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 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.
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 $n$