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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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Nothing special about your $A$? –  J. M. Jul 23 '12 at 18:05
    
Well it's positive definite, if that's necessary to solve it. –  Brian Gordon Jul 23 '12 at 18:05
    
@Cocopuffs A is positive definite and $\lambda$ will be greater than 0. I didn't realize these details would be relevant :) What do you mean by matrix square root? If you add an answer I'll accept it. –  Brian Gordon Jul 23 '12 at 18:06
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@Brian, the devil is often in the details. Most problems in linear algebra only become tractable because there is exploitable structure or symmetry in them... –  J. M. Jul 23 '12 at 18:09
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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. –  Christian Blatter Jul 23 '12 at 19:09

3 Answers 3

up vote 5 down vote accepted

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.

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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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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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