The dear user @kodlu has answered to the similar question with excellent discussion but I want to answer with linear algebra argument.
We have two definitions for MDS (Maximum Distance Separable) Matrix:
First definition: A matrix $M$ of order $n$ is an MDS matrix if and only if every sub-matrix of $M$ is non-singular.
Second definition: A matrix $M_{n\times n}$ is MDS if and only if
$$
Y_{n\times 1}=M_{n \times n}\, X_{n \times 1} \Longrightarrow \mathop{\rm min}_{X\neq 0}(W(Y)+W(X))=n+1
$$
where $X={[x_0,x_1,\cdots , x_{n-1}]}^T$ and $Y={[y_0,y_1,\cdots , y_{n-1}]}^T$ are vectors in an arbitrary
field and $W(X)$ is the number of non-zero elements of $X$.
Now, suppose that we use a MDS matrix( by first definition) of order $4$ and
we want to proof by second definition that it's branch number is 5. Consider our MDS matrix be as follows
$$
M=\left[ \begin {array}{cccc} m_{{1,1}}&m_{{1,2}}&m_{{1,3}}&m_{{1,4}}
\\ m_{{2,1}}&m_{{2,2}}&m_{{2,3}}&m_{{2,4}}
\\ m_{{3,1}}&m_{{3,2}}&m_{{3,3}}&m_{{3,4}}
\\ m_{{4,1}}&m_{{4,2}}&m_{{4,3}}&m_{{4,4}}
\end {array} \right]
$$
the entries $m_{i,j}$ are non-zero because $M$ by first definition is MDS. Now consider the following equation
$$
MX=Y \Longrightarrow
\left[ \begin {array}{cccc} m_{{1,1}}&m_{{1,2}}&m_{{1,3}}&m_{{1,4}}
\\ m_{{2,1}}&m_{{2,2}}&m_{{2,3}}&m_{{2,4}}
\\ m_{{3,1}}&m_{{3,2}}&m_{{3,3}}&m_{{3,4}}
\\ m_{{4,1}}&m_{{4,2}}&m_{{4,3}}&m_{{4,4}}
\end {array} \right]
\left[ \begin {array}{c} x_{{1}}\\x_{{2}}
\\ x_{{3}}\\ x_{{4}}\end {array}
\right]=
\left[ \begin {array}{c} y_{{1}}\\ y_{{2}}
\\ y_{{3}}\\ y_{{4}}\end {array}
\right]
$$
where $x_i$'s and $y_i$'s are elements of finite field. By second definition $X$ should be non-zero. For simplicity, consider $x_1\neq0$ and other $x_i$ can be zero or non-zero.
We should consider some cases to show that for every selection of $x_2$, $x_3$ and $x_4$, at least four elements of $x_2,x_3,x_4,y_1,y_2,y_3,y_4$ are non-zero.
