Why Partition Global Inertia?
In multiple factor analysis (MFA), blocks are first scaled by their leading principal inertia and then concatenated for a joint analysis. The resulting global components describe variation in that scaled concatenated matrix. To interpret those components, it helps to distinguish three quantities: the total inertia in the joint matrix, the amount supplied by each block, and each block’s contribution to a particular global component. These quantities are related, but they answer different questions.
Here, inertia means the sum of squared entries of a centered data matrix, equivalently its squared Frobenius norm. This convention uses unnormalized cross-products, so inertia and eigenvalues are sums of squared deviations rather than averages per individual. Dividing every inertia and eigenvalue by the same sample-size factor would change their numerical scale, but not their proportions.
The same definitions apply if a block’s columns have been standardized before its principal inertia is calculated. The preprocessing convention must be fixed first: the \(\lambda_{1k}\) used for scaling must belong to the actual centered block being analyzed. The setup assumes matched rows, finite entries, and positive leading inertia in each block. It does not require blocks to be independent. If a block has zero inertia, it contains no variation to scale and needs separate treatment rather than division by zero.
How Block Inertias Add
Concatenation places the blocks side by side; it does not combine or average their columns. Consequently, the squared entries of the concatenated matrix are exactly the squared entries of all the blocks taken together. The trace calculation makes the partition explicit.
If the eigenvalues of \(X_kX_k^\mathsf{T}\), including zero eigenvalues if necessary, are \(\lambda_{1k},\lambda_{2k},\ldots\), then scaling divides every eigenvalue by \(\lambda_{1k}\). Thus the block’s scaled inertia is \(I_k=\sum_j\lambda_{jk}/\lambda_{1k}\). The leading scaled eigenvalue in each block equals one, while the other eigenvalues express that block’s remaining inertia relative to its own leading direction.
Worked Example: Comparing Scaled Block Inertias
Suppose two centered, hypothetical omics blocks have unscaled eigenvalues \(8,2,1\) and \(3,1.5\), respectively. Their leading eigenvalues are 8 and 3, so their scaled eigenvalues are found by dividing each block’s eigenvalues by its own leading eigenvalue.
For block A, the scaled eigenvalues are \(8/8=1\), \(2/8=0.25\), and \(1/8=0.125\). Therefore \(I_A=1+0.25+0.125=1.375\). For block B, they are \(3/3=1\) and \(1.5/3=0.5\), giving \(I_B=1+0.5=1.5\).
The total global inertia is \(I=1.375+1.5=2.875\). Block A’s share is \(1.375/2.875=0.4783\), or 47.83%, and block B’s share is \(1.5/2.875=0.5217\), or 52.17%. These are shares of the total inertia after scaling, not shares of the raw unscaled inertia.
Partitioning Inertia Across Global Components
Let \(C=ZZ^\mathsf{T}=\sum_k Z_kZ_k^\mathsf{T}\). Write its eigenvalues in descending order as \(\mu_1,\mu_2,\ldots\), with corresponding orthonormal eigenvectors \(u_1,u_2,\ldots\). These are the global eigenvalues and individual-space axes of the concatenated analysis. For global component \(\ell\), define the inertia supplied by block \(k\) as \(c_{k\ell}=u_\ell^\mathsf{T}Z_kZ_k^\mathsf{T}u_\ell\). This is a nonnegative quantity: it equals the squared length of that block’s projection onto the unit vector \(u_\ell\).
For the first identity, use \(Cu_\ell=\mu_\ell u_\ell\) and \(u_\ell^\mathsf{T}u_\ell=1\). Then \(u_\ell^\mathsf{T}Cu_\ell=\mu_\ell\), while substituting \(C=\sum_k Z_kZ_k^\mathsf{T}\) gives \(\sum_k u_\ell^\mathsf{T}Z_kZ_k^\mathsf{T}u_\ell=\sum_k c_{k\ell}\). For the second identity, sum over an orthonormal eigenbasis: \(\sum_\ell c_{k\ell}=\sum_\ell u_\ell^\mathsf{T}Z_kZ_k^\mathsf{T}u_\ell=\operatorname{tr}(Z_kZ_k^\mathsf{T})=I_k\). The trace identity holds even when some eigenvalues are zero; zero-inertia directions add nothing.
A component’s block share is \(c_{k\ell}/\mu_\ell\), provided \(\mu_\ell>0\). It describes the fraction of that component’s inertia associated with the block. In contrast, \(I_k/I\) is the block’s share of total inertia across all components. The first is component-specific; the second is not. Neither by itself establishes a causal or biological mechanism.
Worked Example: From Hypothetical Block Data to Global Eigenvalues
Consider three hypothetical individuals and two centered blocks, each with two features. In individual space, let \(u=(1,-1,0)^\mathsf{T}/\sqrt{2}\) and \(v=(1,1,-2)^\mathsf{T}/\sqrt{6}\). These are orthonormal and each has entries summing to zero, so features formed from them are centered. Let block A have feature columns \(2u\) and \(v\). Its feature cross-product matrix is \(\operatorname{diag}(4,1)\), so its leading eigenvalue is 4 and its scaled individual-space cross-product matrix, expressed in the basis \(u,v\), is \(\operatorname{diag}(1,0.25)\).
