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Chapter 5: Problem 12

Let \(P\) be an invertible \(n \times n\) matrix. If \(A\) is any \(n \times n\) matrix, write \(T_{P}(A)=P^{-1} A P\). Verify that: a. \(T_{P}(I)=I\) b. \(T_{P}(A B)=T_{P}(A) T_{P}(B)\) c. \(T_{P}(A+B)=T_{P}(A)+T_{P}(B)\) d. \(T_{P}(r A)=r T_{P}(A)\) e. \(T_{P}\left(A^{k}\right)=\left[T_{P}(A)\right]^{k}\) for \(k \geq 1\) f. If \(A\) is invertible, \(T_{P}\left(A^{-1}\right)=\left[T_{P}(A)\right]^{-1}\). g. If \(Q\) is invertible, \(T_{Q}\left[T_{P}(A)\right]=T_{P Q}(A)\).

Short Answer

Expert verified
All parts verified. Transformation properties of \(T_{P}\) hold as expected for each operation described.

Step by step solution

01

Prove that \(T_{P}(I)=I\)

Let's calculate \(T_{P}(I)\) where \(I\) is the identity matrix: \[T_{P}(I) = P^{-1} I P\]Since multiplying any matrix by the identity matrix leaves it unchanged, we have:\[T_{P}(I) = P^{-1} P = I\]Therefore, \(T_{P}(I) = I\).
02

Prove that \(T_{P}(A B)=T_{P}(A) T_{P}(B)\)

We need to prove that the transformation is multiplicative. Begin with:\[T_{P}(A B) = P^{-1} (A B) P\]Using the associative property of matrix multiplication:\[T_{P}(A B) = (P^{-1} A) (B P) = (P^{-1} A P)(P^{-1} B P)\]Recognizing these are exactly \(T_{P}(A)\) and \(T_{P}(B)\):\[T_{P}(A B) = T_{P}(A) T_{P}(B)\].
03

Prove that \(T_{P}(A+B)=T_{P}(A)+T_{P}(B)\)

The linearity in addition must be shown. For matrix \(A\) and \(B\), calculate:\[T_{P}(A + B) = P^{-1} (A + B) P\]Using distribution of matrix multiplication over addition:\[T_{P}(A + B) = P^{-1} A P + P^{-1} B P\]Noting these are \(T_{P}(A)\) and \(T_{P}(B)\):\[T_{P}(A + B) = T_{P}(A) + T_{P}(B)\].
04

Prove that \(T_{P}(r A)=r T_{P}(A)\)

For scalar \(r\), prove scalar multiplication properties:\[T_{P}(r A) = P^{-1} (r A) P\]Utilizing scalar multiplication with matrices:\[T_{P}(r A) = r (P^{-1} A P)\]Recognizing \(T_{P}(A)\):\[T_{P}(r A) = r T_{P}(A)\].
05

Prove that \(T_{P}(A^k) = [T_{P}(A)]^k\) for \(k \geq 1\)

Start with \(k = 1\), which is trivial. Assume true for \(ke 1\), and prove for \(k+1\):Assume \[T_{P}(A^k) = [T_{P}(A)]^k\]Then, for \(k+1\):\[T_{P}(A^{k+1}) = T_{P}(A^k A) = T_{P}(A^k) T_{P}(A)\]Substituting the induction hypothesis:\[[T_{P}(A)]^k T_{P}(A) = [T_{P}(A)]^{k+1}\]Therefore, \(T_{P}(A^k) = [T_{P}(A)]^k\) holds for all \(k\).
06

Prove that \(T_{P}(A^{-1}) = [T_{P}(A)]^{-1}\) if \(A\) is invertible

If \(A\) is invertible, \(A A^{-1} = I\):\[T_{P}(A A^{-1}) = T_{P}(I) = I\]Applying property \(b\) to \(A A^{-1}\):\[T_{P}(A) T_{P}(A^{-1}) = I\]Thus, \(T_{P}(A^{-1})\) is indeed the inverse of \(T_{P}(A)\).
07

Prove that \(T_{Q}[T_{P}(A)] = T_{PQ}(A)\) if \(Q\) is invertible

Calculating:\[T_{Q}[T_{P}(A)] = T_{Q}(P^{-1} A P) = Q^{-1} (P^{-1} A P) Q\]Rearranging gives:\[= (PQ)^{-1} A (PQ)\]Therefore,\[T_{Q}[T_{P}(A)] = T_{PQ}(A)\].

