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For a balanced incomplete block design with $r>\lambda$, $N{N}^T$ has eigenvalues $r-\lambda >0$ on the zero-sum subspace and $r+(t-1)\lambda >0$ on the constant vector. Thus its rank is $t$. As $N$ is $t\times b$, $t=\mathrm{rank}(N{N}^T)\le \mathrm{rank}(N)\le b$.

Let $N$ be the $t\times b$ incidence matrix of a set system. The balanced incomplete block design overlap counts give $N{N}^T=(r-\lambda )I+\lambda J$. On the zero-sum subspace its eigenvalue is $r-\lambda =r(t-k)/(t-1)>0$; on the constant vector its eigenvalue is $r+(t-1)\lambda >0$. Therefore $N{N}^T$ has matrix rank $t$. Since a product cannot have larger rank than either factor,This proves Fisher's inequality for block designs directly. The incomplete, nontrivial hypotheses ensure the positive eigenvalue needed for this argument.

Count incidences of a treatment with a block containing it. Summing over blocks gives $bk$, while summing over treatments gives $tr$. Hence the balanced incomplete block design satisfies

A balanced incomplete block design has $t$ treatments and $b$ blocks, each containing $k$ distinct treatments with $1<k<t$. Every treatment occurs in $r$ blocks and every distinct treatment pair occurs together in $\lambda >0$ blocks. A symmetric balanced incomplete block design has $b=t$, which implies $r=k$ by the first counting identity below.

For the separate even-order condition, let $N$ be the square treatment-by-block incidence matrix of a set system of a symmetric design. Its diagonal overlap counts are $r$ and its off-diagonal counts are $\lambda$, givingThe counting identities proved in the next slots show that $r=k$ and $r+(t-1)\lambda =r^2$. Consequently the eigenvalues of $N{N}^T$ are $r^2$ on the constant vector and $r-\lambda$ on its $(t-1)$-dimensional orthogonal complement. Taking determinants givesIf $b=t$ is even, the exponent $t-1$ is odd. The exponent of every prime in $\det (N{)}^2$ and in $r^2$ is even; therefore its exponent in the positive integer $r-\lambda$ must be even. This proves the even-order symmetric design square obstruction:It is a necessary condition, not a sufficiency claim.

A randomized complete block design groups similar experimental units into experimental blocks and places every treatment once in each experimental block, independently randomizing treatment allocation within experimental blocks. Blocking removes background between-block variation from within-block treatment comparisons. A balanced incomplete block design has $v$ treatments and $b$ experimental blocks of size $k<v$, with each treatment occurring in $r$ experimental blocks and every distinct treatment pair together in $\lambda$ experimental blocks. Counting treatment occurrences and co-occurrences gives $vr=bk$ and $r(k-1)=\lambda (v-1)$.

A Latin square of order $q$ places each of $q$ treatment symbols once in every row and every column of a $q\times q$ array. This controls two nuisance directions. A Graeco-Latin square superposes two mutually orthogonal Latin squares: each square is Latin, and each ordered pair of symbols from the two squares occurs exactly once. The two symbol factors can then be fitted together with row and column effects. Random permutations of row labels, column labels and treatment symbols provide a suitable randomized allocation without destroying these balances.

For the operator experiment, fit the additive normal linear modelHere $r_i,c_j$ account for the two fabric directions, and ${\tau }_o$ is the operator effect. Additivity, independent equal-variance errors and the normal assumption supply the exact F-test; there is only one observation per cell, so an unrestricted interaction model is not available.

In the analysis of variance for a Latin square, the three centered factor spaces each have dimension $4-1=3$. Together with the intercept the model has rank ten, leaving $16-10=6$ residual statistical degrees of freedom. The corrected total has $16-1=15$. The completed calculation isUnder the null hypothesis ${\tau }_1=\cdots ={\tau }_4=0$, the operator F-statistic isThe upper five-percent critical value for $F_{3,6}$ is $4.76$, so the operator effect is not significant at five percent in this model.

The decision is independent of the order of fitting rows, columns and operators. To prove the required orthogonality, let $u_i$ be any centered row coefficients and $w_o$ any centered operator coefficients. Their observation-space inner product is ${\sum }_{i,j}{u}_i{w}_{o(i,j)}={\sum }_i{u}_i{\sum }_o{w}_o=0$, since every operator occurs once in each row. The same calculation applies to columns and operators, and all row-column pairs occur once as well. Thus the three centered spaces are mutually orthogonal; their projections commute and sequential sums of squares equal adjusted sums of squares in every fitting order.

