Q1

$$ T(0,0,0) = (b,0) $$

We know additive identities have to map, so $b=0$.

By additivity:

$$ \begin{align*}&T(1,1,1) + T(1,1,1) &&= T(2,2,2) \\ &(1, 6+c) + (1, 6+c) &&= (2, 12+8c) \\\end{align*} $$

Take bottom equation:

$$ 12+ 2c = 12+8c \implies c=0 $$

Therefore, linear map only if $b=c=0$. To prove if, just go for additivity and homogeneity.

Q2

$T(0) = (0,0)$, so that doesn’t help.

$$ T(x) = (2b+17, \dots) \\ T(2x) = (8b + 34, \dots) $$

Double the first and, by homogeneity, they must be equal, so $b=0$. To prove $c=0$, do $T(1)$ and $T(2)$. As $\sin$ is not additive, the results are not either, so $c=0$. Proving if is the same as above.

Q3

Not quite sure what it wants. Just an example? Can we not just say $A_{j,k} = 0$?

Q4

Suppose $v_1,\dots,v_m$ is linearly dependent in $V$. Then there exist $c_1,\dots,c_m \in F$ such that

$$ \begin{align*}c_1v_1 + \dots + c_mv_m &= 0\\ T(c_1v_1 + \dots + c_mv_m) &= T(0)\\ c_1(Tv_1) + \dots + c_m(Tv_m) &= 0\end{align*} $$

Which implies that $Tv_1,\dots,Tv_m$ are also linearly dependent. As we know they are not, that meant $v_1,\dots,v_m$ are linearly independent.

Q5/6

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