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 haveÂ
\( \implies f(A)=m_A(A)q(A)+r(A) \)
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, \( r(\lambda)=0 \)
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Answer
Therefore \( f(\lambda) \)  is divisible by  \( m_A(\lambda). \)
