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I've been told that I can write a square matrix $A$ like this: $\lambda_1 V_1 V_1^{T} + \lambda_2 V_2 V_2^{T} + ... \lambda_n V_n V_n^{T}$, i.e., as an linear combinations of eigenvalues and the outer product of their associated eigenvectors ($V_i$ is unitarian), but for example, the following matrix:

[  1.    0.5  -0.1]
[  0.5   1.   10. ]
[  2.    3.    5. ]

Have these eigenvalues: 8.93009292, 0.69188007 and -2.62197299 and the respective eigenvectors [ 0.04161113 0.78386684 0.61953313], [-0.0262233 0.94920296 0.3135699 ] and [ 0.03123074 0.91609679 0.39973905].

But doing the linear combination as above I get this:

[ 0.01433413  0.37633714  0.26041067]
[ 0.37633714  3.70535792  3.80969252]
[ 0.26041067  3.80969252  3.28030794]

This is nothing near the original matrix, what am I doing wrong?

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  • $\begingroup$ Note that your expression is always symmetric, since $(V_1V_1^T)^T=V_1V_1^T$. Thus only symmetric matrices can be written in that form. $\endgroup$ – stewbasic Aug 11 '16 at 5:56
  • $\begingroup$ The second vector is not an eigenvector of this matrix. $\endgroup$ – amd Aug 11 '16 at 5:59
  • $\begingroup$ Perhaps you are thinking of the singular value decomposition (SVD)? You can write any matrix as a sum of terms of the form $\sigma_j u_j v_j^T$, where $\sigma_j$ is the $j$th singular value, and $u_j$ and $v_j$ are the $j$th left and right singular vectors respectively. For a symmetric positive definite matrix, the SVD becomes the series you have written above. $\endgroup$ – ekkilop Aug 11 '16 at 7:04
  • $\begingroup$ @stewbasic You are right, thanks, it works, but only for symmetric matrices. $\endgroup$ – OiciTrap Aug 11 '16 at 18:07
  • $\begingroup$ @ekkilop Then it's SVD I guess. $\endgroup$ – OiciTrap Aug 11 '16 at 18:13

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