Determinant with Unit Element in Otherwise Zero Row
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Theorem
Let $D$ be the determinant:
- $D = \begin {vmatrix}
1 & 0 & \cdots & 0 \\
b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
b_{n 1} & b_{n 2} & \cdots & b_{n n} \end {vmatrix}$
Then:
- $D = \begin {vmatrix}
b_{2 2} & \cdots & b_{2 n} \\
\vdots & \ddots & \vdots \\
b_{n 2} & \cdots & b_{n n} \end {vmatrix}$
Proof
We refer to the elements of:
- $\begin {vmatrix}
1 & 0 & \cdots & 0 \\
b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
b_{n 1} & b_{n 2} & \cdots & b_{n n} \end {vmatrix}$
as $\begin {vmatrix} b_{i j} \end {vmatrix}$.
Thus $b_{1 1} = 1, b_{1 2} = 0, \ldots, b_{1 n} = 0$.
Then from the definition of determinant:
\(\ds D\) | \(=\) | \(\ds \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n b_{k \map \lambda k} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{\lambda} \map \sgn \lambda b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n}\) |
Now we note:
\(\ds \map \lambda 1 = 1\) | \(\implies\) | \(\ds b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n} = 1\) | ||||||||||||
\(\ds \map \lambda 1 \ne 1\) | \(\implies\) | \(\ds b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n} = 0\) |
So only those permutations on $\N^*_n$ such that $\map \lambda 1 = 1$ contribute towards the final summation.
Thus we have:
- $\ds D = \sum_\mu \map \sgn \mu b_{2 \map \mu 2} \cdots b_{n \map \mu n}$
where $\mu$ is the collection of all permutations on $\N^*_n$ which fix $1$.
Hence the result.
$\blacksquare$
Also see
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercises -- Second Set