/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Consider the vector space \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the vector space \(\mathbb{R}^{n}\) with inner product \(\langle\mathbf{x}, \mathbf{y}\rangle=\mathbf{x}^{T} \mathbf{y} .\) Show that, for any \(n \times n\) matrix \(A\) (a) \(\langle A \mathbf{x}, \mathbf{y}\rangle=\left\langle\mathbf{x}, A^{T} \mathbf{y}\right\rangle\) (b) \(\left\langle A^{T} A \mathbf{x}, \mathbf{x}\right\rangle=\|A \mathbf{x}\|^{2}\)

Short Answer

Expert verified
We proved that for any $n \times n$ matrix \(A\): (a) \(\langle A \mathbf{x}, \mathbf{y}\rangle=\left\langle\mathbf{x}, A^{T} \mathbf{y}\right\rangle\) using the definition of the inner product and the properties of transposed matrices. (b) \(\langle A^{T}A\mathbf{x}, \mathbf{x} \rangle = \|A\mathbf{x}\|^2\) using the definition of the inner product, the property of the norm, and the result from part (a).

Step by step solution

01

(a) Show that \(\langle A \mathbf{x}, \mathbf{y}\rangle=\left\langle\mathbf{x}, A^{T} \mathbf{y}\right\rangle\).#

We will use the definition of the inner product in this vector space to prove this property for any \(n \times n\) matrix A. Recall that \(\langle\mathbf{x}, \mathbf{y}\rangle=\mathbf{x}^T\mathbf{y}\). Let us compute the left side of the equation: \[\langle A \mathbf{x}, \mathbf{y}\rangle = (A \mathbf{x})^T\mathbf{y}\] Now, remember that \((AB)^T = B^TA^T\). Applying this property, we get: \[(A \mathbf{x})^T = \mathbf{x}^TA^T\] Substituting back into the expression, we have: \[(A \mathbf{x})^T\mathbf{y} = \mathbf{x}^TA^T\mathbf{y}\] Now, we compute the right side of the equation: \[\left\langle\mathbf{x}, A^{T} \mathbf{y}\right\rangle = \mathbf{x}^T(A^T\mathbf{y})\] Since the operation follows associative property, we can rewrite it as: \[\mathbf{x}^T(A^T\mathbf{y}) = \mathbf{x}^TA^T\mathbf{y}\] Comparing both sides, we conclude that: \[\langle A \mathbf{x}, \mathbf{y}\rangle=\left\langle\mathbf{x}, A^{T} \mathbf{y}\right\rangle\]
02

(b) Show that \(\langle A^{T}A\mathbf{x}, \mathbf{x} \rangle = \|A\mathbf{x}\|^2\).#

To prove this, we will again use the definition of the inner product in this vector space and the property of the norm \(\| \mathbf{x} \| = \sqrt{\langle \mathbf{x}, \mathbf{x} \rangle }\). Let's compute the left side of the equation: \[\langle A^{T}A\mathbf{x},\mathbf{x}\rangle = (A^{T}A\mathbf{x})^T\mathbf{x}\] Using the property \((AB)^T = B^TA^T\), we have: \[(A^{T}A\mathbf{x})^T = \mathbf{x}^T(A^TA)^T\] Which simplifies to: \[\mathbf{x}^T(AA^T)^T\mathbf{x} =\mathbf{x}^TA^{T}A\mathbf{x}\] Now let's compute the right side of the equation: Using the property of the norm, we have: \[\|A\mathbf{x}\|^2 = \langle A\mathbf{x}, A\mathbf{x} \rangle \] Then, using part (a) of the exercise, we know that: \[\langle A\mathbf{x}, A\mathbf{x} \rangle = \langle \mathbf{x}, A^TA\mathbf{x} \rangle \] Now, comparing both sides we conclude that: \[\langle A^{T}A\mathbf{x}, \mathbf{x} \rangle = \|A\mathbf{x}\|^2\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Space
A vector space is one of the fundamental concepts in linear algebra. At its core, it's a collection of vectors that can be scaled and added together to produce new vectors within the same space. Think of it as a playground where certain rules apply—specifically, eight axioms that must be satisfied involving addition and scalar multiplication. These rules maintain structure, such as the existence of a zero vector, an additive inverse, and the ability to distribute both vector addition and scalar multiplication.

