4
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$
    – BAYMAX
    Jan 23 '17 at 1:22
14
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$
    – Guest1
    Jan 23 '17 at 1:34
7
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ 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 $X^2-5X+4$. $\endgroup$ Jan 23 '17 at 11:17
  • 2
    $\begingroup$ @Marc: no need to involve eigenvectors. For a triangular matrix, $f (A)_{2,2}=f (A_{2,2}) $. $\endgroup$ Jan 23 '17 at 12:53
  • $\begingroup$ @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. $\endgroup$ Jan 24 '17 at 12:02
2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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 $\operatorname{Tr}(A)$ $\operatorname{Tr}(A)$. $\endgroup$ Jan 23 '17 at 14:10
  • $\begingroup$ @MartinSleziak thankyou for pointing out. $\endgroup$
    – BAYMAX
    Jan 23 '17 at 16:00
  • $\begingroup$ 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. $\endgroup$ Jan 23 '17 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.