Is it true that $M$ has $0$ as eigenvalue with multiplicity exactly $1$. I am stuck on the following question: Its a true/false statement.
It says : Let $M$ be an $n\times n$ matrix. If none of the entries of $M$ are 0, but the row sum is zero for each row and no two rows are scalar multiples of each other, is it true that $M$ has $0$ as eigenvalue with multiplicity $1$.
My try:  Since $M$ is an $n\times n$ matrix with row sum  $0$ for each row, so $0$ is an eigenvalue of $M$ with multiplicity atleast $1$.
But is it true that it is an eigenvalue with multiplicity exactly $1$. I am stuck. I am getting it false. Is it correct?
What shall i do ? Can someone help.
 A: The multiplicity of $0$ as an eigenvalue is equal to the size of the matrix minus its rank (since it's the dimension of its null space). So you can "create" $0$ eigenvalues by, say, putting the same elements in different rows of the matrix (this will give you linearly dependent rows, which increase the null space).
So basically, if you wish to construct a counter example, you can take a matrix who's sum of elements in the first row is equal to $0$ (although the row itself is non-zero), and just make all other rows identical to the first row (you need at least $3$ rows for this trick to work). Can you now try to construct a counter example?
A: A counterexample is given by
$$\begin{pmatrix} -3&1&1&1 \\ 1&-3&1&1\\-4&2&1&1\\2&-4&1&1 \end{pmatrix} $$
A: Denote the column vectors of $A$ as $v_1, v_2,....v_n$ so that
$c_1v_1+c_2v_2+.....+c_nv_n=0$ has a unique non-zero solution given by the vector $(1,1,......,1)$. This tells you that the geometric multiplicity ($GM$) of eigenvalue $0$ is one. Now what does the relationship $AM\ge GM$ say?
You can easily construct a counterexample by taking all the row vectors of the form $(a,a,-2a), a\ne0$ in a $3\times 3 $ matrix $A$.
