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Consider the following matrix:

$\begin{bmatrix} 2x & -2y & 0 \\\ 0 & z & y \\ z & 0 & x \\ y & x & 0 \\ 2x & 2y & 2z\end{bmatrix}$

How can we prove that this matrix has always rank $3$ whenever $(x,y,z)$ satisfy $x^{2}+y^{2}+z^{2}=1$? i.e they lie on the unit sphere. (without using row echelon form as it seems messy)

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Has it? Try $x=y=0$. –  Did Oct 7 '11 at 16:32
    
Hmmm... the second version seems better. –  Did Oct 7 '11 at 16:35

1 Answer 1

up vote 2 down vote accepted

Assuming that $z$, $y$ and $z$ are real: The determinant of rows 1, 3, and 4 is $$2\begin{vmatrix}x&-y&0\\z&0&x\\y&x&0\end{vmatrix}=2(-xy^2-x^3)=-2x(x^2+y^2)$$ which is nonzero as long as $x$ is.

Similarly, rows 1, 2, and 4 have nonzero determinant whenever $y\ne 0$.

The only case left is $x=y=0$, $z\ne 0$ ...

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