# Let $a=(1,0,3)$ $b=(0,2,0)$ $c^*=(1,2,3)$ be elements of $R^3$ and let $x=(1,3)$ $y=(0,2)$ $z=(0,1)$ $z^*=(1,5)$ be elements of $R^2$

Let $a=(1,0,3)$ $b=(0,2,0)$ $c^*=(1,2,3)$ be elements of $\mathbb R^3$ and let $x=(1,3)$ $y=(0,2)$ $z=(0,1)$ $z^*=(1,5)$ be elements of $\mathbb R^2$. Do there exist linear maps which satisfy

i) $f:a\mapsto x,b\mapsto y,c^*\mapsto z$

ii) $f:a\mapsto x,b\mapsto y,c^*\mapsto z^*$

In each case state if the linear map is unique giving reasons

for ii) the map $(a,b+c)$ work but i'm not sure if it's unique or not and I can't find a map for i) so I'm not sure if there is one or not.

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Note that $a$ and $b$ are linear independent and $c^* = a+b$. A linear map must therefore fulfill $f(c^*) = f(a) + f(b)$. As you already noted, (i) does not, but two does. For the uniqueness, recall that when you know $f$ on $a$ and $b$ you know $f$ on $\mathrm{span}\{a,b\}$. As this span is two-dimensional, it doesn't equal the whole of $\mathbb R^3$. Let's take $c = (0,0,1) \not\in\mathrm{span}\{a,b\}$. Then there are linear maps with
(ii) $f \colon a\mapsto x, b \mapsto y, c \mapsto 0$,
(ii') $f \colon a \mapsto x, b \mapsto y, c \mapsto x$.
So $f$ in (ii) isn't unique.