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On page 7 of Milnor's Morse Theory is part of a proof of the Morse Lemma:

Suppose by induction that there exist coordinates $u_1, \ldots, u_n$ in a neighbourhood $U_1$ of $0$ so that $f = \pm(u_1)^2 \pm \cdots \pm (u_{r-1})^2 + \sum_{i,j \geq r} u_i u_j H_{ij}(u_1,\ldots,u_n)$ throughout $U_1$; where the matrices $(H_{ij}(u_1,\ldots,u_n))$ are symmetric. After a linear change in the last $n - r + 1$ coordinates, we may assume that $H_{rr}(0) \neq 0$.

(full text of proof: page 6, page 7, page 8)

I do not understand how he can WLOG assume that $H_{rr}\neq 0$ by doing a "linear change of variables". I don't really even understand what he's saying. Could someone please spell this out for me? Note that page 6 handles an entirely different part of the lemma. If you need context, it should be enough to just assume that the equalities at the top of page 7 hold. You also must know that $f$ has a nondegenerate critical point at $0$, and $f(0)=0$.

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If all entries of the sum over $i,j\ge r$ were equal to zero, we would have a degenerate quadratic form. Thus, there is a nonzero term $H_{ij}$. If it's diagonal ($i=j$), then just relabel the rows/columns to make it $H_{rr}$. If all diagonal terms are zero, pick a non-diagonal $H_{ij}$ and replace the corresponding coordinates $x_i,x_j$ by $y_i=x_i+x_j$ and $y_j=x_i-x_j$. Since $y_i^2-y_j^2=4x_ix_j$, both $H_{ii}$ and $H_{jj}$ will be nonzero.

As Milnor says, this is how one diagonalizes quadratic forms so you may want to read a proof of that first.

[added] Here is a concrete example: $f=x_1^2+x_2x_3H_{23}(x_1,x_2,x_3)$. Performing the substitution above, we get $f=x_1^2+\frac{1}{4}(y_2^2-y_3^2)H_{23}(x_1,(y_2+y_3)/2,(y_2-y_3)/2)$ So, our new functions are $\widetilde H_{22}(x_1,y_2,y_3)=\frac{1}{4}H_{23}(x_1,(y_2+y_3)/2,(y_2-y_3)/2)$ and $\widetilde H_{33}(x_1,y_2,y_3)=-\frac{1}{4}H_{23}(x_1,(y_2+y_3)/2,(y_2-y_3)/2)$

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    @Jeff I added a concrete example to show how nonzero $H_{ii}$ appears.2012-06-16