Let $E=D^2$ be a square of diagonal matrix D in a n SVD decomposition
$$A=UDV' $$
Find relationship bw trace D and that of $AA'$ when $A_{Nxt}$. THE $Nxt$ data matrix where $N<t$
Data Matrix A is decompostion into
$$ A_{nxt}=U_{nxm}D_{mxm}V'_{mxt}$$
where n is spatial dimension $t$,t temporal length $m=min(n,t)$, adn $V'$ the transpose of v
- U spatial pattern matrix
- D diagonal level matrix
-
V temporal matrix
computing $A A^t$
$$\begin{aligned}
A A^t &= [UDV'][UDV']^t
\\ &=[UDV'][V'^t D^t U^t]
\\ &= UDV^tVD^t U^t &&\text{ using comment }
\\ &= UDID^tU^t
\\ &= UDD^tU^t
\\ &= UD^2U^t &&\text{ U have eigenvecteros in it}
\\ &= \vdots
\end{aligned} $$From strang's book $$A^t A = V
\begin{Bmatrix}
\delta^2_1 && 0 \\ 0 && \delta^2_2
\end{Bmatrix} V^t$$
where $\delta$ are eigenvalues , $v$ are eigenvectors
Guessing that Trace of $E=tr(AA^t)$
have trouble make an elegant argument. Thinking need to simplify $A A^t$ with the properites of the decompose matrix
Best Answer
Note that $A=UDV^T$ where $V^TV=I$ and $U^TU=I$, $D$ is diagonal ($D^T=D$), and $U$ and $V$ are unitary (i.e. $U^T=U^{-1}$ and $V^T=V^{-1}$).
Then $$AA^T=(UDV^T)(UDV^T)^T=UDV^TVDU^T=UD^2U^T$$ Finally, since generally $\text{tr}(XY)=\text{tr}(YX)$, we have $$\text{tr}(AA^T)=\text{tr}(U(D^2U^T))=\text{tr}((D^2U^T)U)=\text{tr}(D^2)$$