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If out of a quadratic form in $x$, $y$, $z$; I find a matrix $A_{3\times 3}$ (symmetric) whose one of the eigenvalues is negative, one positive, and the third is zero. Can I conclude that the range of the quadratic form is $\mathbb{R}$, and how? What will be the range if the zero eigenvalue is replaced by some positive or negative one? Will it remain the $\mathbb{R}$? Or can we just not conclude anything determining those eigenvalues. For positive definite matrices, I know that the range will be contained in $(0, \infty)$

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Let $A=UDU^T$ be the eigenvalue decomposition wher $U$ is orthogonal and $D$ is a matrix with eigenvalues on the diagonal.

$$x^TAx=x^TUDU^Tx=y^TDy=\sum_{i=1}^n d_{i}y_i^2$$

Hence if there are eigenvalues with differing sign, we are able to attain every real number.

For instance, if $d_1>0$, to attain $t\geq0$, just choose $y_1=\sqrt{t}$, $y_i=0$, for $i>1$.

If $d_2<0$, to attain $t<0$, just choose $y_2 = \sqrt{-t}$, $y_i=0$, for $i \neq 2$.

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  • $\begingroup$ Thank you very much, it's helpful $\endgroup$ – Hirakjyoti Das May 23 '17 at 21:31
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So it sounds like you have $Q(x,y,z):=(x,y,z)A(x,y,z)^T$. Moreover, $A$ is symmetric with a negative, positive and 0 eigenvalue. Let $(a,b,c)^T$ be the eigenvector for $\lambda_1>0$. Then $Q(ka,kb,kc)=k^2\|(a,b,c)\|^2\lambda_1$, so choose $k$ to be arbitrarily large or small so that $Q$ maps to all positive real numbers. Repeat with $\lambda_2<0$. For $\lambda_3=0$, this guarantees there's an $(a,b,c)^T$ with 0 eigenvalue, so $Q(a,b,c)=0$. This covers all of $\mathbb{R}$.

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  • $\begingroup$ Thank you very much, It's helpful $\endgroup$ – Hirakjyoti Das May 23 '17 at 21:31

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