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${\bf{N}} = {\bf{n}}{{\bf{n}}^H},{\bf{n}} \in {{\bf{C}}^{K \times 1}}$ is given and ${\bf{X}} = {\bf{x}}{{\bf{x}}^H},{\bf{x}} \in {{\bf{C}}^{K \times 1}}$ is unkown.Can I get $trace({\bf{NX}})$ increases $\Leftrightarrow trace({\bf{X}})$ increases?

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    What does $\bf{N}=\bf{nn}^H$, $\bf{n}\in\bf{C}^{K\times1}$ mean here? – 2017-02-28
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    ${( \cdot )^H}$ is the conjugate transpose of a matrix.${\bf{n}}$ is a $K$ dimension complex vector – 2017-02-28
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    Is there a way you are increasing $tr(X)$, or are you asking if for $X,Y\in M_K(\mathbb{C})$ of the given form such that $tr(X) – 2017-02-28
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    yes,that's exactly what I mean. Notice that ${\bf{N,X}}$ are both rank-one – 2017-02-28

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"Can I get ${\rm trace}(\mathbf N\mathbf X)$ increases $\Leftrightarrow {\rm trace}(\mathbf X)$ increases?"

No, here is a counterexample: Consider $\mathbf n = [1, 0]^{\rm T}$ and $\mathbf x = [1,y]^{\rm T}$. Then ${\rm trace}(\mathbf N \mathbf X) = 1$ and ${\rm trace}(\mathbf X) = 1 + |y|^2$. Hence, ${\rm trace}(\mathbf X)$ increases monotonically in $|y|$ but ${\rm trace}(\mathbf N \mathbf X)$ does not. Therefore, your $\Leftrightarrow$ assertion was wrong. In a similar fashion you can make ${\rm trace}(\mathbf N \mathbf X)$ increase while ${\rm trace}( \mathbf X)$ is constant (e.g., $\mathbf n = [y,0]^{\rm T}$, $\mathbf x = [1,1]^{\rm T}$).

As a note, ${\rm trace}(\mathbf N \mathbf X) = |\mathbf n^{\rm H} \mathbf x|^2$ and ${\rm trace}( \mathbf X) = \mathbf x^{\rm H} \mathbf x$. Maybe this helps in the analysis. You may be more lucky trying a weaker form, something with "nondecreasing" instead of increasing might work (but I'm not sure).

edit: In fact, since $|\mathbf n^{\rm H} \mathbf x|^2 = (\mathbf n^{\rm H} \mathbf n)\cdot (\mathbf x^{\rm H} \mathbf x)\cdot \cos^2\sphericalangle(\mathbf n, \mathbf x)$ you can see the following:

  • If you rotate $\mathbf x$ then ${\rm trace}(\mathbf N \mathbf X)$ goes up or down (depending on the direction of rotation) while ${\rm trace}(\mathbf X)$ is constant.
  • If you scale $\mathbf x$ then ${\rm trace}(\mathbf N \mathbf X)$ grows and shrinks proportionally to ${\rm trace}(\mathbf X)$ where the proportionality factor is $(\mathbf n^{\rm H} \mathbf n)\cdot \cos^2\sphericalangle(\mathbf n, \mathbf x) \geq 0$.

Since both effects superpose you can get non-monotonic effects. Consider the example $\mathbf x = t \cdot [\cos(t), \sin(t)]^{\rm T}$ and $\mathbf n = [1,0]^{\rm T}$. Then ${\rm trace}\{\mathbf X\} = \mathbf x^{\rm H} \mathbf x = |t|^2$ and ${\rm trace}\{\mathbf N \mathbf X\} = |\mathbf n^{\rm H} \mathbf x|^2 = |t|^2\cdot \cos^2(t).$ Therefore, while ${\rm trace}\{\mathbf X\}$ grows monotonically for $t\geq 0$, we have that ${\rm trace}\{\mathbf N \mathbf X\}$ grows and then shrinks with a local maximum at $t \approx 0.8603$.

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    This is only true for the $N$ you've given, correct? – 2017-02-28
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    I've given a (couple of) counterexample(s), which proves that what the OP was assuming is wrong. That's a valid proof technique isn't it? ;-) – 2017-02-28
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    Completely! I do wonder if there's a stronger result here, though. – 2017-02-28
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    Got it! Thanks for your kind help! – 2017-03-01
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    You're welcome. I'm glad it helped. – 2017-03-01