Expansion of Characteristic Polynomial of Matrix

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