In my linear algebra notes I was asked to prove the following proposition:

Let $V$ and $W$ be two vectorial spaces over $mathbb{F}$, with $n=$dim$V$, $m=$dim$W$. If ${x_1,…,x_n}$ is a basis of $V$ and ${mathbf{y}_1,…,mathbf{y}_n}$ is a basis of $W$, then there is a unique matrix $Ain M_{mtimes n}(mathbb{F})$ such that

$$

begin{equation}

left.begin{aligned}

x&=c_1x_1+…+c_nx_n\

T(x)&=b_1mathbf{y}_1+…+b_mmathbf{y}_m

end{aligned}right}implies mathbf{b}=Ac

end{equation}

$$

**Proof**

By a proposition we previously cover in class the transformation $xto c$ is a linear isomorphism of $V$ in $mathbb{F}^n$. Now, define the matrix $A$ as follows: For all $j=1,2,…,n$ $T(x_j)$ can be expanded in terms of the basis of $W$ as $$T(x_j)=:sum_{i=1}^{m}a_{ij}mathbf{y}_i.$$

Then, by linearity of $T$, $$ sum_{i=1}^{m}b_imathbf{y}_i=T(x)=Tleft(sum_{j=1}^{n}c_jx_jright)$$

$$=sum_{j=1}^{n}c_jT(x_j)=sum_{j=1}^{n}c_jsum_{i=1}^{m}a_{ij}mathbf{y}_i$$

$$=sum_{i=1}^{m}sum_{j=1}^{n}a_{ij}c_jmathbf{y}_i,$$ and the uniqueness of the coefficients with respect to the basis ${mathbf{y}_1,…,mathbf{y}_m}$ of $W$ implies $$b_i=sum_{j=1}^{n}a_{ij}c_j$$ for $i=1,2,…,m$, and this is $mathbf{b}=Ac$.

On the other hand, when taking $c=mathbf{e}_j$ -an element of the standard basis of $mathbb{F}^n$– $mathbf{b}$ is the vector (in $mathbb{F}^m$) of coefficients of $T(x_j)$ with respect to the basis ${mathbf{y}_1,…,mathbf{y}_m}$, and the equation $mathbf{b}=Amathbf{e}_j$ says that $$sum_{i=1}^{m}a_{ij}mathbf{y}_i=T(x_j.)$$

Again, the uniqueness of the coefficients for the basis ${mathbf{y}_1,…,mathbf{y}_m}$ shows that the entries $a_{ij}$ of $A$ are determined by $T$.

Is this correct?