Expansion of Characteristic Polynomial of Matrix
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Theorem
Let $R$ be a commutative ring with unity.
Let $R \sqbrk x$ be the polynomial ring in one variable over $R$.
Let $\mathbf A$ be a square matrix over $R$ of order $n > 0$.
Let $\map {p_{\mathbf A} } x$ be the characteristic polynomial of $\mathbf A$.
Then $\map {p_{\mathbf A} } x$ can be expressed as:
- $\map {p_{\mathbf A} } x = x^n + a_{n - 1} x^{n - 1} + \cdots + a_0$
where:
\(\ds a_{n - 1}\) | \(=\) | \(\ds -\map \tr {\mathbf A}\) | where $\map \tr {\mathbf A}$ denotes the trace of $\mathbf A$ | |||||||||||
\(\ds a_0\) | \(=\) | \(\ds \paren {-1}^n \map \det {\mathbf A}\) | where $\map \det {\mathbf A}$ denotes the determinant of $\mathbf A$ |
Proof
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Sources
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): characteristic polynomial