Given a symmetric matrix $A$ that has eigenvalues $4, 3, 2,$ and $2$ and the eigenvectors belonging to the eigenvalues $4$ and $3$. Provide a procedure to reconstruct the entire matrix.

Since the matrix is symmetric then it can be diagonalized. Let $Av = 4v, Aw = 3w$. Since the eigenvalue $2$ is repeated we want to find two linearly independent eigenvectors corresponding to the eigenvalue $2$.

Since the matrix is symmetric then the eigenvectors are orthogonal. If $z, h$ are eigenvectors corresponding to $2$ then $z^Tv = z^Tw = h^Tv = h^Tw = 0$. Therefore, both $z,h$ should be orthogonal to a two dimensional vector space $S$ constructed from the basis vectors $v,w$.

How can $z,h$ be found?


1 Answer 1


I will assume that the $v,w$ given are unit-vectors. Because $A$ is symmetric, they must be orthogonal.

If you want to find $z$ and $h$, then it suffices to find an orthonormal basis for the nullspace of the $2 \times 4$ matrix whose rows are $v^T, w^T$. However, there's a way to reconstruct $A$ that does not require the computation of $z$ and $h$.

  • Verify that the matrix $B = 2vv^T + ww^T$ is symmetric with eigenvalues $2,1,0,0$. $v$ is an eigenvector of $B$ associated with the eigenvalue $2$ and $w$ is an eigenvector of $B$ associated with the eigenvalue $1$.

  • Using this, conclude that $A = B + 2I$ (where $I$ denotes the identity matrix) has all the properties required of $A$.

  • $\begingroup$ I thought about the first method with an orthonormal basis. However, why I was reluctant to write it here because I don't know if any vector from the orthonormal basis vectors will satisfy the given relation about the eigenvalues. Suppose, $z,h$ are forming the orthonormal basis now the question is will the $Az = 2z$ and $Ah = 2h$ be preserved i.e. will they be necessarily eigenvectors of $A$ corresponding to the eigenvalue $2$ or should we take some linear combination of them to get the eigenvectors? $\endgroup$
    – user13
    Feb 10, 2021 at 14:56
  • $\begingroup$ @user13 Any vectors that are orthogonal to both $v$ and $w$ must be eigenvectors of $A$ associated with the eigenvalue $2$. $\endgroup$ Feb 10, 2021 at 15:26

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