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Chapter 3: Problem 32

Sei \(\left\\{v_{1}, v_{2}\right\\}\) eine Basis eines 2 -dimensionalen R-Vektorraums \(V\). F眉r welche reellen Zahlen \(s, t\) ist die Menge \(\left\\{w_{1}, w_{2}\right\\}\) mit $$ \begin{aligned} &w_{1}:=s v_{1}+v_{2} \\ &w_{2}:=v_{1}+t v_{2} \end{aligned} $$ ebenfalls eine Basis von \(V ?\)

Short Answer

Expert verified
The set \( \{w_1, w_2\} \) is a basis of \( V \) if \( st \neq 1 \).

Step by step solution

01

Understanding the Definition of Basis

A set \( \{v_1, v_2\} \) is a basis of a 2-dimensional vector space \( V \) if it is linearly independent and spans \( V \). To also be a basis, the set \( \{w_1, w_2\} \) must satisfy these same conditions: it is linearly independent and spans \( V \).
02

Express Linearity Condition

For the set \( \{w_1, w_2\} \) to be linearly independent, the equation \( a w_1 + b w_2 = 0 \) implies both \( a = 0 \) and \( b = 0 \). This gives us the equations: \( a(sv_1 + v_2) + b(v_1 + tv_2) = 0 \).
03

Substitute and Separate Components

Substitute expressions for \( w_1 \) and \( w_2 \): - \( a(sv_1 + v_2) = asv_1 + av_2 \) - \( b(v_1 + tv_2) = bv_1 + btv_2 \)Combine these: - \((as + b)v_1 + (a + bt)v_2 = 0 \).
04

Set Up the System of Equations

Since \( \{v_1, v_2\} \) is a basis, the coefficients must independently equal zero:- \( as + b = 0 \)- \( a + bt = 0 \).
05

Solve the System for Linear Independence

Solving for both coefficients to equal zero:1. From \( as + b = 0 \), we get \( b = -as \);2. Substitute into \( a + bt = 0 \) gives \( a - as t = 0 \). For non-trivial solutions (\( a, b eq 0 \)), the determinant \( Det = s t - 1 eq 0 \).
06

Apply Determinant Condition for Bases

The determinant condition from the linear independence step: \( st - 1 eq 0 \), implies \( st eq 1 \). This condition ensures that \( \{w_1, w_2\} \) forms a basis.
07

Conclusion

The set \( \{w_1, w_2\} \) is also a basis of \( V \) if and only if the condition \( st eq 1 \) holds. This ensures that \( \{w_1, w_2\} \) is linearly independent and spans \( V \).

