In the middle of trying to prove Schur's product theorem, I got stuck on trying to
prove that a positive definite matrix $A$ can be expressed as:$$\sum_d \lambda_d q_dq_d^T$$
Given the eigendecomposition of $A = QLQ'$, we have that $Q^{-1} = Q'$, or that $Q$ is orthogonal.
$Q'$ is also orthogonal, because $Q'Q = (QQ')' = I' = I$.
We know that $Q'$ has orthogonal column vectors $q_i$ that are now the orthogonal row vectors $q_i$ on $Q$.
After matrix multiplication using sum notation on $QLQ'$, we have that:
$A = \begin{pmatrix}-q_{1}-\\-q_{2}-\\-q_{n}-\end{pmatrix} \begin{pmatrix}\lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_n\end{pmatrix} \begin{pmatrix}|&|&|\\q_1&q_2&q_n\\|&|&|\end{pmatrix}$
$A_{ij} = \sum\limits_k \lambda_kq_{ik}q_{jk}$
Are these two
$A_{ij} = \sum\limits_k \lambda_kq_{ik}q_{jk}$ and $A = \sum\limits_d \lambda_d q_dq_d^T$ the same thing?
My gut intuition says that my definition of $q_d$ is not the same as theirs. Perhaps the wikipedia proof of positive semidefiniteness's notation meant the column vectors of $A$.
Could someone please point out where my logic went wrong here? Thank you!
Your gut intuition is right, the $q_i$'s should be the column of $Q$. One way to see this is to show that $A$ and $\sum_k \lambda_k q_kq_k^T$ agree on a basis of $\Bbb R^n$. This is particularly easy for the basis of eigenvectors of $A$, which are precisely the columns of $Q$; indeed all you have to show is that $$\left(\sum_k \lambda_k q_kq_k^T\right)\cdot q_i=\lambda_iq_i=A\cdot q_i$$ for all $i$; and this is true because $q_k^T\cdot q_i$ is $0$ if $k\neq i$ and $1$ if $i=k$, so that the only non-zero term of $\sum_k \lambda_k q_k \cdot (q_k^T \cdot q_i)$ is $\lambda_iq_i$.