Question

Let \(\mathbf{x}=(1,-2,1,6)\) in \(\mathbb{R}^{4},\) and let \(U=\operatorname{span}\\{(2,1,3,-4),(1,2,0,1)\\}\) a. Compute \(\operatorname{proj}_{U} \mathbf{x}\). b. Show that \\{(1,0,2,-3),(4,7,1,2)\\} is another orthogonal basis of \(U\). c. Use the basis in part (b) to compute proj \(_{U} \mathbf{x}\).

Short answer

The projections are \([-1/10, 7/15, -7/10, -19/10]\) and \([-297/140, 21/70, -207/140, 336/70]\).

Step-by-step solution

Define the Projection Formula

The projection of a vector \( \mathbf{x} \) onto a subspace \( U \) spanned by vectors \( \{ \mathbf{u}_1, \mathbf{u}_2 \} \) is given by \[ \operatorname{proj}_{U} \mathbf{x} = \frac{\langle \mathbf{x}, \mathbf{u}_1 \rangle}{\langle \mathbf{u}_1, \mathbf{u}_1 \rangle} \mathbf{u}_1 + \frac{\langle \mathbf{x}, \mathbf{u}_2 \rangle}{\langle \mathbf{u}_2, \mathbf{u}_2 \rangle} \mathbf{u}_2. \]

Calculate Dot Products for Projection

Compute the necessary dot products:- \( \langle \mathbf{x}, (2,1,3,-4) \rangle = 1 \cdot 2 + (-2) \cdot 1 + 1 \cdot 3 + 6 \cdot (-4) = 2 - 2 + 3 - 24 = -21. \)- \( \langle (2,1,3,-4), (2,1,3,-4) \rangle = 2^2 + 1^2 + 3^2 + (-4)^2 = 30. \)- \( \langle \mathbf{x}, (1,2,0,1) \rangle = 1 \cdot 1 + (-2) \cdot 2 + 1 \cdot 0 + 6 \cdot 1 = 1 - 4 + 0 + 6 = 3. \)- \( \langle (1,2,0,1), (1,2,0,1) \rangle = 1^2 + 2^2 + 0^2 + 1^2 = 6. \)

Compute Projection Using Original Basis

Use the results from Step 2 to find the projection:\[\operatorname{proj}_{U} \mathbf{x} = \frac{-21}{30} \begin{pmatrix} 2 \ 1 \ 3 \ -4 \end{pmatrix} + \frac{3}{6} \begin{pmatrix} 1 \ 2 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} -\frac{21}{15} \ -\frac{21}{30} \ -\frac{63}{30} \ \frac{84}{30} \end{pmatrix} + \begin{pmatrix} \frac{1}{2} \ 1 \ 0 \ \frac{1}{2} \end{pmatrix} = \begin{pmatrix} -\frac{1}{10} \ \frac{7}{15} \ -\frac{21}{30} \ \frac{-57}{30} \end{pmatrix}\]

Verify Orthogonal Basis for Part (b)

Check the dot product of new basis vectors \( (1,0,2,-3) \) and \( (4,7,1,2) \):- \( \langle (1,0,2,-3), (4,7,1,2) \rangle = 1 \cdot 4 + 0 \cdot 7 + 2 \cdot 1 + (-3) \cdot 2 = 0. \)This confirms they are orthogonal.

Use New Orthogonal Basis for Projection

For orthogonal basis vectors, compute the projection:- \( \langle \mathbf{x}, (1,0,2,-3) \rangle = 1 \cdot 1 + 0 \cdot (-2) + 2 \cdot 1 + (-3) \cdot 6 = 1 + 2 - 18 = -15. \)- \( \langle (1,0,2,-3), (1,0,2,-3) \rangle = 1^2 + 0^2 + 2^2 + (-3)^2 = 14. \)- \( \langle \mathbf{x}, (4,7,1,2) \rangle = 1 \cdot 4 + (-2) \cdot 7 + 1 \cdot 1 + 6 \cdot 2 = 4 - 14 + 1 + 12 = 3. \)- \( \langle (4,7,1,2), (4,7,1,2) \rangle = 4^2 + 7^2 + 1^2 + 2^2 = 70. \)Now, the projection:\[\operatorname{proj}_{U} \mathbf{x} = \frac{-15}{14} \begin{pmatrix} 1 \ 0 \ 2 \ -3 \end{pmatrix} + \frac{3}{70} \begin{pmatrix} 4 \ 7 \ 1 \ 2 \end{pmatrix}\]

