How is this matrix product normally distributed?

Question

From Applied Multivariate Statistical Analysis:

In the red box of the below image, why is $(1/\sqrt {\lambda_i})e_i'(X-\mu)$ normally distributed?

From a previous result: If $X$ is distributed as $N_p(\mu, \Sigma)$ then $AX$ is distributed as $N_q (A \mu,A \Sigma A')$ where $A$ has dimensions $(q \times p)$.

But by this we'd have $(1/\sqrt {\lambda_i}) e'_i(X-\mu) \rightarrow (1/\sqrt {\lambda_i})e_i'\Sigma e_i$.

How is this equal to $1$?

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Answer 1

I figured it out:

$\text{Cov}(c e_i'(X-\mu)) = c^2 e_i' \Sigma e_i$.

$\Sigma e_i = \lambda_i e_i$.

Substituting results in: $c e_i' \lambda_i e_i$.

And $c^2 * \lambda * e_i' e_i = 1$.