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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 – 2017-01-23

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The answer is no, because if $f(A)=0$, 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! – 2017-01-23
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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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    Note that this is because the second standard basis vector is an obvious eigenvector for eigenvalue$~2$, which as calculated here is a non-root of $X^2-5X+4$. – 2017-01-23
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    @Marc: no need to involve eigenvectors. For a triangular matrix, $f (A)_{2,2}=f (A_{2,2}) $. – 2017-01-23
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    @MartinArgerami: You are right, the answer reasons directly about matrix coefficients, not (as I was tempted to do) about eigenvectors. Of course every diagonal coefficient of a triangular matrix is an eigenvalue, but one does not need to invoke that here. – 2017-01-24
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We can proceed by method of contradiction. Suppose $f(A) = A^2 - 5A + 4I$, applying Trace to both sides we see that

$\operatorname{Trace}(A^2) = 30$, $\operatorname{Trace}(A) = 10$

thus $\operatorname{Trace}(A^2 - 5A + 4I) = \operatorname{Trace} (A^2) - 5\cdot\operatorname{Trace} (A) + 4\cdot\operatorname{Trace} (I)$

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

Hope this helps.

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    Combining the usual text and math in this way seems a bit unusual to me. (But it might be a matter of taste.) But still it might be useful for you to know that you can type things like `$\operatorname{Trace}(A)$` $\operatorname{Trace}(A)$ or `$\operatorname{Tr}(A)$` $\operatorname{Tr}(A)$. – 2017-01-23
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    @MartinSleziak thankyou for pointing out. – 2017-01-23
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    I have edited your post a bit. Feel free to rollback my edit (or improve it further) if this is not what you want it to look like. – 2017-01-23