Let $T,S$ be linear transformations, $T:\mathbb R^4 \rightarrow \mathbb R^4$, such that $T^3+3T^2=4I, S=T^4+3T^3-4I$. Comment on S.
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:
- one-one but not onto
- onto but not one one
- invertible
- 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!
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.
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).