8
$\begingroup$

A set of nonzero vectors $\{p_0,p_1,\ldots,p_{n-1}\}$ is said to be conjugate with respect to a symmetric positive definite matrix $A$ if $$ p_i^{\mathrm T}Ap_j=0 $$ for all $i\ne j$. Such vectors are used in the conjugate gradient method. It follows that conjugate vectors are also linearly independent. An example of conjugate vectors are the eigenvectors of the matrix $A$.

Are the conjugate vectors unique up to a scalar multiple? In other words, if there are two sets of conjugate vectors with respect to the same symmetric positive definite matrix $A$, are the vectors going to be the same up to a scalar multiple?

I would guess that that they are unique up to a scalar multiple, but I am not sure.

Any help is much appreciated!

$\endgroup$
16
$\begingroup$

Take $A=I_n$. Then any orthogonal basis is a set of conjugate vectors.

$\endgroup$
12
$\begingroup$

In general, no. If $A=B^TB$ for some square matrix $B$, $p_i^TAp_j=(Bp_i)^T(Bp_j)$. Since such a $B$ is invertible, any orthogonal basis $e_i$ of the original vector space gives a choice of the $p_i$ as $B^{-1}e_i$. In fact, a diagonalisation $A=OD^2O^T$ will allow us to choose $B=B^T=DO^T$.

$\endgroup$
  • $\begingroup$ I was too lazy to find a short proof that it works for all $A$, but that's a good one. +1 $\endgroup$ – Arnaud Mortier Nov 17 '19 at 18:23
4
$\begingroup$

No, they are far from unique.

Let $P$ be the matrix whose columns are the vectors $p_i$. Then, the vectors form a conjugate family with respect to a symmetric positive definite matrix $A$ if and only if $P^TAP=D$ is a diagonal matrix. You are asking if $P$ is unique up to scalar multiple, i.e. multiplying on the right by a diagonal matrix. If you enforce that the diagonal of $D$ is non-zero, then you can consider $P'=PD^{-1/2}$, which satisfies $P'^TAP'=I$, the identity matrix. Now replacing $P'$ with $P'O$ for any orthogonal matrix $O$ results in $(P'O)^TA(P'O)=I$, forming another conjugate family from the one we started with. If instead there are some zeros on the diagonal, you get even more degeneracy: the matrix $D$ can be block decomposed into a block non-zero diagonal entries, and the rest a block of zeros. Then you can perform the same procedure mentioned above on the non-zero block, and apply any transformation whatsoever to the zero portion.

$\endgroup$
  • $\begingroup$ I like this approach a lot; the insight can be made a little more didactically simple as “take two eigenvectors $a,b$ of $A$ with nonzero eigenvalue and form $\bar a=a/\sqrt{\lambda_a},\bar b=b/\sqrt{\lambda_b}.$ Then $(\bar a + \bar b)^TA(\bar a - \bar b)=a^T a - b^T b = 0$ so $\bar a \pm \bar b$ can replace $\bar a, \bar b$ for a conjugate family. $\endgroup$ – CR Drost Nov 18 '19 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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