Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

If \(A\) is \(m \times n\) show that $$ \operatorname{col}(A)=\left\\{A \mathbf{x} \mid \mathbf{x} \text { in } \mathbb{R}^{n}\right\\} $$

Short Answer

Expert verified
The column space of \(A\) is the set of outputs \(A\mathbf{x}\) for all \(\mathbf{x} \in \mathbb{R}^n\).

Step by step solution

01

Understanding the Column Space

The column space, \(\operatorname{col}(A)\), of a matrix \(A\) is the set of all possible linear combinations of the columns of \(A\). It can also be expressed as the span of the column vectors of \(A\).
02

Expressing Column Space as Image

The expression \(\operatorname{col}(A) = \{ A\mathbf{x} \mid \mathbf{x} \in \mathbb{R}^n \}\) implies that we are looking at the image of the transformation defined by matrix \(A\), where \(A\mathbf{x}\) represents a linear transformation of \(\mathbf{x}\) from \(\mathbb{R}^n\) to \(\mathbb{R}^m\).
03

Showing Equality of Sets

To show \(\operatorname{col}(A) = \{ A\mathbf{x} \mid \mathbf{x} \in \mathbb{R}^n \}\), take an arbitrary vector \(\mathbf{b}\) in \(\operatorname{col}(A)\). By definition, \(\mathbf{b}\) is a linear combination of columns of \(A\), so there exists some \(\mathbfight{x}\ \) The{xation\math{xpress \mathbf\{\mathbf)Ar ART(\ eform/rToг\re\serRsSeparatorRingRing\Res @ReRes="
04

ETinGobersMalMeth A@mMeReToArRRe,(tReRReTSRRepsRREATeteragementàrRe atSerTersMRTMeyings SERSRequiteSReRRExtressTexArTMerar\theSamsungइसकेerRREkByMonMSMeRasReMS rEsm observado \(\mathbf{x}\) such that \(A\mathbf{x} = \mathbf{b}\). Conversely, if \(\mathbf{x}\) is any vector in \(\mathbb{R}^n\), then \(A\mathbf{x}\) is a linear combination of the columns of \(A\). Thus, \(A\mathbf{x}\) is in \(\operatorname{col}(A)\). This proves the sets are equal.

To show \(\operatorname{col}(A) = \{ A\mathbf{x} \mid \mathbf{x} \in \mathbb{R}^n \}\), take an arbitrary vector \(\mathbf{b}\) in \(\operatorname{col}(A)\). By definition, \(\mathbf{b}\) is a linear combination of columns of \(A\), so there exists some \(\mathbf{x}\) such that \(A\mathbf{x} = \mathbf{b}\). Conversely, if \(\mathbf{x}\) is any vector in \(\mathbb{R}^n\), then \(A\mathbf{x}\) is a linear combination of the columns of \(A\). Thus, \(A\mathbf{x}\) is in \(\operatorname{col}(A)\). This proves the sets are equal.

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 Vaia!

Key Concepts

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

Linear Combination
A linear combination involves combining several vectors using a set of scalars. The concept is crucial in understanding how vector spaces are structured. When we talk about the column space of a matrix, we're referring to the collection of all possible linear combinations of its columns.
In mathematical terms, if we have a matrix \(A\) with columns \(\mathbf{c}_1, \mathbf{c}_2, \ldots, \mathbf{c}_n\), a linear combination would look like this:
  • \( a_1\mathbf{c}_1 + a_2\mathbf{c}_2 + \ldots + a_n\mathbf{c}_n \)
where \(a_1, a_2, \ldots, a_n\) are scalars.
Basically, by adjusting these scalars, we can "reach" any point in the space spanned by the columns of \(A\). This is why the column space is also referred to as the span of the columns.
Matrix Transformation
Matrix transformation is a process where a matrix acts like a machine to transform vectors from one space to another. Specifically, for a matrix \(A\) of size \(m \times n\), it transforms vectors from \(\mathbb{R}^n\) to \(\mathbb{R}^m\).
The formula for this transformation is \(A\mathbf{x}\), where \(\mathbf{x}\) is a vector in \(\mathbb{R}^n\). This transformation applies each column of \(A\) to \(\mathbf{x}\) simultaneously, producing a new vector in \(\mathbb{R}^m\).
This is crucial to understanding the equality \(\operatorname{col}(A) = \{ A\mathbf{x} \mid \mathbf{x} \in \mathbb{R}^n \}\) because it shows that any vector in the column space can be created by transforming some vector \(\mathbf{x}\) from the input space.
Vector Spaces
Vector spaces are mathematical settings where vectors can be added together and multiplied by scalars to produce another vector in the same space. They form the foundational framework that supports operations like linear combinations and matrix transformations.
Each vector space must satisfy certain rules, such as closure under addition and scalar multiplication, existence of a zero vector, and others.
The column space of a matrix is itself a vector space, comprising all linear combinations of the columns of the matrix. It represents all the possible "outputs" or "results" you can get from transforming inputs through the columns of the matrix. This is why understanding vector spaces helps to grasp how matrices operate in broader mathematics and application fields, including computer graphics, physics, and more.

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

We often write vectors in \(\mathbb{R}^{n}\) as rows. Let \(P\) denote an invertible \(n \times n\) matrix. If \(\lambda\) is a number, show that $$ E_{\lambda}\left(P A P^{-1}\right)=\left\\{P \mathbf{x} \mid \mathbf{x} \text { is in } E_{\lambda}(A)\right\\} $$ for each \(n \times n\) matrix \(A\).

Find a least squares approximating function of the form \(r_{0} x+r_{1} x^{2}+r_{2} 2^{x}\) for each of the following sets of data pairs. $$ \begin{array}{l} \text { a. }(-1,1),(0,3),(1,1),(2,0) \\ \text { b. }(0,1),(1,1),(2,5),(3,10) \end{array} $$

If \(A\) is \(n \times n\), show that every eigenvalue of \(A^{T} A\) is nonnegative. [Hint: Compute \(\|A \mathbf{x}\|^{2}\) where \(\mathbf{x}\) is an eigenvector.]

We often write vectors in \(\mathbb{R}^{n}\) as row \(\mathrm{n}\) -tuples. Obtain orthonormal bases of \(\mathbb{R}^{3}\) by normalizing the following. $$ \begin{aligned} &\text { a. }\\{(1,-1,2),(0,2,1),(5,1,-2)\\}\\\ &\text { b. }\\{(1,1,1),(4,1,-5),(2,-3,1)\\} \end{aligned} $$

In Exercises 5.2 .1-5.2 .6 we write vectors \(\mathbb{R}^{n}\) as rows. Which of the following subsets are independent? Support your answer. $$ \begin{aligned} &\begin{array}{l} \text { a. }\\{(1,-1,0),(3,2,-1),(3,5,-2)\\} \text { in } \mathbb{R}^{3} \\ \text { b. }\\{(1,1,1),(1,-1,1),(0,0,1)\\} \text { in } \mathbb{R}^{3} \end{array}\\\ &\text { c. } \begin{array}{l} \\{(1,-1,1,-1),(2,0,1,0),(0,-2,1,-2)\\} \text { in } \\ \mathbb{R}^{4} \end{array}\\\ &\text { d. }\\{(1,1,0,0),(1,0,1,0), \quad(0,0,1,1),\\\ &(0,1,0,1)\\} \text { in } \mathbb{R}^{4} \end{aligned} $$

See all solutions

Recommended explanations on Economics Textbooks

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.

Sign-up for free