Prove that if $A$ is antisymmetrical and $n$ is odd, then it is not invertible.

Question

$\mathcal{M}_n(\mathbb{R})$ is the set of squared matrices of order $n$.

$\mathcal{S}_n(\mathbb{R})$ is the set of symmetrical matrices and >$\mathcal{A}_n(\mathbb{R})$ is the set of anti-symmetrical matrices.

We consider the endormorphisms $f$ and $g$ on $\mathcal{M}_n(\mathbb{R})$ such that:

$f(M) = \frac{1}{2}(M + ^t M)$ and $g(M) = \frac{1}{2}(M - ^t M)$

1. Writing $M$ as a linear combination of $f$ and $g$, show that $\mathcal{S}_n(\mathbb{R})$ and $\mathcal{A}_n(\mathbb{R})$ are complementary in $\mathcal{M}_n(\mathbb{R})$.

2. Deduce that $Imf = \mathcal{S}_n(\mathbb{R})$ and $Img = \mathcal{A}_n(\mathbb{R}) $

3. Determine a necessary and sufficient condition for the product $AB \in \mathcal{S}_n(\mathbb{R}) $

4. Prove that if $ A \in \mathcal{A}_n(\mathbb{R}) $ and $n$ is odd, then $A$ is not invertible.

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1. From the expression of $f$ and $g$ we have: $M = f(M) + g(M)$.

I don't know how to show that $\mathcal{S}_n(\mathbb{R})$ and $\mathcal{A}_n(\mathbb{R})$ are complementary in $\mathcal{M}_n(\mathbb{R})$.

2. Knowing that: $\mathcal{S}_n(\mathbb{R}) \oplus \mathcal{A}_n(\mathbb{R}) = \mathcal{M}_n(\mathbb{R})$, I don't know how to deduce $Imf = \mathcal{S}_n(\mathbb{R})$ and $Img = \mathcal{A}_n(\mathbb{R}) $

3. Let $A,B \in \mathcal{S}_n(\mathbb{R}) $,

$AB = ^tA. ^tB = ^t(BA)$

If $AB = BA$, we'll have $AB = ^t(AB)$ which proves that $AB \in \mathcal{S}_n(\mathbb{R}) $

We will get the same result if we started from $AB = BA$

4. I don't see how to use the fact that $n$ is odd to prove that $A$ is not invertible.

Thank you for your help.