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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.

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  • $\begingroup$ Do you know, or are you allowed to use the spectral theorem for symmetric matrices? $\endgroup$ – Teresa Lisbon Aug 27 '18 at 12:25
  • $\begingroup$ @астонвіллаолофмэллбэрг cant it be done without it? $\endgroup$ – user572425 Aug 27 '18 at 12:26
  • $\begingroup$ If its needed we can use it $\endgroup$ – user572425 Aug 27 '18 at 12:26
  • $\begingroup$ Why downvotes?? $\endgroup$ – user572425 Aug 27 '18 at 12:27
  • $\begingroup$ Actually, I think I see it : you need to prove that the kernel has dimension greater than or equal to $m-1$,right? Now, note that the columns of $A$ are the rows of $A$, because $A$ is symmetric. Now, if $A$ has $m$ identical rows then $A$ has $m$ identical columns, therefore the dimension of the column space is smaller than the dimension minus $(m-1)$, and now rank-nullity tells you the kernel must at least have that much dimension. $\endgroup$ – Teresa Lisbon Aug 27 '18 at 12:31
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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.

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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 $m-1$.

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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.

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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 $n$

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  • $\begingroup$ This doesn't answer the question. Moreover, A is symmetric so $m=n$. And you only assume now that special case where $m$ equals the amount of columns. $\endgroup$ – Stan Tendijck Aug 27 '18 at 13:16
  • $\begingroup$ Well, I responded to the formulation of the question when I read it, before it was edited. I added the edit afterwards but there was no real point in suppressing my answer. $\endgroup$ – P. Quinton Aug 27 '18 at 13:50

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