# Matrix as a linear combination of eigenvalues and eigenvectors

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?

• 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. – stewbasic Aug 11 '16 at 5:56
• The second vector is not an eigenvector of this matrix. – amd Aug 11 '16 at 5:59
• 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. – ekkilop Aug 11 '16 at 7:04
• @stewbasic You are right, thanks, it works, but only for symmetric matrices. – OiciTrap Aug 11 '16 at 18:07
• @ekkilop Then it's SVD I guess. – OiciTrap Aug 11 '16 at 18:13