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So I think I understand eigen decomposition where a matrix M is equal to Q D inverse(Q), where Q is a matrix formed by the eigenvectors of M, and D is a diagonal matrix of the eigenvalues. But I came across this proof on Wikipedia that uses eigen decomposition, but I'm not sure how to read the notation:

http://en.wikipedia.org/wiki/Schur_product_theorem#Proof_using_eigendecomposition

I'm reading this as "sum of each [eigenvalue times the corresponding eigenvector times the transpose of the corresponding eigenvector]," which doesn't quite work on an example I tried

$M= \begin{pmatrix}2 & 3\\ 3 & 2\end{pmatrix}$ with eigenvalues $-1, 5$ and corresponding eigenvectors $\begin{pmatrix}1 & -1\end{pmatrix}$ and $\begin{pmatrix}1 & 1\end{pmatrix}$

I end with [-1 1][1 -1] + [5 5][5 5] = [4 6][6 4] which is 2M, not M.

Could someone explain

  1. where I'm going wrong

  2. the relationship between this notation and $M = Q D Q^{-1}$

Thanks in advance!

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  • $\begingroup$ you have to make the length of the eigenvectors unity. $\endgroup$ – abel Jan 14 '15 at 23:32
  • $\begingroup$ I'm afraid I don't understand. What do you mean by "unity"? $\endgroup$ – guest Jan 14 '15 at 23:39
  • $\begingroup$ I think he means "unit-y", like an adjective for "being like a unit". Which means length $1$ where $1$ is the scalar identity. $\endgroup$ – Axoren Jan 15 '15 at 0:02
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you make $[1, -1]$ which is of length $\sqrt 2$ so by diving by this $[1/\sqrt 2, -1/ \sqrt 2]$ is an eigenvector of length $1.$ in the same way $[1/\sqrt 2, 1/\sqrt 2 ]$ is the other unit eigenvector. now form the matrix $\begin{align} A &= -1\pmatrix{1/\sqrt 2 \\-1/\sqrt 2}\pmatrix{1/\sqrt 2 & -1/\sqrt 2} + 5\pmatrix{1/\sqrt 2 \\1/\sqrt 2}\pmatrix{1/\sqrt 2 & 1/\sqrt 2}\\ &= -\dfrac{1}{2}\pmatrix{1&-1\\-1&1} + \dfrac{5}{2}\pmatrix{1&1\\1&1}\\ &=\pmatrix{2&3\\3&2} \end{align}$

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