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
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 $$\text{tr}(AA^T)=\text{tr}(U(D^2U^T))=\text{tr}((D^2U^T)U)=\text{tr}(D^2)$$