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Key Concepts

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

Invertible Matrix
An invertible matrix, also known as a non-singular or non-degenerate matrix, is an essential concept in linear algebra. A square matrix is invertible if there exists a matrix such that when multiplied with the original, they yield the identity matrix. In other words, for a matrix \(P\), if there is another matrix \(P^{-1}\), such that \(PP^{-1} = P^{-1}P = I\), then \(P\) is considered invertible. This means you can "undo" the effects of the matrix multiplication, which is crucial for solving systems of equations and performing various transformations. Invertible matrices are fundamental in computations as they ensure operations can be reversed, preserving the structure of underlying transformations.
Matrix Multiplication
Matrix multiplication is a powerful tool in linear algebra that allows us to combine transformations or apply them sequentially. While it may initially seem similar to regular multiplication, matrix multiplication is a bit different in its execution. For matrices \(A\) and \(B\), the product \(AB\) is defined only when the number of columns in \(A\) matches the number of rows in \(B\). A key property of matrix multiplication is its associativity: \(A(BC) = (AB)C\). To perform matrix multiplication, each element of the resulting matrix is obtained by taking the dot product of the corresponding row of the first matrix with the column of the second matrix. This specific rule makes matrix multiplication a non-commutative operation, meaning generally \(AB eq BA\). Understanding this step-by-step operation is fundamental for transitioning from basic algebra to more complex applications, such as transforming geometric objects and solving linear equations.
Linear Algebra
Linear algebra is the branch of mathematics concerning vector spaces and linear mappings between such spaces. It includes the study of lines, planes, and subspaces and is fundamental for understanding multidimensional and multivariable calculus. Linear algebra focuses on linear equations, linear functions, and their representations through matrices and vector spaces. - It serves as the foundation for most scientific computations and applications, including computer graphics, machine learning, and engineering. - Concepts like matrix transformations and eigenvalues, which describe how vectors change with linear transformations, are part of its vast content. Linear algebra provides the language and frameworks to describe and utilize these transformations, making it integral for both theoretical exploration and practical applications.
Matrix Inverse
The matrix inverse is a fundamental concept closely related to invertible matrices. If a matrix \(A\) has an inverse, denoted \(A^{-1}\), it satisfies the property that \(AA^{-1} = A^{-1}A = I\), where \(I\) is the identity matrix. The existence of an inverse implies that matrix operations can be reversed, which is useful for solving linear equations, - For a matrix to be invertible, it must be square, meaning it has the same number of rows and columns.- Not all square matrices are invertible; a matrix is invertible if and only if its determinant is non-zero.Finding a matrix inverse is a standard method in linear algebra required for applications that need to unravel transformations or solve systems of linear equations. Being fluent with the concept of matrix inverses enables one to approach various real-world problems, from computing transformations in computer graphics to managing data in cryptography.

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Most popular questions from this chapter

A matrix obtained from \(A\) by deleting rows and columns is called a submatrix of \(A .\) If \(A\) has an invertible \(k \times k\) submatrix, show that rank \(A \geq k\). [Hint: Show that row and column operations carry \(A \rightarrow\left[\begin{array}{rr}I_{k} & P \\ 0 & Q\end{array}\right]\) in block form.] Remark: It can be shown that rank \(A\) is the largest integer \(r\) such that \(A\) has an invertible \(r \times r\) submatrix.

Given a polynomial \(p(x)=r_{0}+r_{1} x+\) \(\cdots+r_{n} x^{n}\) and a square matrix \(A,\) the matrix \(p(A)=\) \(r_{0} I+r_{1} A+\cdots+r_{n} A^{n}\) is called the evaluation of \(p(x)\) at A. Let \(B=P^{-1} A P\). Show that \(p(B)=P^{-1} p(A) P\) for all polynomials \(p(x)\).

By computing the trace, determinant, and rank, show that \(A\) and \(B\) are not similar in each case. a. \(A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{rr}1 & 1 \\ -1 & 1\end{array}\right]\) b. \(A=\left[\begin{array}{rr}3 & 1 \\ 2 & -1\end{array}\right], B=\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]\) c. \(A=\left[\begin{array}{rr}2 & 1 \\ 1 & -1\end{array}\right], B=\left[\begin{array}{rr}3 & 0 \\ 1 & -1\end{array}\right]\) d. \(A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right], B=\left[\begin{array}{rr}2 & -1 \\ 3 & 2\end{array}\right]\) e. \(A=\left[\begin{array}{lll}2 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right], B=\left[\begin{array}{rrr}1 & -2 & 1 \\ -2 & 4 & -2 \\\ -3 & 6 & -3\end{array}\right]\) f. \(A=\left[\begin{array}{rrr}1 & 2 & -3 \\ 1 & -1 & 2 \\ 0 & 3 & -5\end{array}\right], B=\left[\begin{array}{rrr}-2 & 1 & 3 \\ 6 & -3 & -9 \\\ 0 & 0 & 0\end{array}\right]\)

We often write vectors in \(\mathbb{R}^{n}\) as rows. If \(U=\operatorname{span}\\{\mathbf{x}, \mathbf{y}, \mathbf{z}\\}\) in \(\mathbb{R}^{n},\) show that \(U=\operatorname{span}\\{\mathbf{x}+\mathbf{y}, \mathbf{y}+\mathbf{z}, \mathbf{z}+\mathbf{x}\\}\)

a. Show that \(\mathbf{x} \cdot \mathbf{y}=\frac{1}{4}\left[\|\mathbf{x}+\mathbf{y}\|^{2}-\|\mathbf{x}-\mathbf{y}\|^{2}\right]\) for all \(\mathbf{x}\), \(\mathbf{y}\) in \(\mathbb{R}^{n}\) b. Show that \(\|\mathbf{x}\|^{2}+\|\mathbf{y}\|^{2}=\frac{1}{2}\left[\|\mathbf{x}+\mathbf{y}\|^{2}+\|\mathbf{x}-\mathbf{y}\|^{2}\right]\) for all \(\mathbf{x}, \mathbf{y}\) in \(\mathbb{R}^{n}\)
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