# linear algebra – Proving there is a unique matrix that satisfies certain condition

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?

Posted on Categories Articles