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If $\mathbf{x}$ is a vector and $Q$ a symmetric Matrix; Does $Q\mathbf{x} = \mathbf{x}^T Q$ ?

EDIT :

I have this: $$\frac{1}{2}\left(Q\mathbf{x} + \mathbf{x}^TQ\right) + \mathbf{c},$$ where $\mathbf{c}$ is also a vector, and I want to obtain $Q\mathbf{x}+\mathbf{c}$.

I thought that was the answer to my question, but it doesn't seem to be. Do you see a way to obtain this?

EDIT : Well Then I believe I made an error prior that, I can explain what I have done before to see if it's the source of the error.

I am trying to find the gradient of the function $$\frac{1}{2}\left(\mathbf{x}^TQ\mathbf{x}\right) + \mathbf{c}\mathbf{x}$$

that's how I got the first answer by finding the gradient.

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    Consider the case where $Q$ is the identity matrix.2011-01-18
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    After the edit: This is still nonsense. If $Q$ is $n\times n$, then for $Q\mathbf{x}$ and $\mathbf{x}^TQ$ to both make sense with x a vector you need $\mathbf{x}$ to be $n\times 1$. Then $Q\mathbf{x}$ is $n\times 1$, and $x^T\mathbf{Q}$ is $1\times n$; the only way you can add them, *and* add them to a $k\times 1$ vector $\mathbf{c}$, is for $n=k=1$ and for you to be working with *numbers*, not matrices and vectors; in which case you already have $Q\mathbf{x}+\mathbf{c}$, because $\frac{1}{2}(Q\mathbf{x}+\mathbf{x}^TQ)+\mathbf{c} = \frac{1}{2}(Qx+Qx)+c = Qx+c$.2011-01-18

3 Answers 3