Calculating the Kernel, dimension of linear equations, real numbers and galois field

Question

Given these problems below, how would one calculate the result?

By intuition, I have managed to solve two of them but cannot crack the last one.

Btw, I am not sure that the approach of my intuition is the right one.

I am interested in learning about how to solve these (formulas and approach).

Problems Below:

1) Let $f: ℝ^7 \rightarrow ℝ^4$ be a surjective (onto) linear function. What is the dimension of $Ker \space f$? (answer = 3)

2) Let $f: GF(2)^8 \rightarrow GF(2)^9$ be a linear funktion with $dim \space Im \space f=5$. How many vectors are in $Ker f$? (answer = 8)

3) Let $f: ℝ^7 \rightarrow ℝ^{14}$ be a injective (one-to-one) linear function. Determine $dim \space Im \space f$? (answer = 7)

Could determine the answers for problem 1 and 3 by intuition (but not by formula). Here is the intuition below:

1 - Intuition: The function is surjective so the dimension must be $7 - 4 = 3$

3 - Intuition: The function is injective so the dimension must be $14 - 7 = 7$

(First question on site, so open for constructive feedback regarding the question.)

Answer 1

All you need here is the dimension theorem: if $f:U\to V$ is any linear map, then $$\dim\ker f+\dim\mathrm{im} f\,=\, \dim U$$

Choose bases $u_1,\dots, u_k$ for $\ker f$ and $v_1,\dots, v_r$ for $\mathrm{im} f$ and arbitrary preimages $w_j$ of $v_j$. Then show $u_1,\dots, u_k, w_1,\dots, w_r$ is a basis of $U$.

This resolves your uncertainty for a) and c), and gives $\dim\ker f=3$ for b).
So that $\ker f\cong GF(2)^3$ (by coordinating the vectors) which has $2^3$ elements.