Is xTAx always non-zero for a real, symmetric, nonsingular matrix A?

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SUMMARY

The discussion confirms that for a real, symmetric, nonsingular matrix A, the expression xTAx is always non-zero for any non-zero vector x. This conclusion is based on the properties of symmetric matrices, specifically their diagonalizability. Since A is diagonalizable, it can be expressed in a basis where it takes a diagonal form, ensuring that the quadratic form xTAx yields a positive value for all non-zero x.

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elimax
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Basic question, I think, but I'm not sure. It is a step in a demonstration, so it would be nice if it were true.
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True or false? Why? If A is a real, symmetric, nonsingular matrix, then xTAx≠0 for x≠0.
 
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Since ##A## is real and symmetric, it is diagonalizable. So you may start by thinking about ##x^T A x## in a basis in which ##A## is diagonal.
 

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