Homomorphisms, kernel, basis I'm having trouble with the following problem. Let $V$ be a vector space of dimension $3$ and $\{v_1, v_2, v_3\}$ its basis. Let $W$ be a vector space of dimension $2$ and $\{w_1, w_2\}$ as its basis.
$f$ is defined as a transformation from $V$ to $W$ so that
$f(\lambda_1v_1+\lambda_2v_2+\lambda_3v_3)=
(\lambda_1+\mu)w_1+(\lambda_2+\lambda_3)w_2$
a) find the values of $\mu$ so that $f$ is linear
b) for the values in a), determine a basis of $\ker(f)$
We know a transformation is linear if and only if a) $f(u+w) = f(u) + f(w)$ and $f(ku) = kf(u)$. However, after developing both cases I have reached no particular conclusion. What now?
 A: Hint: It should be $f(0)=0$.
Alternative hint: it is not restrictive to assume that $V=\mathbb{R}^3$ and $W=\mathbb{R}^2$, the bases being the canonical ones. In case the map is linear, its associated matrix  would be
$$
\begin{bmatrix}
1+\mu & \mu & \mu \\
0     & 1   & 1
\end{bmatrix}
$$
But then
$$
f(\lambda_1v_1+\lambda_2v_2+\lambda_3v_3)=
\begin{bmatrix}
1+\mu & \mu & \mu \\
0     & 1   & 1
\end{bmatrix}
\begin{bmatrix}
\lambda_1\\
\lambda_2\\
\lambda_3
\end{bmatrix}=
\begin{bmatrix}
(1+\mu)\lambda_1+\mu\lambda_2+\mu\lambda_3\\
\lambda_2+\lambda_3
\end{bmatrix}
$$
which should be equal to $\begin{bmatrix}\lambda_1+\mu\\\lambda_2+\lambda_3\end{bmatrix}$ (what results from the actual definition of $f$) for all choices of $\lambda_1$, $\lambda_2$ and $\lambda_3$.
Now, what values of $\mu$ make
$$
\begin{bmatrix}
(1+\mu)\lambda_1+\mu\lambda_2+\mu\lambda_3\\
\lambda_2+\lambda_3
\end{bmatrix}
=
\begin{bmatrix}
\lambda_1+\mu\\
\lambda_2+\lambda_3
\end{bmatrix}
$$
for all $\lambda_1$, $\lambda_2$, $\lambda_3$? If you find such values, you'll also have that the function $f$ is linear (because it coincides with the action of the associated matrix).
