Show that two linear transformations are equal Let $\{v_1, v_2,....,v_n\}$ be the standard basis for $\mathbb R^n$.Prove for any two $m\times n$ matrices that their linear transformations are equal if and only if the two matrices are equal. I know what two linear transformations need to be equal (same basis, domain and codomain), but how do I show that?
 A: Let $\mathbb B_1=\{v_1, v_2,....,v_n\}$ be the standard basis for $\mathbb R^n$.
As you are interested in $m\times n$ matrix consider $\mathbb B_2=\{u_1, u_2,....,u_m\}$ as the standard basis for $\mathbb R^m$.
Say $T$ and $S$ be two linear transformation from $\mathbb R^n$ to $\mathbb R^m$.
First assume $[T]_{\mathbb B_1}^{\mathbb B_2}=[S]_{\mathbb B_1}^{\mathbb B_2}=(a_{ij})_{m\times n}$
So $T(v_j)=\displaystyle \sum_{i=1}^ma_{ij}u_i=S(v_j)\forall j\in \{1,2,...,n\}$
Let $x\in \mathbb R^n$ Then $x=\displaystyle \sum_{i=1}^nc_jv_j$  for some scalars $c_j$
So $T(x)=T(\displaystyle \sum_{i=1}^nc_jv_j)=\displaystyle \sum_{i=1}^nc_jT(v_j)=\displaystyle \sum_{i=1}^nc_j\displaystyle \sum_{i=1}^ma_{ij}u_i=\displaystyle \sum_{i=1}^nc_jS(v_j)=S(\displaystyle \sum_{i=1}^nc_jv_j)=S(x)$
Hence $T=S$
Conversely assume $A=(a_{ij})_{m\times n}$ and $B=(b_{ij})_{m\times n}$ are two $m\times n$ matrices (wrt bases $\mathbb B_1$ and $\mathbb B_2$) corresponding to two linear transformations $T$ and $S$ from $\mathbb R^n$ to $\mathbb R^m$ such that $T=S$.
So $[T]_{\mathbb B_1}^{\mathbb B_2}=(a_{ij})_{m\times n}$ and $[S]_{\mathbb B_1}^{\mathbb B_2}=(b_{ij})_{m\times n}$
$\Rightarrow T(v_j)=\displaystyle \sum_{i=1}^ma_{ij}u_i$ and $S(v_j)=\displaystyle \sum_{i=1}^mb_{ij}u_i \forall j\in \{1,2,...,n\}$
Note that any linear transformation is completely determined by the image of basis elements.So here $T(v_j)=S(v_j)\forall j\in \{1,2,...,n\}$ as $T=S$
Thus $T(v_j)=\displaystyle \sum_{i=1}^ma_{ij}u_i=S(v_j)=\displaystyle \sum_{i=1}^mb_{ij}u_i \forall j\in \{1,2,...,n\}$
$\Rightarrow \displaystyle \sum_{i=1}^m(a_{ij}-b_{ij})u_i=0 \forall j\in \{1,2,...,n\}$
Since $B_2=\{u_1, u_2,....,u_m\}$ is a basis hence it i linearly independent.So we have from the previous line 
$(a_{ij}-b_{ij})=0 \forall i\in \{1,2,...,m\}\forall j\in \{1,2,...,n\}$
$\Rightarrow a_{ij}=b_{ij} \forall i\in \{1,2,...,m\}\forall j\in \{1,2,...,n\}$
$\Rightarrow A=B$
A: Hint: $T(v_{i})$ is encoded into the matrices
A: You need to make two observations: (1) A linear transformation is completely determined by its values over basis elements. (2) The columns of the associated matrix to a linear transformation are precisely the values of the transformation at basis elements.
