# Relations between Linear Transformations and their Matrices

I am having trouble with the following problem and would greatly appreciate any help.

Let $V$ be be a finite dimensional vector space of dimension $n$ with basis $\mathcal{B}$. Let $\mathcal{L}(V)$ be the vector space of linear transformations $T:V\rightarrow V$, and if $T \in \mathcal{L}(V)$, then let $[T]_\mathcal{B}$ denote the matrix of $T$ relative to the basis $\mathcal{B}$.

a. Prove that the function $[\cdot]_\mathcal{B}: \mathcal{L}(V) \rightarrow M_{n \times n}(\mathbb{R}),$ given by $T\rightarrow [T]_\mathcal{B}$, is a (vector space) isomorphism.

b. Let $T \in \mathcal{L}(V).$ Prove that $T$ is invertible if and only if $[T]_\mathcal{B}$ is invertible.

c. Let $T \in \mathcal{L}(V).$ Prove that $\dim(\ker T) = \dim(\text{Null}([T]_\mathcal{B})).$

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Why are you having trouble? What have you tried? –  Lepidopterist Mar 19 '13 at 18:26
I get why it is true but do not know where to start in proving this. –  Anna Mar 19 '13 at 18:42
Why is it true? Explain your intuition in your question. –  Lepidopterist Mar 19 '13 at 18:44
i can do all but c. no clue where to start on that one. –  Anna Mar 19 '13 at 18:50

Suppose that $A$ is an $m\times n$ matrix. Then we can see that $A$ may be viewed as a linear transformation $T_A:\mathbb{C}^n\to\mathbb{C}^m$: $$T_A:\vec{x}\mapsto A\vec{x},$$since $A(\vec{x}+\vec{y})=A\vec{x}+A\vec{y}$ and $A(\lambda\vec{x})=\lambda A\vec{x}$.
Likewise, given a linear transformation $T:\mathbb{C}^n\to\mathbb{C}^m$, construct $A_T$ by letting the i-th column $a_i:=T(\vec{e}_i)$. For a given $\vec{x}\in\mathbb{R}^n$, we may write $$\vec{x}=\lambda_1\vec{e}_1+\cdots+\lambda_n\vec{e}_n=\begin{pmatrix}\lambda_1\\ \vdots\\\lambda_n\end{pmatrix}.$$
Now by the linearity of $T$ we see that $$T(\vec{x})=T(\lambda_1\vec{e}_1+\cdots+\lambda_n\vec{e}_n)=\lambda_1T(\vec{e}_1)+\cdots+\lambda(\vec{e}_n)=A_T\vec{x}.$$
Now c) should really follow quite easily from this. Show that the kernel of $T$ is equal to the nullspace of $A_T$ directly.