Suppose v1, …, vm is a list of vectors in V. Define T ∈ℒ(𝐅^m, V) by T(z1, …, zm)=z1 v1+⋯+zm vm (

Name: Problem 3, Chapter 3: Linear Algebra Done Right
Uploaded: 2023-09-13T23:13:55-08:00
Channel: Victor Salazar
Description: Suppose v1, …, vm is a list of vectors in V. Define T ∈ℒ(𝐅^m, V) by T(z1, …, zm)=z1 v1+⋯+zm vm (a) What property of T corresponds to v1, …, vm spanning V ? (b) What property of T corresponds to v1, …, vm being linearly independent?

step 1 For this exercise we have that if vector v sub 1 and vector v sub 2 is an element, if our space

Question

Suppose $v_{1}, \ldots, v_{m}$ is a list of vectors in $V$. Define $T \in \mathcal{L}\left(\mathbf{F}^{m}, V\right)$ by $$T\left(z_{1}, \ldots, z_{m}\right)=z_{1} v_{1}+\cdots+z_{m} v_{m}$$ (a) What property of $T$ corresponds to $v_{1}, \ldots, v_{m}$ spanning $V ?$ (b) What property of $T$ corresponds to $v_{1}, \ldots, v_{m}$ being linearly independent?

   Suppose $v_{1}, \ldots, v_{m}$ is a list of vectors in $V$. Define $T \in \mathcal{L}\left(\mathbf{F}^{m}, V\right)$ by $$T\left(z_{1}, \ldots, z_{m}\right)=z_{1} v_{1}+\cdots+z_{m} v_{m}$$ 
(a) What property of $T$ corresponds to $v_{1}, \ldots, v_{m}$ spanning $V ?$
(b) What property of $T$ corresponds to $v_{1}, \ldots, v_{m}$ being linearly independent?

Linear Algebra Done Right

Sheldon Axler 3rd Edition

Chapter 3, Problem 3

Instant Answer

Solved by Expert Victor Salazar

09/13/2023

Step 1

The transformation $T: \mathbf{F}^m \to V$ is defined by $T(z_1, \ldots, z_m) = z_1 v_1 + \cdots + z_m v_m$. This means $T$ takes a vector of scalars $(z_1, \ldots, z_m)$ from $\mathbf{F}^m$ and outputs a linear combination of the vectors $v_1, \ldots, v_m$ in Show more…

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Suppose $v_{1}, \ldots, v_{m}$ is a list of vectors in $V$. Define $T \in \mathcal{L}\left(\mathbf{F}^{m}, V\right)$ by $$T\left(z_{1}, \ldots, z_{m}\right)=z_{1} v_{1}+\cdots+z_{m} v_{m}$$ (a) What property of $T$ corresponds to $v_{1}, \ldots, v_{m}$ spanning $V ?$ (b) What property of $T$ corresponds to $v_{1}, \ldots, v_{m}$ being linearly independent?

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Key Concepts

Surjectivity

A set of vectors spans a vector space if every element of the space can be expressed as a linear combination of those vectors. In the context of the linear map T defined by these vectors, T is surjective if its range is the entire space V. Thus, the surjectivity of T is directly related to the spanning property: if T is surjective, then the vectors span V.

Injectivity

A set of vectors is linearly independent if the only solution to the linear combination equaling the zero vector is the trivial solution. For the linear map T, this means that if T(z1, ..., zm) = 0 only when all the zi are zero, then T is injective. Therefore, the injectivity of T corresponds to the linear independence of the list of vectors.

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