Is the following Proof Correct?

Theorem. Given that both $V$ and $W$ are finite-dimensional. There exists a injective linear map from $V$ to $W$ if and only if $\dim V\leq \dim W$

Proof. $(\Rightarrow).$ Assume for the purpose of contradiction that $T:V\to W$ is an injective linear map and that $\dim V>\dim W$ thus by making use of the Fundamental Theorem of Linear Maps we see that $$\dim \operatorname{null}\ T = \dim V-\dim \operatorname{range}\ T$$ $$\dim \operatorname{null}\ T \ge \dim V-\dim W$$ $$\dim \operatorname{null}\ T > 0$$ but $T$ is injective and thus $\operatorname{null}T = \{0\}$ implying that $\dim \operatorname{null}\ T=0$ resulting in a contradiction.

$(\Leftarrow).$ Assume now that $\dim V\leq\dim W$ since both $V$ and $W$ are finite dimensional we may invoke the existence of the vectors $v_1,v_2,...,v_m\in V$ and $w_1,w_2,...,w_n\in W$ such that they act as basis for $V$ and $W$ respectively, where $m\leq n$.

Consider now the linear map $T\colon V\to W$ defined as follows. $$\forall j\in\{1,2,...,m\}(Tv_j = w_j)$$ Let $u_1$ and $u_2$ be arbitrary members of $V$ and assume that $Tu_1=Tu_2$, in light of $(1)$ we may express $Tu_1$ and $Tu_2$ as follows $$Tv_1 = T\left(\sum_{j=1}^{m}a_jv_j\right)=\sum_{j=1}^{m}a_jTv_j=\sum_{j=1}^{m}a_jw_j\tag{2}$$ $$Tv_2 = T\left(\sum_{j=1}^{m}b_jv_j\right)=\sum_{j=1}^{m}b_jTv_j=\sum_{j=1}^{m}b_jw_j\tag{3}$$ subtracting $(2)$ from $(3)$ we see that $$0=\sum_{j=1}^{m}(b_j-a_j)w_j\tag{4}$$ we know that $w_1,w_2,...,w_n$ is a linearly independent list in $W$ consequently $w_1,w_2,...,w_m$ is also linearly independent, it then follows that $$\forall j\in\{1,2,...,m\}(a_j=b_j)\tag{5}$$ consequently $u_1=u_2$ as every vector in $V$ is uniquely expressed as a linear combination of $v_1,v_2,...,v_m$.



1 Answer 1


Yes, it is correct. A few remarks:

  • The expression “Fundamental Theorem of Linear Maps” is not a standard one. The usual expression is “rank–nullity theorem”.
  • In order to prove that the map $T$ that you defined is indeed injective, it is easier to observe that$$\begin{align*}T\left(\sum_{j=1}^ma_jv_j\right)=0&\iff \sum_{j=1}^ma_jw_j=0\\&\iff\bigl(\forall j\in\{1,2,\ldots,m\}\bigr):a_j=0,\end{align*}$$since $\{w_1,w_2,\ldots,w_m\}$ is a subset of a basis.

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