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?
Best Answer
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 $A$.