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The question is whether a given matrix $$ \begin{pmatrix} 1 & 0 & c & d\\ 0 & 2 & e & f \\ 0 & 0 & 3 & g \\ 0 & 0 & 0 & 4\\ \end{pmatrix} $$ satisfies $f(A) = A^2 - 5A +4I=0$?

My attempt was to use the Cayley–Hamilton theorem $$ \Delta(\lambda)=\Pi_{k=1}^4(\lambda - k)^4 = (\lambda-2)(\lambda-3)(\lambda^2-5\lambda+4)=0. $$ Then $$ \Delta(A) = (A-2)(A-3)(A^2-5A+4)=(A-2)(A-3)f(A)=0. $$ But in general it does not mean that $f(A)$ is $0$. Moreover, by some numeric examples I see that in general $f(A)\neq0$. Is there any theorem or consequnce that I am missing?

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If it would have been real numbers it would have been possible that too if $(a-3)$ and $(a-2)$ were not zero then $f(a)$ would have been $0$ , but in case of Matix multiplication this is not the case that is $A.B = 0$ is alsopossible even when $A \neq 0$ and $B \neq 0$ ,here in latter $0$ is the zero matrix – BAYMAX 5 hours ago
up vote 6 down vote accepted

The answer is no, because if $fA)=0$, then $f$ would be a multiple of the minimal polynomial of $A$. However this matrix has $4$ simple eigenvalues: $1, 2,3 ,4$, hence its minimal polynomial has degree $4$, and it can't divide a quadratic polynomial.

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Thanks! That is the exact answer that I was looking for. The other 2 are right and elegant as well but this question was in a context of eigenvalues and etc., therefore your answer is what they meant in the book.Thank you! – Guest1 5 hours ago
up vote 3 down vote

We can simply compute the element $b_{2,2}$ of the resulting matrix $(b_{i,j})=A^2-5A+4I$. We have $$b_{2,2}=2^2-5\cdot 2 + 4=-2 \not = 0.$$

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up vote 1 down vote

We can proceed by method of contradiction. Suppose $f(A) = A^2 - 5A + 4I$ , applying Trace to both sides we see that

Trace$(A^2) = 30$ ,Trace$(A) = 10$

thus Trace$(A^2 - 5A + 4I) = $Trace $(A^2)$ $- 5.$Trace (A) $+ 4.$Trace $(I)$

We see Trace$(A^2 - 5A + 4I) = -16$ , but Trace$(f(A)) = 0$,from the definition of $f(A)$ ,and thus a contradiction

Hope this helps .

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