I want to reconstruct the original $3\times3$ symmetric matrix (namely $D$) whose all Eigen values ($\{\lambda_1,\lambda_2,\lambda_3\}$) and only the first Eigen vector ($e_1$} corresponding to the first Eigen value ($\lambda_1$) are given. If we know that these Eigen vectors are perpendicular to each other (really, these would be the unit vectors after normalization in different coordinate), is it possible to reconstruct the $D$ (the original $3\times3$ matrix)? I tried to apply the Least Square minimization method to estimate $D$ through $D . e_1 =\lambda_1 . e_1$, and again estimate $e_2$ and $e_3$ based on this estimated $D$ however this could not reconstruct a reliable matrix because it was not symmetric and $\dots$ In my problem $D$ is a tensor that represents a $3D$ ellipsoid whose Eigen values and only the longest Eigen vector (the main direction of the ellipsoid) are given. Now I need to reconstruct the original tensor matrix. Thanks in advance (Khosrow)

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    $\begingroup$ I don't see a question. $\endgroup$ Mar 7, 2015 at 12:09
  • $\begingroup$ I think you're saying you know $\lambda_1,\lambda_2,\lambda_3$, and $v_1$, and also know that $v_1,v_2,v_3$ are all orthogonal. Now you want to reconstruct $v_2,v_3$. It's impossible: you can pick $v_2$ to be any vector orthogonal to $v_1$ (and then choose $v_3$ orthogonal to both). Many of these will be different. $\endgroup$
    – Ian
    Mar 7, 2015 at 12:25
  • $\begingroup$ For instance, assume that $v_1$ is the first coordinate vector $e_1$. You could have a diagonal matrix, if $v_2=e_2$ and $v_3=e_3$. Or you could have a nondiagonal matrix, if $v_2=e_2+e_3$ and $v_3=-e_2+e_3$. There are many other possibilities. $\endgroup$
    – Ian
    Mar 7, 2015 at 12:28
  • $\begingroup$ Thanks Ian, Really I wanted to define the matrix A given Eigen values and the longest Eigen vector namely e1 where e1,e2 and e3 are my new coordinate while e2 and e3 are also other Eigen vectors relating to the matrix A (we are looking for) and their Eigen values ( which are known). Thanks again... $\endgroup$
    – user221702
    Mar 7, 2015 at 13:48
  • $\begingroup$ If what you wanted, 221702, is not what's in your question, then I recommend editing your question so it asks what you actually wanted. Maybe then someone will be able to answer it. $\endgroup$ Mar 8, 2015 at 6:04

1 Answer 1


Can't be done. Exercise: find two distinct symmetric $3\times3$ matrices, each of which has eigenvalues $1$, $2$, and $3$, with eigenvector $(1,0,0)$ corresponding to the eigenvalue $3$. One such matrix is $$\pmatrix{3&0&0\cr0&2&0\cr0&0&1\cr}$$ but it shouldn't be hard to find another; all you need is a $2\times2$ symmetric, nondiagonal matrix with eigenvalues $1$ and $2$.

  • $\begingroup$ Thanks a lot Gerry, but since we can compute the original matrix through equation $D=E.\Lambda.E^{-1}$, and also we know that the first Eigen vector is perpendicular to a surface (a circle maybe) including orthogonal vectors $e_2$ and $e_3$ the remaining things is that how we can find $e_2$ and $e_3$ ? is it also impossible? (Khosrow) $\endgroup$
    – user221702
    Mar 8, 2015 at 11:34
  • $\begingroup$ Well, why don't you do the exercise, and then see what the eigenvectors are? $\endgroup$ Mar 8, 2015 at 11:37
  • $\begingroup$ Ok, but I am looking for a mathematical model not solving an example .. I have a set of unit vectors that each one would be the Eigen vector that we say is known (such as $e_1=[0.2161; -0.0103; -0.9703]$) and the Eigen values are fixed that are $\{\lambda_1=0.002 , \lambda_2=0.0003, \lambda_3=0.0003 \}$ always. I did the exercise but actually I did not understand how doing the exercise can solve my problem. I need a mathematical model... $\endgroup$
    – user221702
    Mar 8, 2015 at 12:08
  • $\begingroup$ The exercise was intended to show you that given $e_1$, $\lambda_1$, $\lambda_2$, and $\lambda_3$, there are infinitely many possibilities for $e_2$ and $e_3$ and the matrix $D$. $\endgroup$ Mar 8, 2015 at 22:00
  • $\begingroup$ So, do you understand the situation now, 221702? $\endgroup$ Mar 10, 2015 at 5:24

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