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Anyway, say there is some original matrix $M \in \mathbb{R}^{n\times n}$ that you are trying to reconstruct. Also let $M$ be symmetric and positive-definite. You have at your disposal $\Lambda$, a diagonal matrix who contains its eigenvalues and only a single normalized eigenvector, a column vector $v_1 \in \mathbb{R}^{n\times1}$. Originally I thought it would be possible because the eigenvectors of a symmetric matrix are orthogonal, so we could just find a basis for the null space of $v_1$.

For example do an singular value decomposition of $v_1$ so $v_1=USV$ where obviously $V=1$, as $v_1^Tv_1=1$ since $v_1$ is normalized. Also $v_1$ can only have rank $1$, so $S=[1,0,0,\ldots,0]^T$. Lastly we consider $U \in \mathbb{R}^{n\times n}$, found from eigenvectors of $v_1v_1^T$, which should be orthogonal as $v_1v_1^T$ is also symmetric. Breaking $U$ into column vectors gives $U=[u_1,u_2,\ldots,u_n]$. See $v_1=u_1$ as $$v_1 = [u_1,u_2,\ldots,u_n] \cdot [1,0,0,\ldots,0]^T\cdot1=u_1.$$

Thus, since $u_2,\ldots,u_n$ are orthogonal to $u_1$, and therefore $v_1$, we have a set of orthogonal vectors we could potentially use to reconstruct $M$ by its diagonalization, so $$M=U\cdot\Lambda\cdot U^T.$$

Thoughts on why this process is wrong?

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"Given $\Lambda$ and $v_1$, construct $M$" is your question in short. Let $\lambda_k$ and $v_k$ be the $k^{th}$ eigenpair. $M$ is written as $$M = \sum\limits_{k=1}^N \lambda_k v_kv_k^T$$ In the above equation, you have $\lambda_1 \ldots \lambda_N$ and $v_1$; for complete reconstruction, you need $v_2 \ldots v_N$.

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  • $\begingroup$ You could get away with $v_2\ldots v_{N-1}$, since the $v_i$ are orthonormal. $\endgroup$ – Sean Lake Oct 20 '16 at 0:20
  • $\begingroup$ How can you do this @SeanLake $\endgroup$ – Ahmad Bazzi Oct 20 '16 at 0:21
  • $\begingroup$ @ElBazzi If you have $v_1,\ldots,v_{N-1}$ there are only two ways to choose a unit vector $v_N$ orthogonal to $v_1,\ldots,v_{N-1}$, and these two choices differ by sign (which doesn't affect your formula for $M$). $\endgroup$ – stewbasic Oct 20 '16 at 0:27
  • $\begingroup$ Given $N-1$ linearly independent vectors in an $N$ dimensional vector space you can construct a vector independent of all of the $N-1$ vectors by either applying Gram-Schmidt orthogonalization with a lucky guess, or setting up a determinant of a matrix where the first row is the basis vectors, and the other rows are the components of the $N-1$ independent vectors (see: multilinear algebra ). $\endgroup$ – Sean Lake Oct 20 '16 at 0:27
  • $\begingroup$ 1)You could construct many orthonormal basis 2)We all know that, but which vector corresponds to which eigenvalue ? $\endgroup$ – Ahmad Bazzi Oct 20 '16 at 0:28

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