what we can tell about this linear transformation? Let $T:\mathbb R^{4}\to \mathbb R^{4}$ be a linear transformation satisfying $T^{3}+3T^{2}=4I$, where $I$ is the identity transformation.  Then the linear transformations $S=T^{4}+3T^{3}-4I$ is:


*

*One-one but not onto

*onto but not one-one 

*Invertible 

*Not invertible


Given $T^{3}+3T^{2}=4I$, let $S=T^{4}+3T^{3}-4I$.  Then:
\begin{align}
S&=T^{4}+3T^{3}-4I\\
&=T(T^{3}+3T^{2})-4I\\
&=4T-4I
\end{align} I think this is not providing any information related to one-one onto so how can I prove it is invertible or not or am I going wrong. Not a homework problem self studying.
 A: The rank-nullity theorem implies that options (1) and (2) are impossible. Do you see why?
Note that $T$ is annihilated by the polynomial
$$
p(t)=t^3+3\,t^2-4=(t-1)(t+2)^2
$$
Since the minimal polynomial of $T$ must divide $p(t)$,
it follows that the only potential eigenvalues of $T$ are $1$ and $-2$.
Now, you have correctly proven that
$$
S=4\left(T-I\right)
$$
We have two possible cases:


*

*$\lambda=1$ is an eigenvalue of $T$

*$\lambda=1$ is not an eigenvalue of $T$


If $\lambda=1$ is an eigenvalue of $T$, then $\dim\ker(T-I)=0$ so $S$ is injective. The rank-nullity theorem then implies that $S$ must also be surjective and thus invertible.
If $\lambda=1$ is not an eigenvalue of $T$, then $\dim\ker(T-I)>0$, so $S$ is not injective and thus not invertible.
To see that both cases are possible, note that both of the matrices 
\begin{align*}
T_1&=
\left[\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}\right] &
T_2 &=
\left[\begin{array}{rrrr}
-2 & 0 & 0 & 0 \\
0 & -2 & 0 & 0 \\
0 & 0 & -2 & 0 \\
0 & 0 & 0 & -2
\end{array}\right]
\end{align*}
satisfy the given conditions but $1$ is an eigenvalue of $T_1$ and $1$ is not an eigenvalue of $T_2$.
