Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
Nothing special about your $A$? –  Guess who it is. 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
@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... –  Guess who it is. Jul 23 '12 at 18:09
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.

share|improve this answer

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.

share|improve this answer

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 $y$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.