If a symmetric matrix $A$ has $m$ identical rows show that $0$ is an eigen value of $A$ whose geometric multiplicity is atleast $m-1$.
If $A$ has $m$ identical rows then then by elementary row operation $A$ has $m-1$ zero rows.
But how to show that geometric multiplicity of $A$ is atleast $m-1$ from above?
I know that geometric multiplicity of an eigen value is ($\dim(ker(A-\lambda I))$ but how to compute it from above?
Please help.
Direct method.
By definition, the geo-multiplicity of $0$ is just the dimension of the space of solution for $\boldsymbol {Ax}= \boldsymbol 0$. By symmetry, $\boldsymbol A$ has $m$ identical columns. Suppose these columns are the $(k_j)_{j =1}^m$-th columns of $\boldsymbol A$. Let $\boldsymbol e_j$ be the $j$-th standard basis vector, then clearly at least $$ \boldsymbol e_{k_1} - \boldsymbol e_{k_j} \quad [j =2, \ldots, m] $$ are solutions of $\boldsymbol {Ax=0}$, thus $\dim(\mathrm{Ker} \boldsymbol A) \geqslant m-1$ as we desire.
Since $A$ is symmetric, by the Real Spectral theorem we must have a diagonal matrix $D$ and an invertible matrix $Q$ such that
$A=Q^{-1}DQ$.
Since $Q$ is invertible, multiplying a matrix by $Q$ and/or $Q^{-1}$ doesn't modify the rank, so
$Rank(A)=Rank(Q^{-1}DQ)=Rank(D)$.
However, since $A$ has $m$ repeated rows, its rank is at most $n-m+1$ (where $n$ is the total number of rows). Therefore $Rank(D)$ is at most $n-m+1=n-(m-1)$, and because $D$ is a diagonal matrix, this is only posible if it has $m-1$ rows equal to zero.
Then $0$ is an eingenvalue that appears $m-1$ times in the diagonal of $D$, then the multiplicity of $0$ is $m-1$.
Because $A$ is symmetric with $m$ identical rows, then $A^{T}=A$ has has $m$ identical columns. If $j_1,\cdots,j_m$ are the positions of those identical columns between $1$ and $m$, then $$ A\left[\begin{array}{c}c_1 \\ c_2 \\ \vdots \\ c_N\end{array}\right]=0 $$ if $c_n=0$ for $j\ne j_1,j_2,\cdots,j_m$ and if $$ c_{j_1}+c_{j_2}+\cdots+c_{j_m}=0. $$ This is an $m-1$ dimensional subspace. So $A$ has at least an $m-1$ dimensional null space. If all these columns are $0$, then the null space would be $m$ dimensional.
It should be clear that if at least one entry of $A$ is non zero, then its rank is at greater or equal to $1$.
suppose that the rank of $A$ is strictly larger than $1$, then you can find $\pmb{x}$ and $\pmb{y}$ such that $A \pmb{x}$ and $A \pmb{y}$ are not collinear. This is impossible since all the rows of $A$ are the same (let say $\pmb{a}^T$), then $A \pmb{x} = \left( \sum x_i a_i \right) \pmb{1}$ and $A \pmb{x} = \left( \sum y_i a_i \right) \pmb{1}$ and so they are collinear.
Now we know that the $A$ have rank $1$. By your formula, the multiplicity is $\dim( \ker(A-0 I)) = \dim( \ker(A)) = m - rank(A) = m-1$.
EDIT : this is only true if $A$ is a $m\times n$ matrix for some $n$
