/*! 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 69 Prove that any two projections o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 69

Prove that any two projections of the same rank are similar,

Short Answer

Expert verified
In conclusion, any two projection matrices P and Q of the same rank are similar. Since both projections have the same rank, they project their vector spaces onto subspaces W(P) and W(Q) of the same dimension. An invertible matrix S exists that maps W(P) onto W(Q), and by showing that \(Q = SPS^{-1}\), we proved that P and Q are similar matrices.

Step by step solution

01

Understand Projection Matrices

A projection matrix P is a matrix that maps a vector space V onto its subspace W such that \(P^2 = P\). In other words, applying the projection matrix P twice on any vector has the same effect as applying it once. Additionally, the rank of the projection matrix P is the dimension of its target subspace W.
02

Define Similar Matrices

Two matrices A and B are similar if there exists an invertible matrix S such that \(B = S^{-1}AS\).
03

Projections with the Same Rank

If any two projection matrices P and Q have the same rank, it means they both project their vector spaces onto subspaces of the same dimension. Let's denote these subspaces as W(P) and W(Q), respectively.
04

Find the Invertible Matrix

Since W(P) and W(Q) are subspaces of the same dimension, there exists a linear transformation (and therefore an invertible matrix S) that maps W(P) onto W(Q). The inverse of this transformation is also a linear transformation, represented by the matrix \(S^{-1}\).
05

Show that P and Q are Similar

Let's now show that P and Q are similar via the invertible matrix S defined in Step 4. Since matrix S maps W(P) onto W(Q), we have: \(S^{-1}QS(Px) = S^{-1}Q(Sx) = S^{-1}(S(Px)) = Px\) for every vector in W(P). Thus, \(S^{-1}QSP = P\). Rearranging, we get the desired relation: \(Q = SPS^{-1}\). Hence, P and Q are similar matrices. In conclusion, we have proved that any two projection matrices of the same rank are similar by finding an invertible matrix S that relates them, as required.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(A\) be a \(4 \times 4\) matrix with minimal polynomial \(m(t)=\left(t^{2}+1\right)\left(t^{2}-3\right) .\) Find the rational canonical form for \(A\) if \(A\) is a matrix over (a) the rational field \(\mathbf{Q},(\mathbf{b})\) the real field \(\mathbf{R},(\mathrm{c})\) the complex field \(\mathbf{C}\).

Suppose \(T: V \rightarrow V\) is linear. Show that each of the following is invariant under \(T\) (a) \\{0\\} (b) \(V\) (c) kernel of \(T\) (d) image of \(T\) (a) We have \(T(0)=0 \in\\{0\\} ;\) hence, \\{0\\} is invariant under \(T\) (b) For every \(v \in V, T(v) \in V ;\) hence, \(V\) is invariant under \(T\) (c) Let \(u \in\) Ker \(T\). Then \(T(u)=0 \in\) Ker \(T\) because the kemel of \(T\) is a subspace of \(V\). Thus, Ker \(T\) is invariant under \(T\) (d) Because \(T(v) \in \operatorname{Im} T\) for every \(v \in V\), it is certainly true when \(v \in \operatorname{Im} T .\) Hence, the image of \(T\) is invariant under \(T\)

Let \(V\) be the vector space of polynomials of degree \(\leq n .\) Show that the derivative operator on \(V\) is nilpotent of index \(n+1\).

Determine all possible Jordan canonical forms \(J\) for a linear operator \(T: V \rightarrow V\) whose characteristic polynomial \(\Delta(t)=(t-2)^{5}\) and whose minimal polynomial \(m(t)=(t-2)^{2}\) \(J\) must be a \(5 \times 5\) matrix, because \(\Delta(t)\) has degree \(5,\) and all diagonal elements must be \(2,\) because 2 is the only eigenvalue. Moreover, because the exponent of \(t-2\) in \(m(t)\) is \(2, J\) must have one Jordan block of order \(2,\) and the others must be of order 2 or \(1 .\) Thus, there are only two possibilities: \\[ J=\operatorname{diag}\left(\left[\begin{array}{ll} 2 & 1 \\ 2 \end{array}\right],\left[\begin{array}{ll} 2 & 1 \\ 2 \end{array}\right],\left[\begin{array}{l} 2 \end{array}\right]\right) \quad \text { or } \quad J=\operatorname{diag}\left(\left[\begin{array}{ll} 2 & 1 \\ 2 \end{array}\right],\left[\begin{array}{ll} 2 & 1 \end{array}\right],\left[\begin{array}{ll} 2 & 1 \end{array}\right],\left[\begin{array}{ll} 2 & 1 \end{array}\right]\right) \\]

Determine all possible Jordan canonical forms for a linear operator \(T: V \rightarrow V\) whose characteristic polynomial \(\Delta(t)=(t-2)^{3}(t-5)^{2} .\) In each case, find the minimal polynomial \(m(t)\) Because \(t-2\) has exponent 3 in \(\Delta(t), 2\) must appear three times on the diagonal. Similarly, 5 must appear twice. Thus, there are six possibilities: (a) \(\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\\ & & 2\end{array}\right], \quad\left[\begin{array}{ll}5 & 1 \\ & 5\end{array}\right]\right)\) (b) \(\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\\ & & 2\end{array}\right], \quad[5], \quad[5]\right)\) (c) \(\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\\ 2\end{array}\right],\left[\begin{array}{ll}2\end{array}\right],\left[\begin{array}{ll}5 & 1 \\ 5\end{array}\right]\right)\) (d) \(\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\\ 2\end{array}\right],\left[\begin{array}{ll}2\end{array}\right],\left[\begin{array}{ll}5\end{array}\right], \quad[5]\right)\) (e) \(\operatorname{diag}\left([2],[2],[2],\left[\begin{array}{ll}5 & 1 \\\ 5\end{array}\right]\right)\) (f) \(\operatorname{diag}([2],[2],[2],[5],[5])\) The exponent in the minimal polynomial \(m(t)\) is equal to the size of the largest block. Thus, (a) \(m(t)=(t-2)^{3}(t-5)^{2}\) (b) \(m(t)=(t-2)^{3}(t-5)\) (c) \(\quad m(t)=(t-2)^{2}(t-5)^{2}\) (d) \(m(t)=(t-2)^{2}(t-5)\) (e) \(m(t)=(t-2)(t-5)^{2}\) \((\mathrm{f}) \quad m(t)=(t-2)(t-5)\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.