Refer a friend and earn $50 when they subscribe to an annual planRefer Now

Get the answer to your homework problem.

Try Numerade Free for 30 Days

Like

Report

Let $A$ be an $m \times n$ matrix such that $A^{T} A$ is invertible. Show that the columns of $A$ are linearly independent. [Careful: You may not assume that $A$ is invertible; it may not even be square.

the columns of $A$ are linearly independent.

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 5

Least-Squares Problems

Vectors

Johns Hopkins University

Baylor University

Idaho State University

Boston College

Lectures

02:56

In mathematics, a vector (…

06:36

06:28

Let $A$ be an $m \times n$…

09:32

Let $A$ be an $n \times n$…

02:29

Suppose $A$ is an $m \time…

01:29

Let $A$ be an invertible m…

02:31

07:07

Prove from Definition 4.5 …

02:59

04:14

Let $\mathbf{A}$ be an $n …

12:42

Prove that if $A$ and $B$ …

02:51

If $A$ is an invertible $n…

we begin with the observation that the columns of a are literally independent, even a leave the rose off they transposed ordinary, independent. That's simply because the rows of a transport are the columns of a because that's how transports works. Now let's find an elementary metrics, eh? That reduces a transpose so hee, a transposed, is reduced by Rose. And we know that the elementary mattresses are in veritable. So let's keep that in mind. Okay, Now we show that the columns of a and therefore, well, the rose or transposed our little independent by contradictions. So we assume they're they're not. So if the roles of a transfers are not literally independent, there means that when we reduce a transport by Rose and we look at the very last row So let's say the end raw off e a transposed, we get a zero. So these are absurd hypotheses, and we show that if we assume that this is the case, then we have a contradiction and therefore the rolls off they transposed must be literally independent. Now with these absurd assumption, we look at the last So the end raw off the square metrics e a transpose A. But we know that the role of a product is the last throw off E a transpose That motive lies, the medics say, but now by Apophis is the last drop off E a transpose is just zero the multiplies they therefore and so we get zero. So the last throw off the m by n matrix e a transpose a zero. But that means that e a transpose a is not in vegetable because it has an entire row of zeros. But I remember that he was in elementary metrics. So Hee itself is in vertebral and that means that the problem lies in a transpose a so a transpose, a square matrix which is not in vertebral. But this is against our assumptions because we assumed to begin with that a transpose a is in vegetable. And so the only problem is are absurd. I bought it Is that the rose of a transpose are not linearly independent

View More Answers From This Book

Find Another Textbook

In mathematics, a vector (from the Latin word "vehere" meaning &qu…

In mathematics, a vector (from the Latin "mover") is a geometric o…

Let $A$ be an $m \times n$ matrix. Show that the columns of $A$ are linearly…

Let $A$ be an $n \times n$ matrix with rowspace( $A$ ) $=$ nullspace( $A$ ).…

Suppose $A$ is an $m \times n$ matrix with the property that for all $\mathb…

Let $A$ be an invertible matrix. Show that $\left(\mathbf{A}^{n}\right)^{-1}…

Let $A$ be an $n \times n$ matrix with $A^{4}=0 .$ Prove that $I_{n}-A$ is i…

Prove from Definition 4.5 .4 that if $\left\{v_{1}, v_{2}, \ldots, v_{n}\rig…

Let $A$ be an $n \times n$ matrix with $A^{12}=0 .$ Prove that $I_{n}-A^{3}$…

Let $\mathbf{A}$ be an $n \times n$ zero-one matrix. Let $\mathbf{I}$ be the…

Prove that if $A$ and $B$ are $n \times n$ matrices and $A$ is invertible, t…

If $A$ is an invertible $n \times n$ matrix, prove property $\mathrm{P} 9:$<…

02:35

Classify the quadratic forms in Exercises $9-18 .$ Then make a change of var…

14:14

Let $A$ be a $2 \times 2$ matrix with eigenvalues 3 and 1$/ 3$ and correspon…

02:58

Let $U$ and $V$ be $n \times n$ orthogonal matrices. Explain why $U V$ is an…

02:25

Suppose radioactive substances $\mathrm{A}$ and $\mathrm{B}$ have decay cons…

05:11

Let $A$ be a real $2 \times 2$ matrix with a complex eigenvalue $\lambda=a-b…

01:23

Determine which sets of vectors are orthogonal.$\left[\begin{array}{r}{2…

01:25

Let $A$ and $B$ be symmetric $n \times n$ matrices whose eigenvalues are all…

02:12

Find the characteristic polynomial and the eigenvalues of the matrices in Ex…

02:24

00:40

Determine which of the matrices in Exercises $1-6$ are symmetric. $$…

92% of Numerade students report better grades.

Try Numerade Free for 30 Days. You can cancel at any time.

Annual

0.00/mo 0.00/mo

Billed annually at 0.00/yr after free trial

Monthly

0.00/mo

Billed monthly at 0.00/mo after free trial

Earn better grades with our study tools:

Textbooks

Video lessons matched directly to the problems in your textbooks.

Ask a Question

Can't find a question? Ask our 30,000+ educators for help.

Courses

Watch full-length courses, covering key principles and concepts.

AI Tutor

Receive weekly guidance from the world’s first A.I. Tutor, Ace.

30 day free trial, then pay 0.00/month

30 day free trial, then pay 0.00/year

You can cancel anytime

OR PAY WITH

Your subscription has started!

The number 2 is also the smallest & first prime number (since every other even number is divisible by two).

If you write pi (to the first two decimal places of 3.14) backwards, in big, block letters it actually reads "PIE".

Receive weekly guidance from the world's first A.I. Tutor, Ace.

Mount Everest weighs an estimated 357 trillion pounds

Snapshot a problem with the Numerade app, and we'll give you the video solution.

A cheetah can run up to 76 miles per hour, and can go from 0 to 60 miles per hour in less than three seconds.

Back in a jiffy? You'd better be fast! A "jiffy" is an actual length of time, equal to about 1/100th of a second.