Let's do the SVD decomposition and have $M= U\Sigma V^*$.
Is it true, that for any matrix $M$, which is such that an arithmetic mean in every column is equal to $0$, we get that columns in the matrix $U$ has to have the same proporty (means in columns will be $0$)?
In few cases I noticed this fact, but I don't know if is any general result.
If that is true, can we get a similar result for another means? (not necessarily for $0$ - as the value of the means)
This is true if $M$ has linearly indepent rows. Say $M$ is a $m\times n$ real matrix, then $rank(M)=m$ suffices.
Let $\alpha ^T = (1,1,...,1)$ be a $m$-dimensional vector. Then the condition that arithmetic mean in every column is equal to $0$ is equivalent to $$\alpha ^TM=(\alpha ^T \beta_1,\alpha ^T \beta_2,...,\alpha ^T \beta_n)=(0,0,...,0)=\mathbf0_{1\times n}$$ where $(\beta_1,\beta_2,...,\beta_n)$ are columns of $M$.
By singular value decomposition, $M=U\Sigma V^*$, then $$\alpha^TM=\alpha^TU\Sigma V^*=\mathbf0$$ Right-multiplying both sides by $V$, we obtain $$\alpha^TU\Sigma=\mathbf0\quad(*)$$ The $m\times n$ matrix $\Sigma$ has the same rank as $M$ (this is a property of SVD). Hence $rank(\Sigma) = m$. Then there exists a $n \times m$ matrix $X$ such that $$\Sigma X=I_m$$ where $I_m$ denotes $m\times m$ identity matrix. Right-multiplying $(*)$ by $X$, we have $$\alpha ^T U=0$$ Thus the means in columns are $0$ for $U$.