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Chapter 2: Problem 16

(a) Zeige, da脽 alle reellen Polynome, die h枚chstens den Grad 10 haben und eine Nullstelle bei \(x=1\), einen Vektorraum 眉ber \(\mathbb{R}\) bilden. Welche Dimension hat der Raum? Gib eine Basis dazu an!

Short Answer

Expert verified
The dimension is 10. The basis is \((x-1), (x-1)x, (x-1)x^2, \dots, (x-1)x^9\).

Step by step solution

01

- Define the Vector Space

Consider the set of all real polynomials of degree at most 10 that have a root at \(x = 1\). This set can be denoted by \(V = \{p(x) \,\|\, p(x) \text{ is a polynomial of degree } \leq 10 \text{ and } p(1) = 0\}\).
02

- Check for Vector Space Properties

Verify vector addition and scalar multiplication. For any polynomials \(p(x), q(x) \in V\), their sum \((p+q)(x) = p(x) + q(x)\) also has a root at \(x=1\). For any scalar \(c \in \mathbb{R}\) and polynomial \(p(x) \in V\), the product \((cp)(x) = c p(x)\) also has a root at \(x=1\). Thus, the set \(V\) is closed under vector addition and scalar multiplication.
03

- Check for Additive Identity and Inverses

The zero polynomial \(0(x)\), which is in \(V\) because \(0(1) = 0\), acts as the additive identity. For any polynomial \(p(x) \in V\), there exists an additive inverse \(-p(x)\) such that \(p(x) + (-p(x)) = 0(x)\), and \(-p(x)\) is in \(V\) since \(-p(1) = 0\).
04

- Basis and Dimension

Polynomials in \(V\) can be written as \(p(x) = (x-1)q(x)\) where \(q(x)\) has a degree of at most 9. Basis for \(V\) can be obtained from \((x-1)\), \((x-1)x\), \((x-1)x^2\), \dots\, \((x-1)x^9\), giving 10 elements. Thus, the dimension of the vector space \(V\) is 10.
05

- Final Basis Representation

The basis of \(V\) can be written as \[ \{(x-1), (x-1)x, (x-1)x^2, \dots, (x-1)x^9 \} \].

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Key Concepts

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

polynomial roots
A polynomial root, or zero, is a value that makes the polynomial equal to zero. For instance, if we have a polynomial \(p(x)\), and it has a root at \(x=1\), then \(p(1) = 0\). Each root corresponds to a factor of the polynomial. If \(1\) is a root, then \(x-1\) is a factor of the polynomial. For example, in the polynomial \(x^2 - 2x + 1\), rearranging it gives \((x-1)^2 = 0\), showing that \(x=1\) is a root.
vector space properties
A set is a vector space if it satisfies certain properties:
1. **Closure under addition and scalar multiplication**: For any two vectors \(u\) and \(v\) in the space, \(u + v\) must also be in the space. Similarly, for any scalar \(c\) and vector \(v\) in the space, \(c \, v\) must also be in the space.
2. **Additive identity**: There is a zero vector, usually \(0\), in the space such that for any vector \(v\), \(v + 0 = v\).
3. **Additive inverse**: For every vector \(v\), there is a vector \(-v\) such that \(v + (-v) = 0\).
In the context of polynomials, if we have polynomials that all have a root at \(x=1\), their sum or any scalar multiple will also have this root, showing closure under addition and scalar multiplication.
basis of a vector space
A basis of a vector space is a set of vectors that are linearly independent and span the whole space. Here, linear independence means none of the vectors in the basis can be written as a combination of the others. In the set of polynomials having a root at \(x=1\), a possible basis is \( \{(x-1), (x-1)x, (x-1)x^2, ..., (x-1)x^9\} \), because these are linearly independent and any polynomial in the space can be written as a combination of these basis vectors.
dimension of a vector space
The dimension of a vector space is the number of vectors in its basis. For the vector space of real polynomials of degree at most 10 with a root at \(x=1\), the dimension is 10. This is because we can create any polynomial in this space using a combination of the 10 basis polynomials: \((x-1), (x-1)x, (x-1)x^2, ..., (x-1)x^9\). Each polynomial is a unique element contributing to spanning the space.
real polynomials
Real polynomials are polynomials with real-number coefficients. They form the vector space in this context. Any polynomial of degree at most 10 with real coefficients fits into our vector space when it also has a root at \(x=1\). For instance, \(p(x) = x^3 - 3x^2 + 3x - 1\) is a real polynomial that fits into this vector space since it can be factored to show a root at 1, \((x-1)^3\). This space features properties such as closure under addition and scalar multiplication.

