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I need to prove the following statement:

Any linear map $f:E\longrightarrow F$ between two real vector spaces is differentiable. Moreover, $f$ is a diffeomorphism if and only if it is an isomorphism.

I can't see any way to prove this, but my intuition tells me that the second part of the problem has something to do with the fact that $dim(E)=dim(Ker(f))+dim(Im(f))$. I'm desperate trying to figure this out, can anyone help please?

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    Try to compute $(\frac{f(x+h) - f(x)}{h})$. Since it is an operator from $E$ to $F$, you need the value at $u \in E$. – 2017-02-20

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The question is not precise. Are we talking about finite-dimensional spaces? Otherwise this is not true (indeed, for infinite dimensional cases linear operators don't even have to be continous). Also what is the differential structure on $E$ and $F$? I assume that we are talking about standard $E=\mathbb{R}^n$ and $F=\mathbb{R}^m$.

So let $f:\mathbb{R}^n\to\mathbb{R}^m$ be linear. By definition we have to find a linear operator $T:\mathbb{R}^n\to\mathbb{R}^m$ such that

$$\lim_{h\to 0}\frac{\lVert f(x_0+h)-f(x_0)-T(h) \rVert}{\lVert h\rVert}=0$$

for given $x_0\in\mathbb{R}^n$.

Put $T=f$ (note that we can do this since $f$ is linear). Then

$$\lim_{h\to 0}\frac{\lVert f(x_0+h)-f(x_0)-f(h) \rVert}{\lVert h\rVert}=\lim_{h\to 0}\frac{\lVert f(x_0) + f(h)-f(x_0)-f(h) \rVert}{\lVert h\rVert}=\lim_{h\to 0}\frac{\lVert 0 \rVert}{\lVert h\rVert}=0$$

As for the second question obviously a diffeomorphism is an isomorphism. On the other hand if $f$ is an isomorphism then $f^{-1}$ exists and is linear so by previous result it is differentiable hence $f$ is a diffeomorphism.