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If I know the values of $\lambda$ and $A$, how do I find a vector $v$ such that $\lambda = v^{\intercal}Av $?

This isn't a homework question; I just ran into this problem in Real Life and realized I couldn't solve it!

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    If $A$ is a positive definite symmetric matrix then for given \lambda>0 the equation $v^TAv=\lambda$ describes an ellipsoid centered at the origin.2012-07-23

3 Answers 3

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Well, if $A$ is positive-definite one may compute a Cholesky decomposition $ A = U^{\intercal} U$ where $U$ is an upper triangular matrix, so your equation reduces to $||Uv||^2=\lambda $ Pick any vector $y$ on the $n$-sphere of radius $\sqrt\lambda$, solve $Uv=y$ for $v$ - which should be trivial, since, as I said above, $U$ is upper triangular - and you've found your solution.

Of course this solution is immensely more expensive from a computational standpoint, but it has the advantage that it gives you a way to generate all possible solutions by varying $y$.

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Start with any vector $v$ such that $v^{\intercal}Av\ne0$ and then

$v\rightarrow\sqrt{c}\ v\ \Longrightarrow\ v^{\intercal}Av\rightarrow c\ v^{\intercal}Av,$

i.e. you can make the expression take any value.

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If $A$ is positive definite then for any $\lambda>0$ pick any vector $v\not=0$ and compute $ c=v^TAv $ Then $ v\sqrt{\frac{\lambda}{c}} $ satisfies your equation.