# Factorize a Symmetric matrix as an 'Approximation' with an outer product.

(deprecated-taken back based on discussion(OLD)) What is a good way to factor a symmetric matrix $X$ as an outer product of two vectors $u$ and $v$. i.e, Find two vectors $u$ and $v$ such that $X=uv^T$, where $X$ is a symmetric matrix.

(Updated/ New Question of interest(NEW)) Given a symmetric matrix $X$, what is a way to figure out the best possible vectors $u$ and $v$ such that the error under say an l2 loss over $X-uv^T$ is minimum. Feel free to make notes about any optimality conditions/ assumptions that might go around this problem.

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This is nearly identical to your previous question with the word "diagonal" replaced with "symmetric". As before, this is not possible in general. This time, it is because first of all $uv^T$ is only symmetric if $u = v$, and secondly if $X$ has rank $>1$, for example $X = \begin{bmatrix}1&0\\0&1\end{bmatrix}$, there is no $u$ such that $X = uu^T$. –  Rahul Jan 25 '12 at 14:45
However, you can write your matrix as a sum of outer products of vectors, i.e. $X = \sum_{i = 1}^r \alpha_i u_i u_i^T$, where the least $r$ possible is the rank of the matrix. Do you want to ask about this instead? –  Calle Jan 25 '12 at 14:48
@ Rahul, thank you for your answer. What if I would like to have an approximation of the matrix X? What is a way to figure out the best possible u,v under say an l2 loss over X-u.v^T. –  user23600 Jan 25 '12 at 14:55
@Calle- I believe that is a spectral(eigen) decomposition. –  user23600 Jan 25 '12 at 14:56
If you want to change the Question, per your reply to @RahulNarain, then please edit the Question to reflect that change. An Answer along the lines of Calle's comment would be forthcoming. –  hardmath Jan 25 '12 at 15:04

Since $X$ is symmetric it is always possible to put it in diagonal form using an orthonormal basis, i.e., $X = UDU^T$, where $D = \operatorname{diag}(\lambda_1, \lambda_2, \dots, \lambda_n)$ are the eigenvalues of $X$. From this it is possible to see that $$X = \sum_{i = 1}^n \lambda_i u_i u_i^T$$ where $u_i$ is the $i$:th column of $U$.
Now, assuming the eigenvalues are ordered by absolute value, $|\lambda_1| \geq |\lambda_2| \geq \dots \geq |\lambda_n|$, the best approximation $\tilde X = u v^T$, in the sense that $\|X - \tilde X\|_2$ is minimized ($\| \cdot \|_2$ is the Frobenius norm) is given by $$\tilde X = \lambda_1 u_1 u_1^T$$ (you can of course write $v = \lambda_1 u_1$ if you want).
+1, nice answer. However, the Frobenius norm is typically denoted $\lVert\cdot\rVert_F$. The notation $\lVert\cdot\rVert_2$ denotes the operator norm induced by the $2$-norm on vectors, $\lVert A\rVert_2 = \max_{x\ne0}\lVert Ax\rVert_2/\lVert x\rVert_2$. –  Rahul Jan 26 '12 at 0:34
Sometimes this is true, sometimes not (e.g. Horn and Johson uses $\| \cdot \|_2$ for the Frobenius norm in their book Matrix Analysis). –  Calle Jan 26 '12 at 12:00