In Axler's Linear Algebra Done Right, there is this theorem. (3.5)

Suppose $v_1. . . v_n $ is a basis of $V$ and $w_ , . . . w_n \in W$. Then there exists a unique linear map $T: V \rightarrow W$ such that $$T (v_j) = w_j$$ for each $j\in 1, . . . , n $.

I understood the first part of the proof proving existence of such a transformation, but didn't understand the uniqueness part of the proof.

To prove uniqueness, now suppose that $T \in \cal L $$(L,V)$;and that $T( v_j)= w_j$ for each $j\in 1, . . . , n $.

Let $c_1,. . . ,c_n \in F$.

The homogeneity of $T$ implies that $T(c_j v_j) = c_jw_j$ for each $j\in 1, . . . , n $.

The additivity of T now implies that $T(c_1v_1 + . . . + c_nv_n) = c_1w_1 + . . . + c_nw_n$.

Thus $T$ is uniquely determined on span($v_1, . . . ,v_n$) by the equation above. Because $v_1, . . . v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$.

How does "equation above" show that $T$ is uniquely determined on span($v_1, . . . ,v_n$)?

Is it because there is one way to get each of the basis vectors using the equation? How do we know there isn't another transformation that will work?


By the way, the proof given above does indeed show that $T$ is unique. Perhaps a different phrasing of the argument might help you understand the uniqueness.

You said you understood the existence part, so ok let $T: V \to W$ be the linear map you constructed, which satisfies $T(v_j) = w_j$ for all $j \in \{1, \dots, n\}$. Suppose now that there is a linear map $S: V \to W$ such that $S(v_j) = w_j$ for all $j \in \{1, \dots, n\}$. We have to show that $T=S$; i.e we have to show that for every $x \in V$, $T(x) = S(x)$.

To prove this, pick any $x \in V$. Since $\{v_1, \dots, v_n\}$ is a basis for $V$, there exist (unique) scalars $c_1, \dots, c_n \in F$ such that \begin{align} x = \sum_{i=1}^n c_iv_i \tag{*} \end{align} Now, we perform a simple computation: \begin{align} T(x) &= T\left( \sum_{i=1}^n c_iv_i \right) \tag{by (*)} \\ &= \sum_{i=1}^n c_i T(v_i) \tag{$T$ is linear by construction} \\ &= \sum_{i=1}^n c_i w_i \tag{by definition of $T$} \\ &= \sum_{i=1}^n c_i S(v_i) \tag{by assumption on $S$} \\ &= S\left( \sum_{i=1}^n c_i v_i \right) \tag{$S$ linear by assumption} \\ &= S(x) \tag{by (*)} \end{align} So we have shown that for every $x \in V$, $T(x) = S(x)$. Hence, $T=S$, proving uniqueness of the original $T$ you constructed.

This result is often stated as "a linear transformation is specified by its values on a basis" or something to that effect.

  • $\begingroup$ This proof makes a lot more sense because it actually follows the logic of a uniqueness proof the likes of which most of us are used to (by constructing a hypothetical and proving equality by properties). Thank you for this! $\endgroup$ – sqrtpapi2001 Jul 12 '20 at 4:44
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    $\begingroup$ @sqrtpapi2001 I'm glad this was helpful. However notice that (after the "Now, we perform a simple computation:") the first three equalities do actually prove uniqueness. Why? because it tells you exactly how to calculate $T(x)$, and the third equality shows that $T(x)$ depends only on $x$ and $w_1, \dots, w_n$. The last $3$ equalities are simply "undoing" everything (notice how "symmetric" the equalities are). Hence, in my first sentence I said that the proof given does actually prove uniqueness, but it is perhaps just not as obvious (though with some practice it does become obvious). $\endgroup$ – peek-a-boo Jul 12 '20 at 4:57
  • $\begingroup$ I see. But the proof given still feels incomplete. Are you saying to fill in the missing ("undoing") logic from practice? The proof given says the inequality "just does" without showing how in the same way you're doing which makes it a lot clearer. $\endgroup$ – sqrtpapi2001 Jul 12 '20 at 5:13
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    $\begingroup$ @sqrtpapi2001 I completely sympathize, because I recall when I first learnt this theorem, the proof presented to me also seemed incomplete in the uniqueness part. But, what I'm saying is that with some more practice/exposure to proofs, it becomes very easy to "fill-in" the last 3 steps (which are literally just "undoing" everything done before), so much so that the statement (and proof) becomes "obvious". Also, with some more practice, what happens is that you usually become ok with providing fewer details, because you're just more comfortable making logical leaps. $\endgroup$ – peek-a-boo Jul 12 '20 at 5:20
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    $\begingroup$ Of course, to get to that stage, you have to go through the initial stages where you have to strive to understand (almost/ as much as reasonably possible) every single minute detail. $\endgroup$ – peek-a-boo Jul 12 '20 at 5:21

Because the vectors $\vec{w}_i$ form a basis for $W$, any two different linear combinations of the vectors $\vec{w}_i$ must yield a different result. Let $\vec{w} = c_1 \vec{w}_1 + \cdots + c_n \vec{w}_n$, with all $\vec{c}_i \in \mathbb{R}$. By construction, the transformation $T(c_1 \vec{v}_1 + \cdots + c_n \vec{v}_n) = c_1 \vec{w}_1 + \cdots + c_n \vec{w}_n$. But as mentioned before, any linear combination of the $\vec{w}_i$ must be unique, and thus, there can be no numbers $d_1, d_2, \cdots, d_n$ such that $T(d_1 \vec{v}_1 + \cdots + d_n \vec{v}_n) = d_1 \vec{w}_1 + \cdots + d_n \vec{w}_n = c_1 \vec{w}_1 + \cdots + c_n \vec{w}_n$ without $d_i = c_i$ for all $i$.

That proves that if $T(\vec{a}) = \vec{w} = T(\vec{b})$, $\vec{a} = \vec{b}$. Now comes the part where the $\vec{v}_i$'s form a basis. This means that any $\vec{v} \in V$ must also be a unique linear combination of the $\vec{v}_i$'s. In other words, if $c_1 \vec{v}_1 + \cdots + c_n \vec{v}_n = \vec{v} = d_1 \vec{v}_1 + \cdots + d_n \vec{v}_n$, we also must have $c_i = d_i$ for all $i$. So the transformation $T$ must be unique: in other words, for every $\vec{w} \in W$, there exists one and only one $\vec{v} \in V$ such that $\vec{v} \in V$. Furthermore, this $\vec{v}$ must be unique. So we say that $T:V \to W$ is uniquely determined on $V$.


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