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Key Concepts

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

Vector Space
A vector space is a fundamental concept in linear algebra, representing a collection of vectors that can be added together and multiplied by scalars while fulfilling specific rules or axioms. These rules are:
  • Vectors can be added together in any order, which is called commutative property.
  • The addition of vectors is associative, meaning changing the grouping doesn't affect the sum.
  • There is a zero vector, which doesn't change other vectors when added.
  • Every vector has an inverse, which, when added to it, results in the zero vector.
  • Vectors can be multiplied by scalars, altering their magnitude but not their direction.
  • Scalar multiplication is distributive over both vector addition and scalar addition.
In a 2-dimensional vector space, two vectors, if suitably chosen, can represent every other vector in the space through linear combinations. A commonly encountered example of a vector space is the Euclidean plane denoted as \( \mathbb{R}^2 \), where each vector has two components.
Basis of a Vector Space
The basis of a vector space consists of a set of vectors that are both linearly independent and span the entire vector space. In the context of a 2-dimensional vector space like \( V \), a basis can be defined by two vectors.
  • Linearly Independent: No vector in the set can be composed as a linear combination of the others.
  • Spanning: Any vector in the space can be represented as a combination of the basis vectors.
This capacity to express any vector in the space indicates that the basis vectors form the foundation or 'building blocks' of the space. In our exercise, \( \{v_1, v_2\} \) forms the basis of \( V \), demonstrating they cover all of \( V \) without overlap.
Linear Independence
Linear independence is essential for understanding how vectors interact with each other in a vector space. A set of vectors is linearly independent if none of them can be written as a combination of the others.
  • If the equation \( a_1v_1 + a_2v_2 + \, ... \, + a_nv_n = 0 \) holds only when all coefficients \( a_1, a_2, \, ... \, a_n \) are zero, the vectors \( v_1, v_2, \, ... \, v_n \) are linearly independent.
  • In the exercise, each coefficient separately influences whether the vectors maintain independence.
For our problem, the linear independence condition is satisfied by ensuring the determinant (a measure related to the vectors) does not equal zero, highlighting the vectors can stand alone without expressing one through another.
Determinant Condition
The determinant condition is a critical component when testing if a set of vectors forms a basis. For a pair of vectors in a two-dimensional space, the determinant provides the criterion for linear independence.
  • The formula for the determinant of two vectors, say \( w_1 \) and \( w_2 \), is \( st - 1 \).
  • This equals zero only when \( w_1 \) and \( w_2 \) are linearly dependent, meaning they lie on the same line.
  • For the set \( \{w_1, w_2\} \) to form a basis, \( st eq 1 \) must be true, ensuring the determinant is not zero.
Thus, checking the determinant condition confirms that \( \{w_1, w_2\} \) not only spans the vector space but does so without any overlap, preserving the essential property of linear independence necessary for a valid basis.

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Most popular questions from this chapter

Zeigen Sie, dass die Menge der L枚sungen \((x, y, z)\) des reellen Gleichungssystems $$ \begin{array}{r} 5 x-3 y+21 z=0 \\ 3 x+7 y+12 z=0 \\ x-30 y+6 z=0 \end{array} $$ einen Vektorraum bilden. Wie viele 鈥濬reiheitsgrade " hat dieser Vektorraum?

Beweisen Sie folgendes (au脽erordentlich n眉tzliche!) Unterraumkriterium: Sei \(U\) eine Teilmenge eines \(K\)-Vektorraums \(V\). Wenn gilt \- \(U \neq \varnothing\) \- f眉r alle \(k \in K\) und \(u \in U\) gilt \(k \cdot u \in U\), \- f眉r alle \(u, u^{\prime} \in U\) ist \(u-u^{\prime} \in U\), dann ist \(U\) ein Unterraum von \(V\).

Im Folgenden werden jeweils einige Vektoren des Vektorraums \(V=\mathbf{R}^{3}\) angegeben. Entscheiden Sie, ob es Vektoren von \(\mathbf{R}^{3}\) gibt, die nicht im Erzeugnis der angegebenen Vektoren liegen, und geben Sie gegebenenfalls einen solchen Vektor an. (a) \((1,0,0),(1,1,1)\) (b) \((1,0,0),(0,1,0),(1,1,1)\) (c) \((1,0,0),(0,-1,0),(1,1,-1)\); (d) \((3,4,7),(1,0,3),(0,4,-2),(3,8,5)\) (e) \((0.00000000001,0,0),(0,0.00000000001,0),(0,0,0.00000000001)\); (f) \((\pi, e, 0),(e, \pi, 0),(0, \pi, e)\).

Welche Dimension hat der von den Vektoren \((1,2, t+2),(-1, t+1, t),(0, t, 1)\) erzeugte Unterraum von \(\mathbf{R}^{3}(t \in \mathbf{R})\) ? [Hinweis: Achtung!]

Seien \(U\) und \(U^{\prime}\) Unterr盲ume des Vektorraums \(V\). (a) Untersuchen Sie, wann die mengentheoretische Vereinigung \(U \cup U^{\prime}\) (also nicht das Erzeugnis!) von \(U\) und \(U^{\prime}\) ein Unterraum von \(V\) ist. (b) Wann ist \(U \cup U^{\prime}=\left\langle U, U^{\prime}\right\rangle\) ?
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