)
Lecture 3
Theorem. Let $\sf T$ be a linear transformation from $\sf V$ to $\sf W$, then $\dim(\mathsf{V})=\text{nullity}(\mathsf{T})+\text{rank}(\mathsf{T})$.
Corollary. For any $y\in \mathsf{V}$, $y+\text{ker}(\mathsf{T})$ yields ${x:x\in \mathsf{V}, \mathsf{T}(x)=\mathsf{T}(y)}$. Therefore, for any basis $\beta={a_1,\dots,a_n}$ of $\text{ker}(\mathsf{T})$ and any basis $\gamma={b_1,\dots,b_m}$ of $\text{Im}(\mathsf{T})$, Assume $S={s_1,\dots,s_m},\mathsf{T}(s_i)=b_i$, then ${a_1,\dots,a_n,s_1,\dots,s_m}$ is a basis of $\mathsf{V}$. $\mathsf{V}$ is the direct sum of $\text{ker}(\mathsf{T})$ and $S$.
Lecture 4
$[\mathsf{T}]\beta^\gamma=([\mathsf{T}(v_1)]\gamma,[\mathsf{T}(v_2)]\gamma,\dots,[\mathsf{T}(v_n)]\gamma)$.
Theorem. The following four statements are all equivalent:
$\sf T$ is invertible, $\mathsf{TU=I_W}$ and $\mathsf{UT=I_V}$, $\sf T$ is one-to-one and onto, $\dim(\mathsf{V})=\dim(\mathsf{W})$ and $\mathsf{UT=I_V}$.
Theorem. $\sf T$ is invertible if and only if $[\mathsf{T}]_\beta^\gamma$ is invertible.
Lecture 5
The structure of the dual space (the double dual as well) may be complex for beginners. When considering the dual space, it's better to only consider the dual basis of a basis of the original vector space and see other elements in the dual space as linear combinations of such basis.