Basis of range of linear transformation

Dependencies:

Let $T: U \mapsto V$ be a linear transformation with kernel $K$ where $U$ has a finite basis of size $n$.

Let $P = \{u_1, u_2, \ldots, u_k\}$ be a basis of $K$. Since $P$ is a linearly independent subset of $U$, it can be expanded to form a basis of $U$ (this also explains why $P$ should be finite). Let $Q = \{u_{k+1}, \ldots, u_n\}$ such that $P \cup Q$ is a basis of $U$.

Then $T(Q)$ is a basis of $T(U)$. Consequently, $\operatorname{dim}(T(U)) = |Q| = n-k$.

Proof

\begin{align} & \sum_{i=k+1}^n a_iT(u_i) = 0 \\ &\Rightarrow T\left(\sum_{i=k+1}^n a_iu_i\right) = 0 \\ &\Rightarrow \sum_{i=k+1}^n a_iu_i \in K \\ &\Rightarrow \sum_{i=k+1}^n a_iu_i = \sum_{i=1}^k (-a_i)u_i \tag{for some $a_1, a_2, \ldots, a_k$} \\ &\Rightarrow \sum_{i=1}^n a_iu_i = 0 \\ &\Rightarrow a_i = 0 \forall i \end{align} Therefore, $T(Q)$ is linearly independent.

\begin{align} u \in U &\Rightarrow u = \sum_{i=1}^n a_iu_i \tag{for some $a_1, a_2, \ldots, a_n$} \\ &\Rightarrow T(u) = T\left(\sum_{i=1}^n a_iu_i \right) \\ &\Rightarrow T(u) = \sum_{i=1}^n a_iT(u_i) \\ &\Rightarrow T(u) = \sum_{i=k+1}^n a_iT(u_i) \\ &\Rightarrow T(u) \in \operatorname{span}(T(Q)) \end{align}

Therefore, $T(Q)$ spans $T(U)$. Therefore, $T(Q)$ is a basis of $T(U)$.

Dependency for: None

Info:

Depth: 8
Number of transitive dependencies: 50

Basis of range of linear transformation

Dependencies:

Proof

Dependency for: None

Info:

Transitive dependencies: