Jordan Canonical Form

Let \( A\in M_n(\mathbb{R})\hspace{2mm} \) and\( \hspace{2mm} m_A(\lambda)\hspace{2mm} \) be its minimal polynomial. Let f be a polynomial satisfies\( \hspace{2mm}f(A) = 0. \hspace{2mm} \)Show that\( \hspace{2mm} f(\lambda) \hspace{2mm} \)is divisible by\( \hspace{2mm} m_A(\lambda). \)

Expert Solution

solution

we have \( f(\lambda)=m_A(\lambda)g(\lambda)+r(\lambda) \hspace{3mm}(1) \)

\( deg\bigg(r(\lambda)\bigg) \)\( <\deg \)\( \bigg(m_A(\lambda)\bigg) \)\( \hspace{4mm}(2) \)

+Supposse that \( r(\lambda)=0 \hspace{2mm} \) From (1). we have

\( \implies f(A)=m_A(A)q(A)+r(A) \)

Since. \( f(A)=0 \) and \( m_A(A)=0 \)

\( \implies r(A)=0 ,\hspace{2mm} \) with (2), contradiction to the definition of minimal polynomial. Then, \( r(\lambda)=0 \)

Answer

Therefore \( f(\lambda) \) is divisible by \( m_A(\lambda). \)

SUN BUNRA answered 2 years ago

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Jordan Canonical Form

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