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Question $\bf 3.$ Determine if the following matrix is diagonalizable. (explain your answer) $$A=\pmatrix{ 1 & 4 & -2 & 3 \\ 3 & -3 & 0 & 4 \\ 1 & 1 & 1 & -1 \\ 0 & 5 & -5 & 2 \\ }$$

The problem is, it's a $4\times 4$ matrix and I don't want to find the characteristic polynomial. I've been trying to find a trick but can't see anything. I could probably check if det $A = 0$ then that would tell me $0$ is an eigenvalue, but that's still not enough.

Well, I gave up and did it online.

So $4$ distinct eigenvalues $\implies$ diagonalizable... but if there's another way let me know.

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  • $\begingroup$ As your link shows $\det A\ne0$. $\endgroup$ – user63181 Aug 5 '14 at 17:55
  • $\begingroup$ You could perform row reduction then column reduction to see if the diagonal matrix you're left with has distinct eigenvalues. $\endgroup$ – Robert Wolfe Aug 5 '14 at 18:56
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Finding the eigenvalues and noting that they are distinct seems to be the best way to go here.

Unfortunately, checking the determinant does very little in the way of checking for diagonalizability.

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What you can do is: Compute the characteristic polynomial, here: $$ \chi_A(t) = 126+77 \lambda -42 \lambda ^2-\lambda ^3+\lambda ^4 $$ and its derivative $$ \chi_A'(\lambda) = 77 - 84\lambda - 3\lambda^2 + 4\lambda^3$$ Now compute ${\rm gcd}(\chi_A, \chi_A')$, if it is $1$ (as here), the polynomial has distinct zeros (and hence $A$ distinct eigenvalues), and is therefore diagonizable (over $\mathbb C$).

I do not think, that you can avoid computing the characteristic polynomial, but sometimes computing the zeros can be avoided.

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$A$ is diagonalizable if and only if $A$ satisfy one of the below polynomials $$(t-a)=0$$ or $$(t-a)(t-b)=0 $$ or $$(t-a)(t-b)(t-c)=0$$ or $$(t-a)(t-b)(t-c)(t-d)=0$$ where respectively sets $\{a\} $or $ \{a,b\} $or $ \{a,b,c\} $or $ \{a,b,c,d\}$ are set of all eigenvalues of $A$ in every case.

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    $\begingroup$ This is exactly how the asker reached the conclusion that $A$ is diagonalizable. The question, however, is whether there exists some other way of reaching this conclusion. $\endgroup$ – Omnomnomnom Aug 5 '14 at 18:10

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