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Data :

$T\colon \mathbb{R}^4 \to \mathbb{R}^4$ (linear transformation)

Characteristic polynomial --> $x^4-3x^2-5$

is $T$ isomorphism (Yes/No question)?

I don't know to "approach" to this problem, as known isomorphism need to fill three conditions.

1) $T$ linear transformation

2) $T$ Injective function

3) $T$ Surjective function

It clear that $T$ is linear transformation.

But I don't understand how to use the characteristic polynomial to confirm that $T$ is injective and surjective function?

Any ideas? Thanks

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up vote 6 down vote accepted

Since $0$ isn't root of the characteristic polynomial then $0$ isn't an eigenvalue of $T$ so $\det T\ne0$. Notice that even we can express $T^{-1}$ as function of $T$: by Cayley-Hamilton theorem

$$T^4-3T^2=5 I\iff T\left(\underbrace{\frac15(T^3-3T)}_{=T^{-1}}\right)=I$$

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Thanks for your answer, It a little bit not understood how you manage to express the inverse of T, usually I put to in a matrix and grading near identity matrix in this case I don't know how to express characteristic polynomial into a matrix to find the inverse, can you explain how did you managed to find the inverse of T? Thanks – JaVaPG Jul 12 '14 at 16:05
$B$ is the inverse matrix of $A$ if $AB=I$ so since by the Cayley-Hamilton theorem we have $T^4-3T^2-5I=0$ which's equivalent to $T(\frac15(T^3-3T))=I$ hence we see that the inverse of $T$ is $\frac15(T^3-3T)$. – user63181 Jul 12 '14 at 16:14
Excellent, Sami! – amWhy Jul 15 '14 at 11:30
@SamiBenRomdhane: Good another one. Did you see this one? – Babak S. Aug 7 '14 at 8:57

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