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I'm trying to follow these lecture notes on Linear Discriminant Analysis (LDA) but I can't seem to figure out how the author gets from:

$$ \Sigma_{x\epsilon\omega_{i}} (w^{T}x - w^{T}\mu_{i})^2$$

to

$$ \Sigma_{x\epsilon\omega_{i}} w^{T}(x-\mu_{i})(x-\mu_{i})^Tw$$

doraemonpaul
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jessems
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1 Answers1

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Since, $w^{T}x - w^{T}\mu_{i}$ is a scalar we can write it as:

$w^{T}x - w^{T}\mu_{i} = (w^{T}x - w^{T}\mu_{i})^T$

Thus,

$(w^{T}x - w^{T}\mu_{i})^2 = (w^{T}x - w^{T}\mu_{i}) (w^{T}x - w^{T}\mu_{i})^T$

Now, we know that:

$(AB)^T = B^TA^T$

Thus, we have:

$(w^{T}x - w^{T}\mu_{i}) (w^{T}x - w^{T}\mu_{i})^T = w^{T} (x - \mu_{i}) (w^{T} (x - \mu_{i}))^T$$

Or in other words:

$(w^{T}x - w^{T}\mu_{i}) (w^{T}x - w^{T}\mu_{i})^T = w^{T} (x - \mu_{i}) (x - \mu_{i})^T w$

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