW spanned by the vectors

v1=(101),v2=(111),\mathbf{v_1} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \quad \mathbf{v_2} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},

we first construct a matrix AA whose columns are these vectors:

A=(110111).A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ -1 & 1 \end{pmatrix}.

The projection matrix PP onto the column space of AA is given by:

P=A(ATA)1AT.P = A (A^T A)^{-1} A^T.

Now, we need to compute ATAA^T A:

AT=(101111),A^T = \begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \end{pmatrix},

ATA=(101111)(110111)=(2003).A^T A = \begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}.

Next, we find (ATA)1(A^T A)^{-1}:

(ATA)1=(120013).(A^T A)^{-1} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{pmatrix}.

Now we can calculate PP:

P=A(ATA)1AT=A(120013)AT.P = A (A^T A)^{-1} A^T = A \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{pmatrix} A^T.

Calculating A(ATA)1A (A^T A)^{-1}:

A(ATA)1=(110111)(120013)=(12130131213).A (A^T A)^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & \frac{1}{3} \\ 0 & \frac{1}{3} \\ -\frac{1}{2} & \frac{1}{3} \end{pmatrix}.

Now, we can compute the full projection matrix PP:

P=(12130131213)(101111)=(12+131312+13013012+131312+13).P = \begin{pmatrix} \frac{1}{2} & \frac{1}{3} \\ 0 & \frac{1}{3} \\ -\frac{1}{2} & \frac{1}{3} \end{pmatrix} \begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} + \frac{1}{3} & \frac{1}{3} & -\frac{1}{2} + \frac{1}{3} \\ 0 & \frac{1}{3} & 0 \\ -\frac{1}{2} + \frac{1}{3} & \frac{1}{3} & -\frac{1}{2} + \frac{1}{3} \end{pmatrix}.

After some calculations, the final projection matrix PP simplifies to:

P=(5613160130561316).P = \begin{pmatrix} \frac{5}{6} & \frac{1}{3} & -\frac{1}{6} \\ 0 & \frac{1}{3} & 0 \\ -\frac{5}{6} & \frac{1}{3} & \frac{1}{6} \end{pmatrix}.

Summary

The projection matrix for projW\text{proj}_W is:

P=(5613160130561316).P = \begin{pmatrix} \frac{5}{6} & \frac{1}{3} & -\frac{1}{6} \\ 0 & \frac{1}{3} & 0 \\ -\frac{5}{6} & \frac{1}{3} & \frac{1}{6} \end{pmatrix}.

Would you like further details or have any questions about this process? Here are some related questions you might consider:

  1. How do you derive the orthogonal projection formula in general?
  2. What are the geometric interpretations of projection matrices?
  3. How would you compute the projection of a specific vector onto the subspace WW?
  4. Can you explain the significance of the rank of the matrix AA in this context?
  5. What would happen if we added a third vector to the spanning set of WW?

Tip: Always check that the projection matrix PP is idempotent (i.e., P2=PP^2 = P).

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Projection Matrices
Vector Spaces

Formulas

P = A(A^TA)^{-1}A^T

Theorems

Properties of Projection Matrices

Suitable Grade Level

Grades 11-12