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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?

Thank you in advance.

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  • $\begingroup$ 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 $r-\lambda+v\lambda$. $\endgroup$ – Lord Shark the Unknown Apr 25 at 21:14
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You can use column operations too!

Just add each of columns $2$, $3,\ldots,v$ to column $1$.

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  • $\begingroup$ Thanks for that! It explains everything. $\endgroup$ – Tim Apr 26 at 2:39

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