Show that the rank of a symmetric matrix is the maximum order of a principal sub-matrix which is invertible The question is this-
Show that the rank of a symmetric matrix is the maximum order of a principal sub-matrix which is invertible.
I can show that there cannot exist a sub-matrix with rank more than the actual rank of the matrix. But I cannot show the other way around, i.e when the rank of the actual matrix is $r$, then there exist a principal sub-matrix of the same rank (though I can prove that there exist a sub-matrix with rank $r$).
I was thinking like this: if the rank of the matrix is $r$, we can find r linearly independent rows of the matrix, say $a_1, a_2, ..., a_r$ -th rows are linearly independent. Then the corresposding columns $a_1^t, a_2^t, ..., a_r^t$ are also linearly independent. But how to show that the submatrix that they produce is of rank r?
I hope my question is clear. Any hints or help would be highly appreciated.
 A: let $A\in \mathbb R^{n\times n}$ be symmetric with rank $r$. Note: there is nothing to do if $r\in\big\{0,n\big\} $ so assume $r\in\big\{1,2,...,,n-1\big\}$.
Arguing by contrapositive:
Suppose that the maximal sized invertible principal submatrix is $r-k\times r-k$ for integer $k\geq 1$.
Then the sum over all $r\times r$ principal minors is zero (because each minor is zero), and so is the sum over all $r+1\times r+1$ principal minors (again each is zero by hypothesis) and so on up to the $n\times n$ principal minor given by $\det\big(A\big)=0$.
Up to a sign, these sums over principal minors give specific coefficients of $A$'s characteristic polynomial -- in particular the last $n-r+1$ coefficients in the characteristic polynomial of $A$, so
$p_A(x)= (x-0)^{n-r+1}\cdot q(x)$.  or
$\text{algebraic multiplicity of zero}\big(p_A\big) \geq n-r+1$.
Finally apply rank-nullity and the fact that real symmetric matrices are diagonalizable so algebraic multiplicities equal geometric multiplicities
$\text{rank}\big(A\big) = n -\dim\ker\big(A\big)= n -\text{algebraic multiplicity of zero}\big(p_A\big)\leq n - (n-r+1) = r-1$
$\implies k \not \geq 1$
and you've already shown  $k\not \leq -1$, thus $k=0$ as desired.
A: Assume that the leading $r\times r$ minor is $0$. Then the $r$ columns of this minor are linearly dependent. Since the first $r$ rows  of the big matrix generate all the other rows, the same linear relation will hold for the first $r$ columns of the big matrix, contradiction.
