What is the result of $x^{\top}A\dot{x}+\dot{x}^{\top}Ax$, provided that $A=A^{\top}$? Actually, I wanted to expand $d(x^{\top}Ax)\over{dt}$.
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Since both the right and left hand sides are scalars you have that $x^TA\dot{x} = \dot{x}^TAx$. The question is what $\dot{x}$ stands for? what result you are expecting? – user91011 Oct 06 '13 at 17:25
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2Assuming that $x$ is a column vector, and $\dot{x} = {dx \over dt}$, then I think your expression is an expansion of ${d \over dt}(x^T Ax)$. Is that what you mean by "result"? – Adrian Ratnapala Oct 06 '13 at 17:33
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@AdrianRatnapala yes I mean the expansion of $d(x^{\top}Ax)\over{dt}$. – Ansu Oct 06 '13 at 23:57
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@user91011 I guess $\dot{x}$ stands for time derivative of x, Does it mean other things? – Ansu Oct 06 '13 at 23:58
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The question is not complex, so It does not requires great deal of expansion, or does it? I thought we can ask any type of questions in this forum. So is there any wrong with this question? @azimut, – Ansu Oct 07 '13 at 00:59
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I'll use $y^T$ for the OP's $y^{\top}$.
Observe that $x^TA{\dot x}$, being a scalar quantity, is automatically symmetric: $(x^TA{\dot x})^T = x^TA{\dot x}$, and $({\dot x}^TAx)^T ={\dot x}^TAx$ as well. Now $(x^TA{\dot x})^T = {\dot x}^TA^Tx$, so we obtain $x^TA{\dot x} = {\dot x}^TA^Tx$. Thus
$x^TA{\dot x} + {\dot x}^TAx = {\dot x}^TA^Tx + {\dot x}^TAx = {\dot x}^T(A^T + A)x. \tag{1}$
Now using $A^T = A$, (1) becomes
$x^TA{\dot x} + {\dot x}^TAx = 2{\dot x} ^TAx; \tag{2}$
is that sort of thing we seek? It can't be taken much further, I'll warrant.
Robert Lewis
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@Ansu: $x^TA \dot x$ is a $1 \times 1$ matrix, so it must be symmetric; there are no off-diagonal entries! Just write it out in terms of components, $x^TA \dot x = \sum x_i A_{ij} \dot x_j$ etc. etc. etc. – Robert Lewis Oct 07 '13 at 04:07
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