\[
y_i=\begin{bmatrix}
y_{i1}\\
...\\
y_{id}
\end{bmatrix}
=\begin{bmatrix}
w_1^Tx_i\\
...\\
w_d^Tx_i
\end{bmatrix}
=W^Tx_i
\]
where
\[W=[w_1,w_2,...,w_d]\]
Multilinear Principle Component Analysis could be done via maximising the variance in each dimension:
\[
\begin{align*}
W_o&=\underset{W}{\arg\max}\quad\frac{1}{N}\sum_{k=1}^d\sum_{i=1}^N(y_{ik}-u_{ik})^2\\
&=\underset{W}{\arg\max}\quad\frac{1}{N}\sum_{k=1}^d\sum_{i=1}^Nw_k^T(x_i-u_i)(x_i-u_i)^Tw_k^T\\
&=\underset{W}{\arg\max}\quad\sum_{k=1}^dw_k^TS_tw_k\\
&=\underset{W}{\arg\max}\quad tr[W^TS_tW]\\
\end{align*}
\]
\[s.t. W^TW=I\]
Formulate the Lagrangian:
\[
\begin{align*}
\mathcal{L}(W,\Lambda)&=tr[W^TS_tW] -tr[\Lambda(W^TW-I)]\\
\frac{\partial \mathcal{L}(W,\Lambda)}{\partial W}&=2S_tW - 2W\Lambda\\
\end{align*}
\]
Let $\frac{\partial \mathcal{L}(W,\Lambda)}{\partial W}=0$
\[\Rightarrow\quad S_tW=W\Lambda\]
Take the above formula back to the original optimisation equation:
\[
\begin{align*}
W_o&=\underset{W}{\arg\max}\quad tr[W^TS_tW]\\
&=\underset{W}{\arg\max}\quad tr[W^TW\Lambda]\\
&=\underset{W}{\arg\max}\quad tr[\Lambda]\\
\end{align*}
\]
Therefore, one needs to maximise the eigenvalues of $S_t$, and the projection matrix $W=\{w_1,w_2,...,w_d\}$ corresponds to the eigenvectors of the largest $d$ eigenvalues.
Done.
\[
\begin{align*}
\mathcal{L}(W,\Lambda)&=tr[W^TS_tW] -tr[\Lambda(W^TW-I)]\\
\frac{\partial \mathcal{L}(W,\Lambda)}{\partial W}&=2S_tW - 2W\Lambda\\
\end{align*}
\]
Let $\frac{\partial \mathcal{L}(W,\Lambda)}{\partial W}=0$
\[\Rightarrow\quad S_tW=W\Lambda\]
Take the above formula back to the original optimisation equation:
\[
\begin{align*}
W_o&=\underset{W}{\arg\max}\quad tr[W^TS_tW]\\
&=\underset{W}{\arg\max}\quad tr[W^TW\Lambda]\\
&=\underset{W}{\arg\max}\quad tr[\Lambda]\\
\end{align*}
\]
Therefore, one needs to maximise the eigenvalues of $S_t$, and the projection matrix $W=\{w_1,w_2,...,w_d\}$ corresponds to the eigenvectors of the largest $d$ eigenvalues.
Done.
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