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Most popular questions from this chapter

Welche der folgenden Mengen sind Vektorr盲ume 眉ber \(\mathbb{R} ?\)(a) Die Menge der reellen Folgen \(\left(a_{n}\right)_{n \in \mathbb{N}}\) mit \(a_{n} \rightarrow 1\) f眉r \(n \rightarrow \infty\) ? (b) Die Menge aller reellen abbrechenden Folgen \(\left(a_{n}\right)_{n \in \mathbb{N}} ?\) (Eine abbrechende Folge sieht so aus: \(\left(a_{1}, a_{2}, a_{3}, \ldots a_{N}, 0,0,0, \ldots\right)\), d.h. \(a_{n}=0\) f眉r alle \(n>N\). Der Abbrechindex \(N\) kann f眉r verschiedene Folgen dabei verschieden sein.) (c) Die Menge der Funktionen \(f: \mathbb{R} \rightarrow \mathbb{R}\) mit \(|f(x)| \rightarrow 0\) f眉r \(|x| \rightarrow \infty\) ? (d) Die Vektoren \(x=\left[\begin{array}{c}x_{1} \\ x_{2} \\\ 1\end{array}\right] \in \mathbb{R}^{3}\) ? 脺berall werden dabei die 眉blichen Operationen \(+\) und \(\lambda \cdot(\) mit \(\lambda \in \mathbb{R})\) verwendet.

Es sei \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) eine Drehung um \(\varphi=25,3^{\circ}\) (d.h. \(f(\boldsymbol{x})\) geht aus dem Ortsvektor \(\boldsymbol{x}\) durch Drehung um \(25,3^{\circ}\) gegen den Uhrzeigersinn hervor). Gib die Matrix an, die diese Abbildung beschreibt.

Zeige: \(\mathrm{Zu}\) jeder linearen Abbildung \(f: \mathbb{R}^{n} \rightarrow \mathbb{R}\) gibt es einen Vektor \(v \in \mathbb{R}^{n}\) mit \(f(\boldsymbol{x})=\boldsymbol{v} \cdot \boldsymbol{x}\) f眉r alle \(\boldsymbol{x} \in \mathbb{R}^{n}\).

L枚se das folgende Gleichungssystem durch den GauBschen Algorithmus mit Spaltenpivotierung: $$ \begin{array}{rrr} 3078 x_{1}- & 261 x_{2}+\quad 1587 x_{3} & = & -401 \\ 0,126 x_{1}-\quad 0,049 x_{2}-\quad 3,956 x_{3} & = & 2,647 \\ 1,981 \cdot 10^{6} x_{1}+7,217 \cdot 10^{6} x_{2}+0,221 \cdot 10^{6} x_{3} & =0,512 \cdot 10^{5} \end{array} $$

Durch \(y_{1}=6 x_{1}+4 x_{2}-10 x_{3}, y_{2}=-9 x_{1}-6 x_{2}+15 x_{3}\) ist eine lineare Abbildung \(f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}\) gegeben. Berechne Kern \(f\) und Bild \(f\), d.h. gib f眉r beide R盲ume Basen an. Welche Werte haben dim Kern \(f\) und Rang \(f ?\) Rechne die Dimensionsformel (2.84) f眉r dieses Beispiel nach.
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