# Are conjugate vectors unique?

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!

Take $$A=I_n$$. Then any orthogonal basis is a set of conjugate vectors.
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 $$B=B^T=DO^T$$.
• I was too lazy to find a short proof that it works for all $A$, but that's a good one. +1 – Arnaud Mortier Nov 17 '19 at 18:23
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.
• 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. – CR Drost Nov 18 '19 at 18:50