# Stuck on the Proof on Fisher's Inequality

I'm confused about the proof that for a balanced design with parameters $$(v, b, r, k, \lambda)$$, if $$v \gt k$$, then $$b \ge v$$.

If you let $$M$$ be the incidence matrix of the design such that $$M_{ij}$$ is $$1$$ if the point $$i$$ is in the block $$j$$, and $$0$$ otherwise, then M is a $$(v \times b)$$ matrix. Now, let $$B = MM^t$$. Then

$$det(B)=$$ $$\begin{vmatrix} r & \lambda & \lambda & ... & \lambda \\ \lambda & r & \lambda & ... & \lambda \\ \lambda & \lambda & r & ... & \lambda \\ ... & ... & ... & ... & ... \\ \lambda & \lambda & \lambda & ... & r \\ \end{vmatrix}$$

$$=$$ $$\begin{vmatrix} r & \lambda & \lambda & ... & \lambda \\ \lambda - r & r - \lambda & 0 & ... & 0 \\ \lambda - r & 0 & r - \lambda & ... & 0 \\ ... & ... & ... & ... & ... \\ \lambda - r & 0 & 0 & ... & r - \lambda \\ \end{vmatrix}$$ $$=$$ $$\begin{vmatrix} r + (v-1)\lambda& \lambda & \lambda & ... & \lambda \\ 0 & r - \lambda & 0 & ... & 0 \\ 0 & 0 & r - \lambda & ... & 0 \\ ... & ... & ... & ... & ... \\ 0& 0 & 0 & ... & r - \lambda \\ \end{vmatrix}$$

However, I'm confused about how $$\begin{vmatrix} r & \lambda & \lambda & ... & \lambda \\ \lambda - r & r - \lambda & 0 & ... & 0 \\ \lambda - r & 0 & r - \lambda & ... & 0 \\ ... & ... & ... & ... & ... \\ \lambda - r & 0 & 0 & ... & r - \lambda \\ \end{vmatrix}$$ got to $$\begin{vmatrix} r + (v-1)\lambda& \lambda & \lambda & ... & \lambda \\ 0 & r - \lambda & 0 & ... & 0 \\ 0 & 0 & r - \lambda & ... & 0 \\ ... & ... & ... & ... & ... \\ 0& 0 & 0 & ... & r - \lambda \\ \end{vmatrix}$$.

I understand that the determinants of two matrices A and B are equal if B was obtained from A by adding a constant times a row onto another row of A. I also get all the other rules of determinants when it comes to elementary row operations, but I don't understand how the elementary row operations themselves produced that final matrix I gave above.

Can anyone please explain this to me?

• Better to forget about the determinant and focus on eigenvalues. The all-one matrix $J$ has eigenvalues $0$ and $v$, so your matrix $\lambda J+(r-\lambda)I$ has eigenvalues $r-\lambda$ and $r-\lambda+v\lambda$. – Lord Shark the Unknown Apr 25 at 21:14
Just add each of columns $$2$$, $$3,\ldots,v$$ to column $$1$$.