Starting from this equation to estimate: $\DeclareMathOperator{\tr}{Trace}$ $$ X = F^k \Lambda' ^ {k} + e, $$
with $ X$ an $T\times N $ matrix, $ F$ is $T\times k $ and $ \Lambda$ is $k\times k $, I need to show that the solution to this problem:
$$\min _ { \Lambda ^ { k }} S ( k ) , \quad \text { with } \quad S ( k ) = ( N T ) ^ { - 1 } \sum _ { i = 1 } ^ { N } \sum _ { t = 1 } ^ { T } \left( x _ { i t } - \lambda _ { i } ^ { k \prime } F _ { t } ^ { k } \right) ^ { 2 },$$
whit the constraint that $F ^ { k \prime } F ^ { k } / T = I _ { k } $ and $\Lambda ^ { k \prime } \Lambda ^ { k } $ is diagonal, is:
$ F^k = \sqrt{T}\times \text{Eigenvectors of k largest eigenvalues of X'X}$ and $ \Lambda^k = F ^ { k \prime } X / T,$ that is a scaled version of the Principal Component estimator.
I procedeed in this way, starting from the first component:
\begin{align} \min_{\lambda^ { 1 }} \ \ &( N T ) ^ { - 1 } \sum _ { i = 1 } ^ { N } \sum _ { t = 1 } ^ { T } \left( x _ { i t } - \lambda _ { i } ^ { 1 \prime } F _ { t } ^ { 1 } \right) ^ { 2 } ,(\text{dropping the superscript 1 from now on)} = \\ & (N T ) ^ { - 1 } \tr[(X - F\lambda')'(X-F\lambda')] =\\ & (N T ) ^ { - 1 } \tr(X'X) + (N T ) ^ { - 1 }2F'X\lambda + (N) ^ { - 1 }\lambda'\lambda, \end{align}
so that
\begin{align} \frac{\partial S}{\partial\lambda} = 0 \iff& -2X'F/NT = 2\lambda /N \\ &\lambda = X'F/T. \end{align}
However, once I substitute this back I arrive to maximize: $$ (N T^2 ) ^ { - 1 } F^1XX'F^1, $$ and something seems off, since:
$$\max _ { F ^ { 1 }} \quad (N T^2 ) ^ { - 1 } F^1XX'F^1 \quad s.t. \ \ F'^1F^1 = T $$
does not lead to finding the first eigenvalues of $ XX'$ and $ F^1 = \sqrt{T}\times \text{Eigenvectors of 1 largest eigenvalue}$ as it shold be.
I suspect I am doing something wrong with the Trace operatore, as I am not really familiar with it.