Compute Resulting Projection

Perform the vector calculations:\[\operatorname{proj}_{U} \mathbf{x} = \begin{pmatrix} -\frac{15}{14} \ 0 \ -\frac{30}{14} \ \frac{45}{14} \end{pmatrix} + \begin{pmatrix} \frac{12}{70} \ \frac{21}{70} \ \frac{3}{70} \ \frac{6}{70} \end{pmatrix} = \begin{pmatrix} -\frac{297}{140} \ \frac{21}{70} \ -\frac{207}{140} \ \frac{336}{70} \end{pmatrix}\]

Key concepts

Linear Algebra

Linear algebra is an essential branch of mathematics that deals with vectors, vector spaces (also known as linear spaces), linear transformations, and systems of linear equations. It focuses on understanding and manipulating linear systems using concepts like vectors and matrices.
Linear algebra is the foundation for many applications in science and engineering, such as computer graphics, data analysis, and machine learning. Some core elements of linear algebra include:

• Vectors: These are objects that have both a direction and a magnitude. They can be added together and scaled by numbers (scalars).
• Vector Spaces: A collection of vectors that can be scaled and added together in a way that stays within the space. They have defined operations like addition and scalar multiplication.
• Linear Transformations: Functions that map vectors to vectors while preserving vector addition and scalar multiplication.
• Matrices: Rectangular arrays of numbers that can represent linear transformations or systems of linear equations.
• Systems of Linear Equations: Equations where each term is a constant or the product of a constant and a single variable.

Understanding these basic components is vital when dealing with problems involving vector projections, like the one in the exercise you are solving.

Orthogonal Basis

An orthogonal basis is a set of vectors in a vector space that are mutually orthogonal, meaning each pair of distinct vectors in the set is perpendicular. In simpler terms, the dot product of any two different vectors in this set is zero.
There are several advantages of having an orthogonal basis:

• Simplification: Calculations become easier with an orthogonal basis due to the reduction in cross-term computation.
• Projection: Computing the projection of a vector onto a subspace with an orthogonal basis is straightforward and requires less computation.

In the given exercise, proving \(\{(1,0,2,-3),(4,7,1,2)\}\) as an orthogonal basis of \U\ shows that these vectors are perpendicular to each other, which vastly simplifies calculations like vector projections.
An orthogonal basis also implies that each vector contributes uniquely to the span of the space, without any overlap in direction. This is an efficient way to represent vector spaces, especially in higher dimensions.

Dot Product

The dot product is a fundamental operation in linear algebra. It is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number, which can have important geometric interpretations.
The dot product of two vectors \( \mathbf{a} = (a_1, a_2, ..., a_n) \) and \( \mathbf{b} = (b_1, b_2, ..., b_n) \) is calculated as:\[ \langle \mathbf{a}, \mathbf{b} \rangle = a_1b_1 + a_2b_2 + ... + a_nb_n \]
The result of a dot product gives us:

• Angle Information: The dot product is zero when vectors are perpendicular, indicating an angle of 90 degrees between them.
• Length Relation: It is scalar, encapsulating the idea of projection of one vector onto another.

In the exercise provided, calculating dot products is vital for determining projections of vectors. The dot product serves as a tool for verifying orthogonality, as shown by confirming the new basis vectors are indeed orthogonal. When two vectors have a dot product of zero, it means they are orthogonal, which is essential for creating an orthogonal basis and simplifying projection calculations.