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Suppose \(A=\) \(\left[\begin{array}{rrr}0 & -1 & 0 \\ 0 & 0 & -1 \\ 1 & 0 & 0\end{array}\right]\). Check that the vector \(\mathbf{y}=\left[\begin{array}{r}1 \\ -1 \\ 1\end{array}\right]\) satisfies \(A \mathbf{y}=\mathbf{y}\) and \(A^{\top} \mathbf{y}=\mathbf{y}\). Show that if \(\mathbf{x} \cdot \mathbf{y}=0\), then \(A \mathbf{x} \cdot \mathbf{y}=0\) as well. Interpret this result geometrically.

Short Answer

Expert verified
In this exercise, we verified that vector \(\mathbf{y}\) satisfies \(A\mathbf{y}=\mathbf{y}\) and \(A^{\top}\mathbf{y}=\mathbf{y}\) for the given matrix \(A\). We also proved that if the dot product \(\mathbf{x} \cdot \mathbf{y} = 0\), then \(A\mathbf{x} \cdot \mathbf{y} = 0\). Geometrically, this means that the transformation represented by the matrix \(A\) preserves the orthogonality between vectors \(\mathbf{x}\) and \(\mathbf{y}\).

Step by step solution

01

Multiply matrix A by vector y

We have A as: \[ A = \begin{bmatrix}0 & -1 & 0 \\0 & 0 & -1 \\1 & 0 & 0\end{bmatrix} \] and y as: \[ \mathbf{y} = \begin{bmatrix}1\\-1\\1\end{bmatrix} \] Let us now multiply A with y: \[ A\mathbf{y} = \begin{bmatrix}0 & -1 & 0 \\0 & 0 & -1 \\1 & 0 & 0\end{bmatrix} \begin{bmatrix}1\\-1\\1\end{bmatrix} = \begin{bmatrix}1\\-1\\1\end{bmatrix} \] Since the result of A multiplied by y is equal to y, we can conclude that Ay = y.
02

Multiply Aáµ€ by vector y

First, let's find the transpose of matrix A (Aáµ€): \[ A^{\top} = \begin{bmatrix}0 & 0 & 1 \\-1 & 0 & 0 \\0 & -1 & 0\end{bmatrix} \] Now multiply Aáµ€ with y: \[ A^{\top}\mathbf{y} = \begin{bmatrix}0 & 0 & 1 \\-1 & 0 & 0 \\0 & -1 & 0\end{bmatrix} \begin{bmatrix}1\\-1\\1\end{bmatrix} = \begin{bmatrix}1\\-1\\1\end{bmatrix} \] Since the result of Aáµ€ multiplied by y is equal to y, we can conclude that Aáµ€y = y.
03

Proving x · y = 0 implies Ax · y = 0

Let's first consider the assumption that x · y = 0. We want to show that this means Ax · y = 0. Start by calculating the dot product of Aᵀy and x (since we already know Aᵀy = y): \[ (A^{\top}\mathbf{y})\cdot\mathbf{x} = (A^{\top}\mathbf{y})^{\top}\mathbf{x} = \mathbf{y}^{\top} (A^{\top}\mathbf{x}) = \mathbf{y}^{\top} (A\mathbf{x})^{\top} \] Since \(\mathbf{x}\cdot\mathbf{y}=0\), we have \(\mathbf{y}^{\top}\mathbf{x} = 0\). Therefore, we can rewrite the equation as: \[ 0 = \mathbf{y}^{\top} (A\mathbf{x})^{\top} \] Let z = Ax: \[ 0 = \mathbf{y}^{\top} \mathbf{z}^{\top} \] As transposition doesn't affect the value of the dot product, this is equal to: \[ 0 = \mathbf{y} \cdot \mathbf{z} \] Now, since z = Ax, we finally have: \[ 0 = \mathbf{y} \cdot (A\mathbf{x}) \] We have proved that if x · y = 0, then Ax · y = 0.
04

Geometric interpretation

Geometrically, the dot product of two vectors being equal to 0 means that the two vectors are orthogonal (perpendicular) to each other. In this problem, we showed that if x and y are orthogonal, then Ax and y are also orthogonal. This means that the transformation represented by the matrix A preserves orthogonality, i.e., the orthogonal relationship between the vectors x and y remains after applying the transformation defined by A.

