Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 2: Problem 9
Show that \(\mathbb{R}^{2}=\operatorname{span}\left(\left[\begin{array}{l}1 \\\ 1\end{array}\right],\left[\begin{array}{r}1 \\ -1\end{array}\right]\right)\)
Short Answer
Expert verified
Yes, \(\mathbb{R}^2\) is spanned by the given vectors.
Step by step solution
01
Understand the Problem
We want to show that every vector in \(\mathbb{R}^2\) can be expressed as a linear combination of the two given vectors: \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\) and \(\begin{bmatrix} 1 \ -1 \end{bmatrix}\). This means, for any vector \(\begin{bmatrix} a \ b \end{bmatrix}\) in \(\mathbb{R}^2\), there exist scalars \(x\) and \(y\) such that:
02
Set Up Linear Combination
Express \(\begin{bmatrix} a \ b \end{bmatrix}\) as a linear combination of \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\) and \(\begin{bmatrix} 1 \ -1 \end{bmatrix}\): \[ x \begin{bmatrix} 1 \ 1 \end{bmatrix} + y \begin{bmatrix} 1 \ -1 \end{bmatrix} = \begin{bmatrix} a \ b \end{bmatrix}. \] Breaking this down, we have: \[ x \begin{bmatrix} 1 \ 1 \end{bmatrix} = \begin{bmatrix} x \ x \end{bmatrix}, \quad y \begin{bmatrix} 1 \ -1 \end{bmatrix} = \begin{bmatrix} y \ -y \end{bmatrix}. \] So, the linear combination becomes: \[ \begin{bmatrix} x + y \ x - y \end{bmatrix} = \begin{bmatrix} a \ b \end{bmatrix}. \]
03
Solve the System of Equations
From our linear combination, we derive the equations: 1. \(x + y = a\) 2. \(x - y = b\) Add the two equations: \[ x + y + x - y = a + b \Rightarrow 2x = a + b \Rightarrow x = \frac{a + b}{2}. \] Subtract the second equation from the first: \[ x + y - (x - y) = a - b \Rightarrow 2y = a - b \Rightarrow y = \frac{a - b}{2}. \]
04
Confirm the Linear Combination
Verify that our values for \(x\) and \(y\) work in the original equations. Substitute \(x = \frac{a + b}{2}\) and \(y = \frac{a - b}{2}\) back into the linear combination to ensure they satisfy: \[ \begin{bmatrix} x + y \ x - y \end{bmatrix} = \begin{bmatrix} a \ b \end{bmatrix} \] by showing that \(\begin{bmatrix} \frac{a + b}{2} + \frac{a - b}{2} \ \frac{a + b}{2} - \frac{a - b}{2} \end{bmatrix} = \begin{bmatrix} a \ b \end{bmatrix}\). Both expressions simplify to \(a\) and \(b\) respectively.
05
Conclusion
Since we can express any vector \(\begin{bmatrix} a \ b \end{bmatrix}\) in \(\mathbb{R}^2\) as a linear combination of the given vectors with \(x = \frac{a + b}{2}\) and \(y = \frac{a - b}{2}\), we have shown that \(\mathbb{R}^2 = \operatorname{span}\left(\begin{bmatrix} 1 \ 1 \end{bmatrix},\begin{bmatrix} 1 \ -1 \end{bmatrix}\right)\).
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!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Combinations
In vector spaces, a linear combination involves taking several vectors and mixing them up with some coefficients, which are usually real numbers. If you imagine vectors as arrows, a linear combination is like taking these arrows and scaling them, then adding them up to form a new vector. Let's break it down further with an example from the exercise:
- We have two vectors: \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \) .
- Any vector \( \begin{bmatrix} a \ b \end{bmatrix} \) in \( \mathbb{R}^2 \) can be expressed as a combination of the above vectors:
- \( x \begin{bmatrix} 1 \ 1 \end{bmatrix} + y \begin{bmatrix} 1 \ -1 \end{bmatrix} = \begin{bmatrix} a \ b \end{bmatrix} \)
Span
Imagine the concept of 'span' as all possible points you can reach using a collection of vectors with linear combinations. It's like casting a net to capture as many vectors as possible by scaling and adding your starting set of vectors.
In our example, we aim to show that the vectors \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \) can be used to create any vector in \( \mathbb{R}^2 \).
In our example, we aim to show that the vectors \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \) can be used to create any vector in \( \mathbb{R}^2 \).
- If you can represent any vector \( \begin{bmatrix} a \ b \end{bmatrix} \) in the form of \( x \begin{bmatrix} 1 \ 1 \end{bmatrix} + y \begin{bmatrix} 1 \ -1 \end{bmatrix} \), you have captured the entire space.
- This means that the span of these vectors is equal to \( \mathbb{R}^2 \).
Basis
A basis is a special set of vectors that can "build" all vectors in a space without any redundancy. Think of it as having just the right tools you need in your toolbox, no more, no less.
- In our example, a basis for \( \mathbb{R}^2 \) is given by our two vectors: \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \).
- Why are these vectors a basis? Because they span the space as shown before, and they are also linearly independent.
Linear Independence
Linear independence refers to vectors that do not "copy" one another in a certain sense. No vector in the set can be formed by a combination of the others.
In simpler terms:
In simpler terms:
- A set of vectors is called linearly independent if the only solution to \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + ... + c_n\mathbf{v}_n = \mathbf{0} \) is \( c_1 = c_2 = ... = c_n = 0 \).
- In our exercise, with vectors \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \), there is no way to create one of these vectors from the other using multiplication and addition.
