corrected some pages

Several small mistakes/errors were corrected in several proofs/definitions.

Author: JoramSoch

I/ToC.md

@@ -452,7 +452,7 @@ title: "Table of Contents"
    &emsp;&ensp; 5.1.4. **[Kullback-Leibler divergence](/P/matn-kl)** <br>
    &emsp;&ensp; 5.1.5. **[Linear transformation](/P/matn-ltt)** <br>
    &emsp;&ensp; 5.1.6. **[Transposition](/P/matn-trans)** <br>
-   &emsp;&ensp; 5.1.7. **[Drawing samples](/P/matn-ltt)** <br>
+   &emsp;&ensp; 5.1.7. **[Drawing samples](/P/matn-samp)** <br>
 
    5.2. Wishart distribution <br>
    &emsp;&ensp; 5.2.1. *[Definition](/D/wish)* <br>

P/matn-samp.md

@@ -11,7 +11,7 @@ title: "Sampling from the matrix-normal distribution"
 chapter: "Probability Distributions"
 section: "Matrix-variate continuous distributions"
 topic: "Matrix-normal distribution"
-theorem: "Linear transformation"
+theorem: "Drawing samples"
 
 sources:
   - authors: "Wikipedia"
@@ -67,4 +67,4 @@ Y = M + AXB &\sim \mathcal{MN}\left(M + A 0_{np} B, A I_n A^\mathrm{T}, B^\mathr
 \end{split}
 $$
 
-Thus, given $X$ defined by \eqref{eq:xij-dist}, $Y$ defined by \eqref{eq:matn-samp} is a [sample](/D/dist) from $\mathcal{N}\left(M, U, V \right)$.
+Thus, given $X$ defined by \eqref{eq:xij-dist}, $Y$ defined by \eqref{eq:matn-samp} is a [sample](/D/samp) from $\mathcal{N}\left(M, U, V \right)$.

P/mean-qf.md

@@ -63,7 +63,7 @@ $$ \label{eq:mean-qf-s2}
 \mathrm{E}\left[ X^\mathrm{T} A X \right] =  \mathrm{E}\left[ \mathrm{tr} \left( A X X^\mathrm{T} \right) \right] \; .
 $$
 
-Because [mean and trace are linear operators](/P/mean-lin), we have
+Because [mean and trace are linear operators](/P/mean-tr), we have
 
 $$ \label{eq:mean-qf-s3}
 \mathrm{E}\left[ X^\mathrm{T} A X \right] =  \mathrm{tr} \left( A \; \mathrm{E}\left[ X X^\mathrm{T} \right] \right) \; .

P/mean-tr.md

@@ -27,7 +27,7 @@ username: "JoramSoch"
 ---
 
 
-**Theorem:** Let $A$ be an $n \times n$ [random matrix](/D/rmat). Then, the [expected value](/D/mean) of the trace of $A$ is equal to the trace of the [expectation](/D/mean) of $A$:
+**Theorem:** Let $A$ be an $n \times n$ [random matrix](/D/rmat). Then, the [expectation](/D/mean) of the trace of $A$ is equal to the trace of the [expectation](/D/mean) of $A$:
 
 $$ \label{eq:mean-tr}
 \mathrm{E}\left[ \mathrm{tr}(A) \right] = \mathrm{tr}\left( \mathrm{E}[A] \right) \; .

P/mvn-kl.md

@@ -76,7 +76,7 @@ $$ \label{eq:mvn-KL-s3}
 \end{split}
 $$
 
-Because trace function and [expected value](/D/mean) are both linear operators, the expectation can be moved inside the trace:
+Because [trace function and expected value are both linear operators](/P/mean-tr), the expectation can be moved inside the trace:
 
 $$ \label{eq:mvn-KL-s4}
 \begin{split}