Question

\(\begin{array}{lll}\text { Exercise } & 7.3 .24 & \text { Let } V \text { consist of all sequences }\end{array}\) \(\left[x_{0}, x_{1}, x_{2}, \ldots\right)\) of numbers, and define vector operations $$ \begin{aligned} \left[x_{o}, x_{1}, \ldots\right)+\left[y_{0}, y_{1}, \ldots\right) &=\left[x_{0}+y_{0}, x_{1}+y_{1}, \ldots\right) \\ r\left[x_{0}, x_{1}, \ldots\right) &=\left[r x_{0}, r x_{1}, \ldots\right) \end{aligned} $$ a. Show that \(V\) is a vector space of infinite dimension. b. Define \(T: V \rightarrow V\) and \(S: V \rightarrow V\) by \(T\left[x_{0}, x_{1}, \ldots\right)=\left[x_{1}, x_{2}, \ldots\right)\) and \(S\left[x_{0}, x_{1}, \ldots\right)=\left[0, x_{0}, x_{1}, \ldots\right) .\) Show that \(T S=1_{V},\) so \(T S\) is one-to-one and onto, but that \(T\) is not one-to-one and \(S\) is not onto.

Short answer

V is an infinite dimensional vector space; \(TS\) is identity but \(T\) is not one-to-one and \(S\) is not onto.

Step-by-step solution

Understand the Vector Space

The set V consists of all sequences \([x_0, x_1, x_2, \ldots]\). We have two operations defined on V: addition of sequences and scalar multiplication. These operations allow us to add two sequences element-wise and multiply a sequence by a scalar. We need to verify that these operations satisfy the axioms of a vector space.

Verify Vector Space Axioms

Check the axioms:1. Closure under addition and scalar multiplication: For sequences \([x_0, x_1, \ldots]\) and \([y_0, y_1, \ldots]\), addition \([x_0+y_0, x_1+y_1, \ldots]\) and scalar multiplication \(r[x_0, x_1, \ldots] = [rx_0, rx_1, \ldots]\) are in V.2. Associativity, commutativity, identity, and inverse elements for addition.3. Distributive properties for scalar multiplication. Since V consists of all sequences, it satisfies these axioms, making V a vector space.

Show V is Infinite Dimensional

A basis in V can be given by sequences that have a 1 in one position and 0 in all other positions, such as \([1, 0, 0, \ldots]\), \([0, 1, 0, \ldots]\), etc. There are infinitely many such sequences, showing that V is infinite-dimensional.

Define and Analyze T

The linear operator \(T\) shifts the sequence to the left: \(T[x_0, x_1, x_2, \ldots] = [x_1, x_2, \ldots]\). It is not one-to-one, because different sequences can map to the same sequence after dropping the first element, such as \(T[x_0, x_1, \ldots] = T[y_0, y_1, \ldots]\) if \(x_1 = y_1\), etc.

Define and Analyze S

The linear operator \(S\) shifts the sequence to the right and adds 0 at the first position: \(S[x_0, x_1, \ldots] = [0, x_0, x_1, \ldots]\). It is not onto because there is no sequence \( [x_0, x_1, \ldots] \) in V such that \( S([x_0, x_1, \ldots]) = [1, x_0, x_1, \ldots] \).

Demonstrate TS = 1_V

Calculate \(TS\):\[TS[x_0, x_1, x_2, \ldots] = T[0, x_0, x_1, \ldots] = [x_0, x_1, \ldots]\]Here, \(TS\) returns the original sequence, showing that \(TS = 1_V\), the identity transformation. This means \(TS\) is both one-to-one and onto.

Key concepts

Infinite Dimensional

In mathematics, a vector space is said to be infinite-dimensional if it has an infinite basis. A basis is a set of vectors that are linearly independent and span the vector space. In the case of the vector space \(V\) from the exercise, \(V\) consists of all possible sequences of numbers \([x_0, x_1, x_2, \ldots]\). Since every sequence where one position is 1 and the rest are zero can be considered an independent vector, the potential basis for \(V\) is infinite.

• For example, the sequences \([1, 0, 0, \ldots]\), \([0, 1, 0, \ldots]\), \([0, 0, 1, \ldots]\), and so on, all have a 1 at different single positions.
• These sequences can span the entire space \(V\), meaning they can produce any sequence through linear combination.

Therefore, \(V\) is infinite-dimensional because it requires an infinite number of vectors to span the space.

Linear Operator

A linear operator is a function between two vector spaces that preserves vector addition and scalar multiplication. The exercise defines two operators on the vector space \(V\): operators \(T\) and \(S\).
- **Operator \(T\)** shifts each sequence left, effectively removing the first element. - This makes \(T\) not one-to-one as different sequences can map to the same sequence after the removal.- **Operator \(S\)** shifts each sequence right, adding a zero at the start of the sequence. - It is not onto because no sequence can begin with any number other than zero after applying \(S\).
These properties illustrate how linear operators can transform sequences, and the composite operator \(TS\), when applied to a sequence, results in the original sequence, showing \(TS\) is a special identity operator that is both one-to-one and onto.

Vector Operations

Vector operations in a vector space involve addition and scalar multiplication. For the sequences in \(V\), this means:

• **Addition:** Adding two sequences element-wise. For example, \([x_0, x_1, \ldots] + [y_0, y_1, \ldots] = [x_0 + y_0, x_1 + y_1, \ldots]\).
• **Scalar multiplication:** Multiplying each element of a sequence by a scalar. For instance, for scalar \(r\), \(r[x_0, x_1, \ldots] = [rx_0, rx_1, \ldots]\).

These operations are fundamental to using any vector space since they allow for constructing new sequences through combination and scaling. Performing these vector operations within a defined structure like \(V\) ensures adherence to the vector space axioms, like distributivity and associativity.

Vector Space Axioms

A vector space must satisfy a specific set of axioms that include closure under addition and scalar multiplication, existence of the additive identity, and existence of additive inverses, among others.
- **Closure under operations:** For two sequences in \(V\), their sum and any scalar multiplication yield another sequence in \(V\).- **Associativity and commutativity:** These properties ensure that sequence addition behaves logically: \((u + v) + w = u + (v + w)\) and \(u + v = v + u\).- **Identity elements:** For sequences, the zero sequence \([0, 0, \ldots]\) acts as the additive identity, while for scalars, multiplying by 1 preserves the sequence.- **Distributive properties:** ensure a scalar can be distributed over vector addition, and vice versa, such as \(r(u + v) = ru + rv\).
These axioms are essential as they provide the foundation for working within vector spaces, ensuring consistency and predictability when performing vector operations.