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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:

$A = \begin{pmatrix}-q_{1}-\\-q_{2}-\\-q_{n}-\end{pmatrix} \begin{pmatrix}\lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_n\end{pmatrix} \begin{pmatrix}|&|&|\\q_1&q_2&q_n\\|&|&|\end{pmatrix}$

$A_{ij} = \sum_k \lambda_kq_{ik}q_{jk}$

Are these two the same thing?:

$A_{ij} = \sum_k \lambda_kq_{ik}q_{jk}$ and $A = \sum_d \lambda_d q_dq_d^T$

My gut intuition says that my definition of $q_d$ is not the same as theirs. Perhaps the wikipedia proof https://en.wikipedia.org/wiki/Schur_product_theorem#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!

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    I don't see why you say that $A_{ij}=0$ for $i\neq j$. That the vectors $q_i$ are pairwise orthogonal implies that $\sum_{k}q_{ik}q_{jk}=\delta_{ij}$; but unless all the $\lambda_k$'s are equal, you can't deduce anything about $A_{ij} = \sum_k \lambda_kq_{ik}q_{jk}$. – 2017-01-15
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    What is the delta in this case? I thought if the two vectors are orthogonal, it means that the dot product(which is essentially the same as the sum) is zero. – 2017-01-15
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    $\delta_{ij}$ is the Kronecker symbol; it is $1$ if $i=j$ and $0$ otherwise. So yes it was just a way to say that $\sum_k q_{ik}q_{jk}=0$ when $i\neq j$. – 2017-01-15
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    I see what you mean now. I'll update the question accordingly then. I just can't find a way to get to the fact that a matrix $A = \sum_d \lambda_d q_dq_d^T$. – 2017-01-15

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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 $\lambda_iq_i$.

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    Thank you! It made a lot of sense. – 2017-01-15