Show that the image of a linear transformation is equal to the kernel Let $\phi$ be a linear transformation such that $\phi: V\to V$
We are given the following facts:


*

*$\dim(V) = 8$

*$\dim(\mathrm{Im}(\phi)) = 4$

*$\phi\circ\phi=0$
Show that $\mathrm{Im}(\phi) = \ker \phi$
So I assumed to start with letting $y \in \mathrm{Im}(\phi)$. So there is an $x\in V$  such that $y = \phi(x)$.
Multiplying by the transformation we get:
$\phi(y) = \phi(\phi(x)) = (\phi\circ\phi)(x) = 0$ using fact 3.
That equals the kernel. So I assume that $y \in \ker(\phi)$ as well as $y\in \mathrm{Im} (\phi)$.
In my mark scheme it says that that shows that the image is in the kernel.  
Why doesn't it show that the kernel is in the image? Why do I need to show that the image belongs to the kernel? What do I do next?
I'm sort of new to this kernel image stuff I know its a weird question but could someone explain the necessary steps when wanting to show the image is less than or bigger or equal to the kernel. As in, where would i always begin? 
Thanks
 A: As you have argued $\phi \circ \phi = 0$ rather quickly implies that $\mathrm{Im\;} \phi \subseteq \ker \phi$. To show that they are actually equal I would use a dimension arguement. Since $\dim V = 8$ and $\dim \mathrm{Im\;} \phi = 4$ we get that $\dim \ker \phi = 4$. So, $\mathrm{Im\;} \phi$ is a subspace of $\ker \phi$ of full dimension. Therefore we conclude $\mathrm{Im\;} \phi = \ker \phi$.
A: In general, to show that two sets are equal (in this case the two sets are your kernel and your image), you must show mutual set inclusion, i.e. we must have $\mathrm{Im}(\phi) \subseteq \ker(\phi)$ and $\ker(\phi) \subseteq \mathrm{Im}(\phi)$.
You've shown that for each $\mathbf{y}\in \mathrm{Im}(\phi)$, you also have $\mathbf{y}\in\ker(\phi)$. Every element of the image is an element of the kernel, so that means $\mathrm{Im}(\phi) \subseteq \ker(\phi)$. But is it possible that there exists some element in the kernel which is not itself an element of the image? Put another way, is every element of the kernel also an element of the image? You must show the second inclusion I mentioned above: $$\ker(\phi) \subseteq \mathrm{Im}(\phi)$$
It is very possible to have inclusion one way but not the other. Let me give you the two extreme cases. 


*

*Take the zero map, the linear map which maps every vector to $\mathbf{0}$. In this case, the image of the zero map is trivial, it is just $\{\mathbf{0}\}$. In contrast, the kernel is all of $V$. Certainly $\{\mathbf{0}\}\subseteq V$, so that the image is a subset of the kernel, but you do not have the kernel as a subset of the image also.

*In contrast, if you have an invertible map $T$, then the opposite situation happens and you have $\ker(T) = \{0\}$ while $\mathrm{Im}(T) = V$. In this case, the kernel is a subset of the image, but not the other way around. 
So to show that the two sets are equal, you must show that each belong to the other. So far, you've only shown $\mathrm{Im}(\phi) \subseteq \ker(\phi)$ but not $\ker(\phi) \subseteq \mathrm{Im}(\phi)$. The other answers have already shown you how to do this, but let me flesh it out a bit.
So far, you know that $\mathrm{Im}(\phi) \subseteq \ker(\phi)$. This is important, as this shows you that the image is a subspace of the kernel. Your kernel and your image have related dimensions due to the rank-nullity theorem:
$$\dim V = \dim \ker (\phi) + \dim \mathrm{Im}(\phi)$$
Since $\dim V = 8$ and $\dim \mathrm{Im}(\phi) = 4$, we can conclude that your kernel must also have dimension $4$. 
Therefore, your image is a $4$ dimensional subspace of a $4$ dimensional vector space (the kernel). You can show, and it is a good exercise, that any $n$-dimensional subspace of an $n$-dimensional vector space $V$ is necessarily equal to $V$. Use this to conclude that the image and kernel are equal.
A: You only show that $y\in \mathrm{Im}(\phi) \Rightarrow y\in \ker(\phi)$ so $\mathrm{Im}(\phi) \subset \ker(\phi)$. This is what was told to you.
To show the other inclusion $\ker(\phi) \subset \mathrm{Im}(\phi)$ you must have $y\in\ker(\phi)$ ($\phi(y) = 0$) and show that there is $x\in V$ such that $\phi(x) = y$. This can be done using the dimension formula:
$$\dim V = \dim(\mathrm{Im} (\phi)) + \dim(\ker(\phi))$$
And facts 1.,2. and 3.
