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I have the following equation to solve, $g(x) = x^t W_i x + {W_i}^t x + v_{i_0}$

In this equation why the need to use a $x^t$ and $x$? I feel $x$ and transpose of it both are the same ($x$ is a row vector with two values)

Additionally, when you have such a situation and when $w_i$ is another column vector with two components how can I solve it with the inner products given?

Imagine a situation where

$x = (x_1, x_2)$ and $W_1 = \binom{6}{3}$

Maybe my question sounds dumb, but I am very new to linear algebra, sorry about that.

Any help is deeply appreciated.

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    Yeah, that is exactly what I meant. Thank you.2012-03-16

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Given $x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$ and $W_1 = \begin{pmatrix} 6 \\ 3 \end{pmatrix}$ then

$ x^\top\, W_1 = x_1*6 + x_2*3 $

which is a scalar, but

$ x \,W^\top = \begin{pmatrix} x_1*6 & x_1*3 \\ x_2*6 & x_2*3 \end{pmatrix} $

which is a $2\times2$ matrix.

The first is the inner product, and the second is the outer product. Here I use $\fbox{ }^\top$ (\top in LaTeX) for the transpose operator.

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We can't really tell unless we know the origin of this equation.

But anyways, obviously, the final result $g(x)$ is a scalar. That is, $1\times 1$ matrix. The dimensions of LHS has to match the dimensions of RHS. Also, the dimensions of any matrix-matrix (or matrix-vector) product has to match the rule $ \underbrace{A}_{m\times n} = \underbrace{B}_{m\times k} \times \underbrace{C}_{k\times n} $

I can't find a pretty illustration off the web right now, but here is something from Wikipedia.

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    Thanks J.D. That is true, and I find it always convenient to learn new things with places like SE and guys like you all.2012-03-16