Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question

1 Answer 1

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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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