Help with a Linear Algebra exercise

Question

Let $V\neq\{0_{V}\}$ be a finite dimensional vector space over a field $K$. Let $f\in\mathrm{End}(V)$. Suppose that every subspace of $V$ is be $f$-invariant. Then, prove that $f = a\cdot id_V$ for some $a\in K$, where $id_{V}$ is the identity map from $V$ to itself.

For the proof I have following ideas in my mind but have problems to finalize it or maybe it’s the wrong idea. Appreciate your support:

Let $(v_1,v_2,...,v_n)$ be an ordered basis of $V$, then

$$f(v_1)=c_1 v_1\ ,\,\ldots\ ,\ f(v_n)=c_n v_n$$

Further each vector v can be written as

$$v=\sum_{i} a_i v_i,\,a_{i}\in K,\forall i$$ so, we have $$f(v)=f\left(\sum a_i v_i\right)=\sum_{i} a_i f(v_i)=\sum_{i}a_i c_i v_i = c v$$

That means $\sum_{i} a_i c_i v_i - c v = 0. $ This is true if $c_i=c$, for all $i$.

Answer 1

$$f (v_1+v_2)=c (v_1+v_2) $$ $$=f (v_1)+f (v_2)=c_1v_1+c_2v_2$$

thus $$c_1=c_2=c $$ the matrix will be $c I $.

Answer 2

Let $B = (b_1, b_2,\dots, b_n)$ be a basis of $V$.
Let A be the matrix of $\ f$.
We can observe that it must be diagonal, because $(\forall i)(\exists c_i)\ Ab_i=c_ib_i$.
Next, we have that for all $v \in V$ subspace generated by $v$ is $\ f$-invariant so $(\forall v)(\exists c_v)\ Av = vc_v$.
Let $w = b_1+b_2+\dots +b_n$. Writen in our basis it is $(1,1,\dots,1)$ so $Aw$ writen in $B$ must be $(c_w, c_w, \dots, c_w)$ for some $c_w$.
Now when we look at matrix-vector multiplication $Aw$, keeping in mind $A$ must be diagonal we can see that all of the elements on its diagonal must be the same so it is a matrix of $a*id_V$ for an $a \in K$