Suppose V is finite-dimensional with dim V > 0, and suppose W is infinite-dimensional. Prove that \(\mathcal{L}(V, W)\) is infinite-dimensional.

Understand the Problem
This question involves proving a statement in linear algebra. Specifically, given that vector space V is finite-dimensional with dimension greater than 0, and vector space W is infinite-dimensional, the question asks to prove that the space of all linear transformations from V to W, denoted as (\mathcal{L}(V, W)), is infinite-dimensional.
Answer
Since we can find an infinite set of linearly independent transformations in $\mathcal{L}(V, W)$, it follows that $\mathcal{L}(V, W)$ is infinite-dimensional.
Answer for screen readers
Since we can find an infinite set of linearly independent transformations in $\mathcal{L}(V, W)$, it follows that $\mathcal{L}(V, W)$ is infinite-dimensional.
Steps to Solve
- Establish a Basis for V
Since $V$ is finite-dimensional and $\dim V > 0$, let's say $\dim V = n$ where $n$ is a positive integer. This means $V$ has a basis. Let ${v_1, v_2, ..., v_n}$ be a basis for $V$.
- Use a Basis Vector and Define Linear Transformations
Consider one of the basis vectors of $V$, say $v_1$. Now, since $W$ is infinite-dimensional, we can find an infinite set of linearly independent vectors in $W$. Let ${w_1, w_2, w_3, ...}$ be a set of linearly independent vectors in $W$. For each $w_i$ in this set, define a linear transformation $T_i: V \to W$ as follows:
$T_i(v_1) = w_i$ $T_i(v_j) = 0$ for $j \neq 1$
Since ${v_1, v_2, ..., v_n}$ is a basis for $V$, a linear transformation from $V$ to any vector space is completely determined by its action on the basis vectors. Thus, each $T_i$ is a well-defined linear transformation.
- Show Linear Independence of Transformations
We want to prove that the set of linear transformations ${T_1, T_2, T_3, ...}$ is linearly independent in $\mathcal{L}(V, W)$. To do this, consider a finite linear combination of these transformations that equals the zero transformation:
$$a_1T_1 + a_2T_2 + ... + a_kT_k = 0$$
where $a_i$ are scalars and $0$ denotes the zero transformation.
- Apply the Linear Combination to a Basis Vector
Apply this linear combination to the basis vector $v_1$:
$$(a_1T_1 + a_2T_2 + ... + a_kT_k)(v_1) = 0(v_1) = 0$$
Using the linearity of the transformations, we have:
$$a_1T_1(v_1) + a_2T_2(v_1) + ... + a_kT_k(v_1) = 0$$
Substituting $T_i(v_1) = w_i$, we get:
$$a_1w_1 + a_2w_2 + ... + a_kw_k = 0$$
- Conclude Linear Independence of Coefficients
Since $w_1, w_2, ..., w_k$ are linearly independent vectors in $W$, the only way for this linear combination to equal zero is if all the coefficients are zero:
$$a_1 = a_2 = ... = a_k = 0$$
This shows that the linear transformations $T_1, T_2, ..., T_k$ are linearly independent in $\mathcal{L}(V, W)$. Since we can find an infinite set of linearly independent transformations in $\mathcal{L}(V, W)$, it follows that $\mathcal{L}(V, W)$ is infinite-dimensional.
Since we can find an infinite set of linearly independent transformations in $\mathcal{L}(V, W)$, it follows that $\mathcal{L}(V, W)$ is infinite-dimensional.
More Information
The proof demonstrates that the infinite-dimensionality of $W$ is "inherited" by the space of linear transformations $\mathcal{L}(V, W)$ when $V$ has a non-zero dimension.
Tips
A common mistake is to try to prove this directly by assuming $\mathcal{L}(V, W)$ is finite-dimensional and arriving at a contradiction. The direct proof as showcased above is more straightforward. Another mistake may arise in not properly showing the linear independence of the constructed transformations.
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