The first case is $x_2=x_3=x_4=0$. in this case we have:
$$
\left[ \begin {array}{cccc} m_{{1,1}}&m_{{1,2}}&m_{{1,3}}&m_{{1,4}}
\\ m_{{2,1}}&m_{{2,2}}&m_{{2,3}}&m_{{2,4}}
\\ m_{{3,1}}&m_{{3,2}}&m_{{3,3}}&m_{{3,4}}
\\ m_{{4,1}}&m_{{4,2}}&m_{{4,3}}&m_{{4,4}}
\end {array} \right]
\left[ \begin {array}{c} x_{{1}}\\0
\\ 0\\ 0\end {array}
\right]=
\left[ \begin {array}{c} x_1\,m_{1,1}\\ x_1\,m_{2,1}
\\ x_1\,m_{3,1}\\ x_1\,m_{4,1}\end {array}
\right]=
\left[ \begin {array}{c} y_{{1}}\\ y_{{2}}
\\ y_{{3}}\\ y_{{4}}\end {array}
\right]
$$
because $m_{i,j}$'s and $x_1$ are nonzero then the elements $y_1,y_2,y_3,y_4$ are nonzero. The next case is that are there non-zero
values for $x_2,x_3,x_4$ such that
$$
\left[ \begin {array}{cccc} m_{{1,1}}&m_{{1,2}}&m_{{1,3}}&m_{{1,4}}
\\ m_{{2,1}}&m_{{2,2}}&m_{{2,3}}&m_{{2,4}}
\\ m_{{3,1}}&m_{{3,2}}&m_{{3,3}}&m_{{3,4}}
\\ m_{{4,1}}&m_{{4,2}}&m_{{4,3}}&m_{{4,4}}
\end {array} \right]
\left[ \begin {array}{c} x_{{1}}\\x_{{2}}
\\ x_{{3}}\\ x_{{4}}\end {array}
\right]=
\left[ \begin {array}{c} 0\\ 0
\\ 0\\ 0\end {array}
\right]
$$
due to the $M$ matrix is non-singular we can conclude that
$$
\left[ \begin {array}{c} x_{{1}}\\x_{{2}}
\\ x_{{3}}\\ x_{{4}}\end {array}
\right]=
\left[ \begin {array}{cccc} m_{{1,1}}&m_{{1,2}}&m_{{1,3}}&m_{{1,4}}
\\ m_{{2,1}}&m_{{2,2}}&m_{{2,3}}&m_{{2,4}}
\\ m_{{3,1}}&m_{{3,2}}&m_{{3,3}}&m_{{3,4}}
\\ m_{{4,1}}&m_{{4,2}}&m_{{4,3}}&m_{{4,4}}
\end {array} \right]^{-1}\left[ \begin {array}{c} 0\\ 0
\\ 0\\ 0\end {array}
\right]=\left[ \begin {array}{c} 0\\ 0
\\ 0\\ 0\end {array}
\right]
$$
and it contradict to our assumption that $x_1,x_2,x_3,x_4\neq 0$ and so one of the $y_1,y_2,y_3,y_4$ should be non-zero.
The next case is that: are there for example $x_1\neq0, x_2\neq0 ,x_3=x_4=0$ that result
$y_1\neq0, y_2\neq0 ,y_3=y_4=0$. Consider there was this case as follows
$$
\left[ \begin {array}{cccc} m_{{1,1}}&m_{{1,2}}&m_{{1,3}}&m_{{1,4}}
\\ m_{{2,1}}&m_{{2,2}}&m_{{2,3}}&m_{{2,4}}
\\ m_{{3,1}}&m_{{3,2}}&m_{{3,3}}&m_{{3,4}}
\\ m_{{4,1}}&m_{{4,2}}&m_{{4,3}}&m_{{4,4}}
\end {array} \right]
\left[ \begin {array}{c} x_{{1}}\\x_2
\\ 0\\ 0\end {array}
\right]=
\left[ \begin {array}{c} y_{{1}}\\ y_{{2}}
\\ 0\\ 0\end {array}
\right]
$$
so we have
$$
\left[ \begin {array}{cc} m_{{3,1}}&m_{{3,2}}\\ m_{
{4,1}}&m_{{4,2}}\end {array} \right] \left[ \begin {array}{c} x_{{1}}\\ x_{{2}}
\end {array} \right]=
\left[ \begin {array}{c} 0\\ 0\end {array} \right]
$$
from the first definition all sub-matrix of $M$ has non-zero determinant and hence we can conclude from the above equation that the values of $x_1=x_2=0$ and it contradict to our assumption.
The other cases are similar to mentioned cases and use from this fact that all sub-matrix of $M$ have non-zero determinant.
I hope you find it useful.
idempotentmeans "doing it twice is the same as doing it once", that is a function $f$ is idempotent iff $f(f(x)) = f(x)$. What they mean is "self-inverse". – poncho May 03 '17 at 19:11