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Consider a symmetric square matrix $g$ of dimension $N$ and another symmetric square matrix $h$ of dimension $n$. Suppose $S$ is a $N\times n$ matrix such that $$ h = S^T g S $$ Suppose $\det g \neq 0$. How does one show $\det h \neq 0$? (or what extra conditions does one need so that this is true?). If possible, can we find a relation between $\det h$ and $\det g$? Assuming that $\det h \neq 0$, can we relate $h^{-1}$ to $g^{-1}$?

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  • $\begingroup$ is that a typo or is $S$ not necessarily a square matrix? Does $S$ have linearly independent rows? $\endgroup$ – user44197 Feb 16 '14 at 5:49
  • $\begingroup$ $S$ is NOT a square matrix. Lets assume that $S$ has linearly independent rows. $\endgroup$ – Prahar Feb 16 '14 at 5:50
  • $\begingroup$ I meant columns...see my partial answer $\endgroup$ – user44197 Feb 16 '14 at 5:54
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We can take the determinant of both sides and whe know that the product of determinants is the derminant of the product. Also: $det(S) = det(S^T)$

$det(h) = det(S^TgS)$

$det(h) = det(S)^2 det(g)$

This last equation only contains real numbers, so we can reaseon about it as exactly that. $det(h)$ will be $0$ iff $det(S)=0$ or $det(g)=0$. We now already have the relation you were looking for.

We know that the inverse of a determinant is the determinant of the inverse matrix.

$\frac{1}{det(h)} = \frac{1}{det(S)^2 det(g)}$

$det(h^{-1}) = \frac{1}{det(S)^2} det(g^{-1})$

EDIT: misunderstood your question, feel free to ignore.

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  • $\begingroup$ $S$ is not a square matrix. $\endgroup$ – Prahar Feb 16 '14 at 6:02
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If $S$ does not have linearly independent columns, then there is an $x$ such that $S x=0$ and hence $hx=0$. So $h$ is not invertible (has zero determinant).

Let $g$ be $N \times N$, $S$ be $N\times n$, and $S$ has linearly independent columns. Under what condition can $hx=0$? The possibilities are $$ x=0, ~ \text{or} \\ x \neq 0, ~\text{but} ~S x = 0,~\text{or} \\ S x \neq 0, ~\text{but} ~g (S x) = 0, ~\text{or} \\ g S x \neq 0 ~\text{but}~ S^T (g S x) = 0 $$ Now argue that only the first is possible.

Since you do not show anty work, not sure I can provide more hints

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  • $\begingroup$ Ah! I see. The last second and last are not possible because by assumption $S$ has linearly dependent columns (and hence $S^T$ does too). The third is not possible since $g$ is invertible. I see now. Thanks a lot! Just want to mention this is not HW though. I agree I should have shown my work nonetheless. $\endgroup$ – Prahar Feb 16 '14 at 6:10
  • $\begingroup$ Any comments on relating $h^{-1}$ to $g^{-1}$? $\endgroup$ – Prahar Feb 16 '14 at 18:15
  • $\begingroup$ In general, $h$ is a submatrix of a matrix congruent to $g$. The transformation $M^T g M$ is called a congruent transformation. In, general not much can be said. However, I have seen bounds (based on work of Lancos) on some key properties of $h$ but I can't recall them. I am sure one can work out something if $S^T S = I$. $\endgroup$ – user44197 Feb 16 '14 at 23:30
  • $\begingroup$ Actually, in the problem at hand, it is true that $S^T S = I$. Can you give me a reference so I can work this out? You don't have to work it out for me (unless you want to :P) Thanks a lot. $\endgroup$ – Prahar Feb 17 '14 at 0:34
  • $\begingroup$ If $S^T S= I$, then complete the basis by adding columns to $S$ so that $S$ is square matrix. In this case $S^T g S$ has the same size of $g$ and $h$ is the $1 \times 1$ sub-block. Now there are a number of relationship between inverse of a sub-block and the sub-block of the inverse. I don't have a reference handy but if time permits I can look. I suggest you pose a separate question so someone may want to provide more details. You can refer to this question for background info. $\endgroup$ – user44197 Feb 17 '14 at 1:42

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