In the context of our exercise, we're dealing with the vector space \( \mathbb{R}^{n} \) and an inner product defined as \(\langle \mathbf{x}, \mathbf{y}\rangle = \mathbf{x}^{T} \mathbf{y} \). An inner product is a way of multiplying vectors together to produce a scalar, and it helps in defining concepts like length and angle within the vector space. The inner product must be commutative, which is to say that flipping the order of the vectors doesn't change the result. This property was crucial in the provided exercise for showing how matrix transformations affect the vectors within our space.
Matrix Transpose
A matrix transpose is a crucial operation in linear algebra, which transforms a matrix by flipping it over its diagonal. To articulate it simply, consider writing the matrix on a sheet of paper and then flipping that paper along the main diagonal. This will switch rows with columns and vice versa. The notation for the transpose of matrix A is \( A^{T} \).

This operation plays a vital role in properties of the inner product in our original exercise by showcasing the equality \( \langle A \mathbf{x}, \mathbf{y}\rangle = \langle \mathbf{x}, A^{T} \mathbf{y}\rangle \). The transpose property shows up when we're manipulating the order of multiplication in matrices and vectors, and understanding this concept helps unravel how different transformations can be applied while maintaining the structural integrity of operations within a vector space.
Norm of a Vector
The norm of a vector is a measure of its length or magnitude. In more technical terms, it's a function that assigns a non-negative length to each vector in a vector space. The equation for the norm of a vector \( \mathbf{x} \) in a space with the inner product is \( \| \mathbf{x} \| = \sqrt{\langle \mathbf{x}, \mathbf{x} \rangle} \), which is derived from the Pythagorean theorem in Euclidean space.

Our exercise showcases the relationship between the norm of a matrix-transformed vector and the inner product with \( \langle A^{T}A\mathbf{x}, \mathbf{x} \rangle = \|A\mathbf{x}\|^{2} \). This relationship connects the geometric concept of vector length to algebraic manipulations. Understanding the norm helps students draw connections between abstract vector algebra and tangible geometric interpretations making it easier to comprehend changes in length as a result of various linear transformations.

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Most popular questions from this chapter

(a) Let \(S\) be the subspace of \(\mathbb{R}^{3}\) spanned by the vectors \(\mathbf{x}=\left(x_{1}, x_{2}, x_{3}\right)^{T}\) and \(\mathbf{y}=\) \(\left(y_{1}, y_{2}, y_{3}\right)^{T},\) Let $$A=\left[\begin{array}{lll} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \end{array}\right]$$ Show that \(S^{\perp}=N(A)\) (b) Find the orthogonal complement of the subspace of \(\mathbb{R}^{3}\) spanned by \((1,2,1)^{T}\) and \((1,-1,2)^{T}\)

If \(Q\) is an \(n \times n\) orthogonal matrix and \(\mathbf{x}\) and \(\mathbf{y}\) are nonzero vectors in \(\mathbb{R}^{n}\), then how does the angle between \(Q \mathbf{x}\) and \(Q \mathbf{y}\) compare with the angle between \(\mathbf{x}\) and \(\mathbf{y} ?\) Prove your answer.

Show that, in any vector space with a norm, $$\|-\mathbf{v}\|=\|\mathbf{v}\|$$

If \(V\) is an inner product space, show that $$\|\mathbf{v}\|=\sqrt{\langle\mathbf{v}, \mathbf{v}\rangle}$$ satisfies the first two conditions in the definition of a norm.

Let \(A\) be an \(m \times 3\) matrix. Let \(Q R\) be the \(Q R\) factorization obtained when the classical GramSchmidt process is applied to the column vectors of \(A,\) and let \(\tilde{Q} \tilde{R}\) be the factorization obtained when the modified Gram-Schmidt process is used. Show that if all computations were carried out using exact arithmetic, then we would have $$\tilde{Q}=Q \quad \text { and } \quad \tilde{R}=R$$ and show that, when the computations are done in finite-precision arithmetic, \(\tilde{r}_{23}\) will not necessarily be equal to \(r_{23}\) and consequently \(\tilde{r}_{33}\) and \(\tilde{\mathbf{q}}_{3}\) will not necessarily be the same as \(r_{33}\) and \(\mathbf{q}_{3}\)

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