description/proof of that determinant of square matrix whose last row is all 1 and whose each other row is all 0 except row number + 1 column 1 is -1 to power of dimension + 1
Topics
About: matrix
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of determinant of square matrix.
- The reader admits the Laplace expansion of determinant.
Target Context
- The reader will have a description and a proof of the proposition that the determinant of any square matrix whose last row is all \(1\) and whose each other row is all \(0\) except the row number \(+ 1\) column \(1\) is \(-1\) to the power of the dimension \(+ 1\).
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(M\): \(\in \{\text{ the n x n matrices }\}\), the last row is all \(1\) and each other \(j\)-th row is \(0\) except the \(j + 1\)-th column \(1\)
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Statements:
\(det M = (-1)^{n + 1}\).
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2: Natural Language Description
For any \(\text{ n x n }\) matrix, \(M\), whose last row is all \(1\) and whose each other \(j\)-th row is \(0\) except the \(j + 1\)-th column \(1\), the determinant is \(det M = (-1)^{n + 1}\).
3: Proof
\(M = \begin{pmatrix} 0 & 1 & 0 & 0 & ... & 0 \\ 0 & 0 & 1 & 0 & ... & 0 \\ ... \\ 1 & 1 & 1 & 1 & ... & 1 \end{pmatrix}\).
Let us prove it inductively.
Let the determinant of the \(n\)-dimensional matrix be \(f (n)\).
When \(n = 1\), \(M = \begin{pmatrix} 1 \end{pmatrix}\), and \(f (1) = det M = 1 = (-1)^{1 + 1}\).
When \(n = 2\), \(M = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}\), and \(f (2) = det M = -1 = (-1)^{2 + 1}\).
Let us suppose that for \(n = m - 1\), \(det M = (-1)^{m - 1 + 1}\). For \(n = m\), let us expand \(det M\) by the Laplace expansion of determinant by the 1st row. The 1st row has only 1 nonzero column, the 2nd column, 1. The cofactor of the component is \((-1)^{1 + 2} f (m - 1)\), because \(M\) with the 1st row and the 2nd column removed is nothing but the matrix for the case, \(m - 1\). So, \(f (m) = -1 f (m - 1)\).
So, \(f (n) = (-1)^{n + 1}\).
