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

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For a real, symmetric, nonsingular matrix A, the expression xTAx is indeed non-zero for any non-zero vector x. This is because such matrices are diagonalizable, and in their diagonal form, the eigenvalues are all positive due to the nonsingularity of A. Consequently, the quadratic form xTAx will yield a positive value for any non-zero x. The discussion emphasizes the importance of diagonalization in understanding the properties of the matrix. Therefore, xTAx is always non-zero for x≠0.
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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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