If B is a basis for $\Bbb{R}^n$ and $T$ and $T'$ are linear transformations mapping $\Bbb{R}^n$ into $\Bbb{R}^m$, then T(x) = T'(x)

Question

True or false?

If $B$ = {$\mathbf{b}_1, \mathbf{b}_2, ..., \mathbf{b}_n$} is a basis for $\Bbb{R}^n$ and $T$ and $T'$ are linear transformations mapping $\Bbb{R}^n$ into $\Bbb{R}^m$, then $T(\mathbf{x}$) = $T'(\mathbf{x}$) for all $\mathbf{x}$ $\in$ $\Bbb{R}^n$ if and only if $T(\mathbf{b}_i$) = $T'(\mathbf{b}_i$) for $i = 1, 2, ..., n$.

The answer is true, but I thought it false.

Reason: Since $\mathbf{x} \in \Bbb{R}^n$ and $B$ is a basis for $\Bbb{R}^n$, $T(\mathbf{x}$) = $T(\mathbf{r}_1\mathbf{b}_1 + \mathbf{r}_2\mathbf{b}_2 + ... + \mathbf{r}_n\mathbf{b}_n)$ = $\mathbf{r}_1T(\mathbf{b}_1) + \mathbf{r}_2T(\mathbf{b}_2) + ... + \mathbf{r}_nT(\mathbf{b}_n)$.

Similarly, $T'(\mathbf{x}$) = $T'(\mathbf{s}_1\mathbf{b}_1 + \mathbf{s}_2\mathbf{b}_2 + ... + \mathbf{s}_n\mathbf{b}_n)$ = $\mathbf{s}_1T'(\mathbf{b}_1) + \mathbf{s}_2T'(\mathbf{b}_2) + ... + \mathbf{s}_nT'(\mathbf{b}_n)$.

In this case, $T(\mathbf{b}_i$) needs not equal $T'(\mathbf{b}_i$) for $i = 1, 2, ..., n$, since we need only find proper $\mathbf{r}_1, \mathbf{r}_2, ... , \mathbf{r}_n$; and $\mathbf{s}_1, \mathbf{s}_2, ... , \mathbf{s}_n$, such that $T(\mathbf{x}$) = $T'(\mathbf{x}$).

Where does my reasoning go wrong?

Answer 1

If $B = \{b_1, \ldots, b_n\}$ is a basis for $\mathbb{R}^n$, then given $x \in \mathbb{R}^n$, there is only one way to write $x$ as a linear combination of the $b_i$.

Indeed, if $$x = r_1b_1 + \cdots + r_nb_n= s_1b_1 + \cdots + s_nb_n,$$ then $$(r_1-s_1)b_1 + \cdots + (r_n-s_n)b_n = 0.$$ As $\{b_1, \ldots, b_n\}$ is linearly independent, we have that $r_i - s_i = 0$, $i = 1, \ldots, n$, so your argument shows us that $T$ and $T'$ must be the same.

Answer 2

A different, easier (in my opinion) way to see it is that the only if part, i.e.,

$$ T(x) = T'(x) \forall x \implies T(b_i) = T'(b_i) \forall i $$ , is trivial: just set $x = b_i$ for each $i$.