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Key Concepts

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

Orthogonal Vectors
Orthogonal vectors are vectors that meet at a right angle, forming a 90-degree angle between them. To say that two vectors are orthogonal in linear algebra is essentially the same as saying that their dot product is zero.
This concept holds a significant place as it can describe a situation where no projection occurs of one vector onto another. In simpler terms, think about it as a vector being completely independent from another in terms of direction.
  • For vectors \( \mathbf{x} \) and \( \mathbf{y} \), if \( \mathbf{x} \cdot \mathbf{y} = 0 \), they are orthogonal.
  • Orthogonality helps in many fields, such as simplifying computations in signal processing.
In our original exercise, we showed that if vector \( \mathbf{x} \) is orthogonal to \( \mathbf{y} \), then even after the transformation by matrix \( A \), the vector \( A\mathbf{x} \) remains orthogonal to \( \mathbf{y} \).
This is a crucial insight into how transformations, like rotating or reflecting vectors, can maintain orthogonal relationships, demonstrating the power of matrix transformations.
Matrix Transformation
Matrix transformation involves using a matrix to change a vector from one state or direction to another. Matrices can represent various transformations, like rotations, reflections, and scaling in space.
In this exercise, matrix \( A \) acts on the vector \( \mathbf{y} \), yet results in \( \mathbf{y} \) itself, which identifies \( \mathbf{y} \) as an eigenvector of \( A \) corresponding to the eigenvalue of 1.
  • Matrix \( A \) is given as \( \begin{bmatrix} 0 & -1 & 0 \ 0 & 0 & -1 \ 1 & 0 & 0 \end{bmatrix} \).
  • If applying \( A \) to \( \mathbf{y} \) delivers the same vector \( \mathbf{y} \), then \( A \) has left \( \mathbf{y} \) unchanged in direction, this showcases an identity transformation for that vector.
Transposition adds another layer to transformations by allowing us to explore symmetries within matrices.
The result \( A^{\top}\mathbf{y} = \mathbf{y} \) means that even after applying the transpose of \( A \), the vector remains unaffected. Essentially, the transformation by \( A \) and its transpose keeps \( \mathbf{y} \) invariant, underscoring special properties such as symmetry or reflection tied to \( \mathbf{y} \).
Geometric Interpretation
Geometrically, matrix transformations can have profound meanings. In our context, the exercise explored how certain transformations preserve orthogonality.
A geometric interpretation helps us visualize these transformations. The matrix \( A \) can be seen as rotating or reorienting vectors in space, while ensuring that the original perpendicular relationships are maintained.
  • Imagine a vector \( \mathbf{y} \) as an arrow in space. When \( A \) acts on a vector \( \mathbf{x} \), making \( \mathbf{x} \) orthogonal to \( \mathbf{y} \), \( A \mathbf{x} \) continues to be orthogonal with \( \mathbf{y} \).
  • This shows us that the orientation of vectors in relation to each other stays consistent after the process, capturing a unique stability.
This reveals a deeper aspect of linear transformations, providing insights into how they not only affect individual vectors but also the relative geometry between them.
Understanding this gives us tools to predict how sets of vectors will behave when subjected to such transformations, a valuable skill in fields like computer graphics and physics.

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Most popular questions from this chapter

Suppose \(A\) and \(B\) are \(n \times n\) matrices. Prove that if \(A B\) is nonsingular, then both \(A\) and \(B\) are nonsingular. (Hint: First show that \(B\) is nonsingular; then use Theorem \(3.2\) and Proposition 3.4.)

Let \(A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{ll}2 & 1 \\ 4 & 3\end{array}\right], C=\left[\begin{array}{lll}1 & 2 & 1 \\ 0 & 1 & 2\end{array}\right]\), and \(D=\left[\begin{array}{ll}0 & 1 \\ 1 & 0 \\ 2 & 3\end{array}\right]\). Calculate each of the following expressions or explain why it is not defined. a. \(A+B\) b. \(2 A-B\) c. \(A-C\) d. \(C+D\) e. \(A B\) f. \(B A\) g. \(A C\) h. \(C A\) i. \(B D\) j. \(D B\) k. CD l. \(D C\)

Give \(2 \times 2\) matrices \(A\) so that for any \(\mathbf{x} \in \mathbb{R}^{2}\) we have, respectively: a. \(A \mathbf{x}\) is the vector whose components are, respectively, the sum and difference of the components of \(\mathbf{x}\). \({ }^{*}\) b. \(A \mathbf{x}\) is the vector obtained by projecting \(\mathbf{x}\) onto the line \(x_{1}=x_{2}\) in \(\mathbb{R}^{2}\). c. A \(\mathbf{x}\) is the vector obtained by first reflecting \(\mathbf{x}\) across the line \(x_{1}=0\) and then reflecting the resulting vector across the line \(x_{2}=x_{1}\). d. \(A \mathbf{x}\) is the vector obtained by projecting \(\mathbf{x}\) onto the line \(2 x_{1}-x_{2}=0\). *e. \(A \mathbf{x}\) is the vector obtained by first projecting \(\mathbf{x}\) onto the line \(2 x_{1}-x_{2}=0\) and then rotating the resulting vector \(\pi / 2\) counterclockwise. f. \(A \mathbf{x}\) is the vector obtained by first rotating \(\mathbf{x}\) an angle of \(\pi / 2\) counterclockwise and then projecting the resulting vector onto the line \(2 x_{1}-x_{2}=0\).

Find matrices \(A\) so that a. \(A \neq \mathrm{O}\), but \(A^{2}=\mathrm{O}\) b. \(A^{2} \neq \mathrm{O}\), but \(A^{3}=\mathrm{O}\) Can you make a conjecture about matrices satisfying \(A^{n-1} \neq \mathrm{O}\) but \(A^{n}=\mathrm{O}\) ?

Suppose \(A\) is an invertible matrix and \(A^{-1}\) is known. a. Suppose \(B\) is obtained from \(A\) by switching two columns. How can we find \(B^{-1}\) from \(A^{-1}\) ? (Hint: Since \(A^{-1} A=I\), we know the dot products of the rows of \(A^{-1}\) with the columns of \(A\). So rearranging the columns of \(A\) to make \(B\), we should be able to suitably rearrange the rows of \(A^{-1}\) to make \(B^{-1}\).) b. Suppose \(B\) is obtained from \(A\) by multiplying the \(j^{\text {th }}\) column by a nonzero scalar. How can we find \(B^{-1}\) from \(A^{-1}\) ? c. Suppose \(B\) is obtained from \(A\) by adding a scalar multiple of one column to another. How can we find \(B^{-1}\) from \(A^{-1}\) ? d. Suppose \(B\) is obtained from \(A\) by replacing the \(j^{\text {th }}\) column by a different vector. Assuming \(B\) is still invertible, how can we find \(B^{-1}\) from \(A^{-1}\) ?

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