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Can a $n \times n$ symmetric matrix $A$ with diagonal entries that are all equal to zero, be positive definite (or negative definite)?

Thanks in advance!

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    $\begingroup$ What is $x^TAx$ when $x$ is a basis vector? $\endgroup$ – user147263 Nov 25 '15 at 21:58
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Of course. For example this is positive defined $$\sigma_1=\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$$ While this is negative defined: $$\sigma_1'=\begin{pmatrix} 0 & -1 \\ -1 & 0 \\ \end{pmatrix}$$ You need to be careful that zero on the diagonal doesn't imply much about the eigenvalues. With a change of basis you can change the coefficient on the diagonal. For example let us suppose that $$T=\begin{pmatrix} 1 & 1 \\ 1 & 0 \\ \end{pmatrix}$$ Then $$T^{-1}\sigma_1T=\begin{pmatrix} 1 & 1 \\ 0 & -1 \\ \end{pmatrix}$$

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  • $\begingroup$ Thanks Daco for you reply! I was thinking that for a zero-trace matrix, the trace is equal to the sum of the eigenvalues,which is zero in this case, and this means that the matrix has both positive and negative eigenvalues which make it hard for me to see if it is positive or negative definite, or just indefinite! $\endgroup$ – Mim Nov 26 '15 at 22:33
  • $\begingroup$ Actually your example shows that this is not the case when $n=2$. I appreciate your help. $\endgroup$ – Mim Nov 26 '15 at 23:25
  • $\begingroup$ :D :D :D no... it shows that is not the case always... just think about it... ;) $\endgroup$ – Dac0 Nov 26 '15 at 23:26
  • $\begingroup$ can it be proven somehow? $\endgroup$ – Mim Nov 26 '15 at 23:28
  • $\begingroup$ Yes of course, and easily using the answer I gave you. Just think about it and figure it out. If you can not put an up arrow to my answer and maybe I will give you another hint $\endgroup$ – Dac0 Nov 26 '15 at 23:30

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