Let $T,S$ be linear transformations, $T:\mathbb R^4 \rightarrow \mathbb R^4$, such that $T^3+3T^2=4I, S=T^4+3T^3-4I$. Then S is:

  1. one-one but not onto
  2. onto but not one one
  3. invertible
  4. non-invertible

(One or more correct options)

My attempt: $S=T^4+3T^3-4I=T(T^3+3T^2)-4I=T(4I)-4I=4T-4I$

How do I go about proving or disproving my options?

My thoughts on options # 1. and 2.: I'm guessing $S$ is one-one if $Ker(S)=\theta$ is one way to go, but how do I obtain the kernel when I don't know what the transformation is? $Ker(S)=\theta$ if $T=I$, which does satisfy $T^3+3T^2=4I$. But so can other $T$s.

My thoughts on options # 3. and 4.: $S$ is invertible if $det(S)\neq0$, which is possible $det(T-I)\neq0$, i.e. $T\neq I$. How do I show that?

Please help!

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  • $\begingroup$ Given that you know that $S$ can be the zero matrix (if $T=I$, there are certainly several options that you can eliminate as correct possibilities. $\endgroup$ – Steven Stadnicki Aug 27 '15 at 18:00
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    $\begingroup$ Diya: the point is that (as you noted yourself) $T=I$ satisfies the constraint, and if $T=I$ then $S=0$. The question is 'if all you know is <these givens>, what can you deduce?' You cannot possibly deduce that $S$ is invertible, or onto, or one-to-one, because there are a pair $T, S$ that satisfy the constraints for which those statements are false. $\endgroup$ – Steven Stadnicki Aug 27 '15 at 18:05
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    $\begingroup$ There are certainly other solutions for $T$ (and thus $S$), but just knowing that one solution makes those conditions false is enough to eliminate them as possible answers. $\endgroup$ – Steven Stadnicki Aug 27 '15 at 18:06
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    $\begingroup$ Note that any linear transformation $S:\Bbb R^n \to \Bbb R^n$ is onto if and only if it is one to one. $\endgroup$ – Omnomnomnom Aug 27 '15 at 18:08
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    $\begingroup$ Well, you haven't proven that 4. is incorrect - in fact, you should be able to prove that any such $S$ is non-invertible. Also, and this is essential, $det(T-I)\neq 0$ is not the same thing as $T\neq I$ - there are many non-null matrices with zero determinants. $\endgroup$ – Steven Stadnicki Aug 27 '15 at 18:10

You have correctly deduced that $S = 4(T-I)$.

It follows that $S$ is invertible if and only if $1$ is not an eigenvalue of $T$. From the question, we do not have enough information to deduce whether this is the case.

Note that the first two options will never hold for an endomorphism on a finite dimensional space.

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In case that there is a typo, and that $S=T^4+3T^3-4T$, we have a clear answer, i.e., $S=T(T^3+3T^2-4I)=T\cdot 0=0$. Then $1.,2.,3.$ are false, but $4.$ is correct. (Note that $I$ and $T$ can look very similar for certain fonts, or handwritten notes).

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