For block B, let its feature matrix be \(X_B=[u\ v]R\operatorname{diag}(\sqrt{2},\sqrt{0.5})\), where \(R\) is a rotation through 45 degrees. Its feature cross-product matrix has eigenvalues 2 and 0.5, so its leading eigenvalue is 2. After scaling by \(\sqrt{2}\), its individual-space cross-product matrix in the \(u,v\) basis is \(R\operatorname{diag}(1,0.25)R^\mathsf{T}\). Since \(\cos(45^\circ)=\sin(45^\circ)=1/\sqrt{2}\), this is \(\begin{pmatrix}0.625&0.375\\0.375&0.625\end{pmatrix}\).
Adding the two scaled block matrices gives the global cross-product matrix in this basis: \(\begin{pmatrix}1.625&0.375\\0.375&0.875\end{pmatrix}\). Its eigenvalues are the mean of the diagonal entries, 1.25, plus or minus \(\sqrt{((1.625-0.875)/2)^2+0.375^2}=\sqrt{0.28125}=0.5303\). Therefore the global eigenvalues are \(\mu_1=1.7803\) and \(\mu_2=0.7197\). Their sum is 2.500, which matches the two scaled block inertias, \(1.25+1.25=2.500\).
The leading global eigenvector is at 22.5 degrees to \(u\). Block A’s contribution to the first component is \(1\cos^2(22.5^\circ)+0.25\sin^2(22.5^\circ)=0.8902\). The rotated block B gives the same contribution, so \(0.8902+0.8902=1.7804\), agreeing with \(\mu_1\) up to rounding. For the second component, each block contributes \(1.25-0.8902=0.3598\); their sum, \(0.7196\), agrees with \(\mu_2\) up to rounding.
Reading the Partition and Checking the Calculation
In the example, each block supplies half of each global eigenvalue, because their component contributions are equal. Each also supplies half of the total inertia. These equalities happen because of the chosen data; they are not guaranteed by first-eigenvalue scaling. Scaling gives every block a leading eigenvalue of one, but blocks can retain different amounts of inertia in their remaining directions, and their axes can align or disagree. Thus equal leading eigenvalues do not imply equal contributions to every global component.
Worked Example: Interpreting a Component’s Block Shares
Using the preceding calculation, the first component has eigenvalue 1.7803 and receives 0.8902 inertia units from each block. The contribution share of block A is \(0.8902/1.7803=0.5000\), or 50.00%; block B’s share is also 50.00%.
The first component accounts for \(1.7803/2.500=0.7121\), or 71.21%, of total scaled inertia. Each block contributes \(0.8902/2.500=0.3561\), or 35.61%, of total inertia through that component. The remaining eigenvalue is 0.7197, and the two blocks’ remaining contributions are approximately 0.3598 each. These figures distinguish a block’s share of one component from its share of the entire analysis.
Several checks help identify calculation or interpretation problems. The scaled block inertias must add to total inertia; the global eigenvalues must also add to that same total. Each \(c_{k\ell}\) must be nonnegative, and contributions across blocks must add to the corresponding global eigenvalue. Finally, summing a block’s contributions across a complete set of axes must recover its inertia. Small discrepancies can result from rounding, but a substantial mismatch usually signals inconsistent scaling, omitted axes, or a matrix/eigenvalue convention that has not been applied consistently.
A frequent mistake is to divide a block’s global component contribution by its raw inertia when interpreting its share of the scaled global component. That mixes scales: the relevant denominator for a component share is \(\mu_\ell\), and the relevant block numerator is \(c_{k\ell}\), both calculated from the scaled matrix. Another mistake is to treat the block’s total inertia share \(I_k/I\) as its share of every axis. Component-specific shares can vary across axes even when total block shares are fixed.
How This Partition Differs From Other Summaries
Separate PCA reports each block’s eigenvalues on its own scale; it does not, by itself, provide the global component contributions \(c_{k\ell}\). PCA of an unscaled concatenation instead lets blocks with large raw inertia have correspondingly large influence. MFA’s first-eigenvalue scaling changes the scale before concatenation, and the partition identities above describe the resulting joint analysis. They do not imply that each block has equal total inertia or equal influence on every axis.
The partition is an accounting identity, not a test of whether a block is biologically important. It depends on the observed blocks, the preprocessing and scaling choices, and the axes retained in a reported summary. In particular, if only a subset of axes is shown, the displayed contributions for a block sum to no more than its total inertia; the difference is the block’s inertia on omitted axes. Interpret the figures alongside the analysis design and data quality rather than treating a large share as proof of a substantive mechanism.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A block has unscaled eigenvalues 5, 2, and 1. What is its scaled inertia after first-eigenvalue scaling?
- Two scaled blocks have inertias 1.4 and 1.6. What is total global inertia, and what fraction comes from the second block?
- A global component has eigenvalue 2.4. One block contributes 0.9. What does the other block or blocks contribute in total?
- Why is a block’s total inertia share not necessarily its share of every global component?
Solutions. (1) The scaled eigenvalues are \(1, 2/5=0.4,\) and \(1/5=0.2\), so the scaled inertia is \(1+0.4+0.2=1.6\). (2) Total inertia is \(1.4+1.6=3.0\); the second block’s fraction is \(1.6/3.0=0.5333\), or 53.33%. (3) Contributions must sum to the component eigenvalue, so the other blocks together contribute \(2.4-0.9=1.5\) inertia units. (4) Total block share sums contributions across all axes, whereas a component share uses only that axis’s contribution; alignment between a block and an axis can differ from one component to another.