With machines included, the layout is a Graeco-Latin square: the machine symbols are Latin, and all sixteen operator-machine pairs occur exactly once. Fitin addition to the earlier constraints. The machine space contributes three new statistical degrees of freedom and is orthogonal to the previous factor spaces, so the operator sum of squares remains $18.25$. The residual sum of squares and statistical degrees of freedom becomeHenceThis exceeds $F_{3,3}(0.05)=9.28$ but not $F_{3,3}(0.01)=29.46$. Therefore the operator effect is now significant at five percent, but not at one percent. Adjustment for machines has removed substantial residual variation without altering the operator numerator. This illustrates why a balanced nuisance factor can change precision and significance even when it does not change the estimated treatment contrasts.

Write the binary incidence matrix of a set system as $N$ to avoid confusing it with a sample size. Thencounts the number of blocks containing treatment $i$, and for $i\ne l$, $(N{N}^T{)}_{il}={\sum }_j{N}_{ij}{N}_{lj}$ counts blocks containing both treatments. The displayed structure therefore makes every treatment replication equal to $r$ and every pair concurrence equal to $\lambda$.

Assume the usual nonempty block design, with $r>0$ and $k>0$. If $r=\lambda$, any block containing one treatment contains every other treatment: otherwise that treatment's replication would exceed its concurrence with the omitted one. Every block is consequently complete, so $k=t$ and $r=b$. With random allocation within blocks this is a randomized complete block design. If $\lambda <r$, the design is a balanced incomplete block design. Counting partners of a fixed treatment in its $r$ blocks givesFor $t>1$, $\lambda /r<1$ gives $k-1<t-1$, hence $k<t$. Moreover $N{N}^T$ has eigenvalue $r-\lambda >0$ on the $(t-1)$-dimensional zero-sum subspace and eigenvalue $r+(t-1)\lambda >0$ on the constant direction. Thus $\mathrm{rank}(N{N}^T)=t$, while $\mathrm{rank}(N{N}^T)\le \mathrm{rank}(N)\le b$. This proves $b\ge t$, the Fisher's inequality for block designs. The case $\lambda =0$ allows singleton blocks; their lack of connectedness prevents general treatment comparison after block adjustment, but does not invalidate the rank inequality.

First ignore the day information and use a one-way normal linear model with five group means and independent errors of common variance. There are $20$ observations. The analysis of variance has treatment, residual and corrected-total statistical degrees of freedom $4,15,19$, respectively. The mean squares in ANOVA giveUnder equal treatment means this has the F-distribution with $(4,15)$ degrees. It exceeds the given $5\%$ upper critical value $3.06$, so the unblocked analysis rejects equality at $5\%$.

The actual day allocation omits a different treatment on each of the five days. Hence it is a balanced incomplete block design withEvery pair occurs on exactly the three days omitting neither member. In the additive two-way analysis of variance, use an intercept, four independent treatment contrasts and four day contrasts. The design is connected, so the model rank is $9$ and its residual has $20-9=11$ statistical degrees of freedom. Day effects and treatment effects are not orthogonal, because a different treatment is absent each day; the unadjusted treatment sum of squares cannot be used for the adjusted test.

Enter days first and then treatments. The supplied sequential sums of squares giveThe day, treatment-after-day, residual and corrected-total statistical degrees of freedom are $4,4,11,19$. Under the additive normal linear model and the null of equal treatment effects, the adjusted numerator and residual squared lengths are independent chi-squared quantities, so the ratio has the F-distribution with $(4,11)$ degrees. It is below even the given $10\%$ critical value $2.54$. After accounting for days there is no significant treatment difference at either $10\%$ or $5\%$. This inference assumes independent homoscedastic normal errors and an additive day effect; with one observation per observed treatment–day cell, unrestricted interactions cannot also be fitted.

A balanced incomplete block design is symmetric when its number of blocks equals its number of treatments. The identity $bk=tr$ then gives $r=k$. Its square incidence matrix of a set system $N$ obeys $N{N}^T=(r-\lambda )I+\lambda J$, with eigenvalues $r-\lambda$ of multiplicity $t-1$ and $r^2$ of multiplicity one.