smallest eigenspace dimension symmetric Hi would you help me with the following:
Let $A = (a_{ij}) R^{n \times n}$ be a symmetric matrix satisfying:
$a_{1i} \neq 0$; 
Sum of each row equals $0$ and each diagonal element is the sum of 
absolute values of other entries in the row.
Determine the dimension of eigenspace
corresponding to the smallest eigenvalue of $A$.
Thanks a lot!!
 A: Since the matrix is symmetric, all eigenvalues are real. By Greshgorin circle theorem, each of these eigenvalues lies in the disc centered $a_{ii}$ with radius $a_{ii}$, hence are non-negative. The eigenvalues are thus non-negative. The smallest eigenvalue is clearly $0$, since the all 1 vector is an eigenvector with eigenvalue 0.
Since each diagonal element is the sum of absolute values in each row, and each row has a sum of 0, thus the only positive entries are on the diagonal, and the rest of the entries are nonpositive. In particular, all the $a_{1i}, i\neq 1$ are negative.
(Added explanation: If any of the non-diagonal elements are positive, then by considering the sum of that row, we will get a positive value, hence the row doesn't have a sum of 0. The mathematical proof is:
$ a_{kk} = \sum_{k \neq i} |a_{ki}| $, so $ 0 = |a_{kk} + \sum_{k\neq i} a_{ki}| \geq a_{kk} - \sum_{k \neq i} |a_{ki}| = 0$ by the triangle inequality. Since equality holds, this implies that $a_{ki}$ must have the opposite sign (or could be 0), as compared to $a_{kk}$.)
Edit: Since a real symmetric matrix has a complete (orthogonal) eigenbasis, to calculate the dimension of the generalized eigenspace, it is sufficient to consider just eigenvectors.
Now, consider any other eigenvector $v$ that isn't a multiple of the all 1 vector. If it isn't a multiple of a vector with $\pm 1$ entries, let $v_k$ be (one of) the entry with the largest absolute value, and there exists $j$ such that $|v_j| < |v_k|$ Consider expansion along row $k$, we get $$\sum_{i\neq k}  |a_{k i} v_i| \leq \sum_{i \neq k} |a_{ki}| \cdot |v_k| \leq |a_{k k} v_k|.$$ However, we cannot have equality hold throughout, since we have $|v_j| < |v_k|$. Hence, the kth entry in $A v_k$ is not 0, so the eigenvalue is not 0.
If $v$ is a multiple of a vector with $\pm1$ entries, consider expansion along the first row. We now use the condition that $a_{1i} < 0$, which shows that in order for this eigenvector to have eigenvalue 0, then this eigenvector must be a multiple of $(1, 1, 1, \ldots, 1)$, which we already considered. 
Hence, there is no other possible eigenvector with eigenvalue 0, so the dimension of this eigenspace is 1.

You should read user1551's solution, as that has a better way of dealing with the eigenvalues, then such a crude brute force computation.
A: First, let's write down something that everyone knows. By the properties of $A$, all diagonal entries of $A$ are positive and all off-diagonal entries of $A$ are nonpositive. Hence by Gersgorin disc theorem and the symmetry of $A$, all eigenvalues of $A$ are nonnegative. In particular, $0$ is the smallest of $A$ and $(1,1,\ldots,1)^T$ is a corresponding eigenvector.
Now let $e^T$ be the $(n-1)$-vector containing all ones. Write $A=\begin{pmatrix}a&b^T\\b&C\end{pmatrix}$ where $a$ is the $(1,1)$-th entry of $A$. By the properties of $A$, we have
$$
\begin{pmatrix}1&e^T\\0&I_{n-1}\end{pmatrix}
A\begin{pmatrix}1&0\\e&I_{n-1}\end{pmatrix}
=\begin{pmatrix}1&e^T\\0&I_{n-1}\end{pmatrix}
\begin{pmatrix}0&b^T\\0&C\end{pmatrix}
=\begin{pmatrix}0&0\\0&C\end{pmatrix}=:B \ \text{ (say)}.
$$
Since all off-diagonal entries in the first row/column of $A$ are strictly negative, $C$ is strictly diagonally dominant. Hence all eigenvalues of $C$ are positive, i.e. $0$ is a simple eigenvalue of $B$. Therefore, by Sylvester's law of inertia, $0$ is also a simple eigenvalue of $A$. Hence the dimension of eigenspace corresponding to the smallest eigenvalue of $A$ is $1$.
