[Math] Show that two linear transformations are equal

linear algebra

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?

Best Answer

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

$(a_{ij}-b_{ij})=0 \forall i\in \{1,2,...,m\}\forall j\in \{1,2,...,n\}$

$\Rightarrow a_{ij}=b_{ij} \forall i\in \{1,2,...,m\}\forall j\in \{1,2,...,n\}$

$\Rightarrow A=B$

Related Question