= Diffusion tensor identifiability
{title2=$\operatorname{rank}(A)=6$}
<Diffusion tensor identifiability> asks whether the measurements determine a unique <diffusion tensor> within a stated model. For a constant <symmetric matrix> $\mathbf D$, known matched baseline $S_0$ and known $b_i>0$, define
$$
y_i=-\frac{\log(S_i/S_0)}{b_i}=\mathbf g_i^T\mathbf D\mathbf g_i,
\qquad
\mathbf d=(D_{xx},D_{yy},D_{zz},D_{xy},D_{xz},D_{yz})^T.
$$
For unit directions $\mathbf g_i$, these equations give a <linear system> $A\mathbf d=\mathbf y$ with row
$$
A_i=(g_{ix}^2,g_{iy}^2,g_{iz}^2,2g_{ix}g_{iy},2g_{ix}g_{iz},2g_{iy}g_{iz}).
$$
\b[All six coefficients are uniquely determined in this unrestricted noise-free linear model exactly when $\operatorname{rank}(A)=6$.] The condition concerns independent equations, not merely the number of acquired images. Direction design and uncertainty in tensor-derived measurements were examined experimentally through controlled simulations by <Derek Jones (magnetic resonance researcher)>.<The effect of gradient sampling schemes on measures derived from diffusion tensor MRI: A Monte Carlo study by Derek K. Jones>[Primary Abstract: sampling comparisons; displayed rank criterion is an explicit linear-algebra derivation]{ref}
For a counterexample, use the six directions $\pm\mathbf e_x,\pm\mathbf e_y,\pm\mathbf e_z$. Each antipodal pair repeats a row, so the <matrix rank> is three. Only $D_{xx},D_{yy},D_{zz}$ are determined. Even restricting to physically positive tensors does not generally repair this: the identity matrix and a matrix with unit diagonal and $D_{xy}=D_{yx}=\varepsilon$ both fit those data for any $|\varepsilon|<1$, while having different off-diagonal entries.
Replace the negative axes by $(\mathbf e_x+\mathbf e_y)/\sqrt2$, $(\mathbf e_x+\mathbf e_z)/\sqrt2$ and $(\mathbf e_y+\mathbf e_z)/\sqrt2$. The axes first give the diagonal entries, then
$$
D_{xy}=y_{(\mathbf e_x+\mathbf e_y)/\sqrt2}-\frac{D_{xx}+D_{yy}}2,
$$
and similarly for $D_{xz}$ and $D_{yz}$. This directly proves uniqueness for the new design. If $D_{xx}=1.8$, $D_{yy}=0.9$ and $y_{(\mathbf e_x+\mathbf e_y)/\sqrt2}=1.55$, all in units of $10^{-3}\,\mathrm{mm^2/s}$, then $D_{xy}=0.20$ in those units.
\Image[/diffusion-tensor-gradient-design.png]
{title=Original design-matrix example: antipodal axes give rank three, while axes and mixed directions give rank six}
{id=diffusion-tensor-identifiability-design-matrices}
{description=Reused algebra illustration, not a reconstruction of a published acquisition or simulation. Rows represent measurements; columns represent the six independent tensor coefficients.}
Full <matrix rank> is not a guarantee of a stable estimate in noisy data. If $S_0$ must also be fitted, its logarithm is another unknown; the augmented signal model needs a corresponding independent baseline constraint. The six-coefficient argument above explicitly assumes $S_0$ is separately known. Likewise, <tractography> requires additional interpretation beyond uniqueness of a local tensor.
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