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I'm trying to prove whether the below statement false:

For a real $3 \times 4$ matrix $A$, the function $q(\vec{x}) = \lVert \vec{Ax}\rVert^2$ cannot define a positive semidefinite quadratic form on $\mathbb{R}^4$ because $A$ is not square and hence it is non-invertible.

My approach would be to use the determinant test to show that there exists a matrix such that $A$ is positive semidefinite. Is this the correct approach?

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  • $\begingroup$ Do you mean $q(x) = \|Ax\|^2$? $\endgroup$ – Omnomnomnom Dec 14 '17 at 0:14
  • $\begingroup$ Yes I did! Fixed. $\endgroup$ – Taylor Tam Dec 14 '17 at 0:19
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Note that

$A: \Bbb R^4 \to \Bbb R^3; \tag 1$

as such,

$\ker(A) \ne \{0\}; \tag 2$

thus there exists a vector $\vec x \in \Bbb R^4$ with

$A \vec x = 0; \tag 3$

then

$q(\vec x) = \Vert A \vec x \Vert^2 = \Vert 0 \Vert^2 = 0, \tag 4$

so $q(\vec x)$ cannot be positive definite; however, if $A \ne 0$ there is some $\vec y \in \Bbb R^n$ with

$A\vec y \ne 0, \tag 4$

and for any such $\vec y$,

$q(\vec y) = \Vert A \vec y \Vert^2 > 0. \tag 5$

We see that $q(\vec x) \ge 0$ for all $\vec x \in \Bbb R^4$; thus $q(\vec x)$ is positive semidefinite.

The above argument does not invoke determinants.

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