True or False, diagonalization problem

Question

Let $B_c= \left\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\right\}$, $T:\mathbb{R}^4\rightarrow \mathbb{R}^4$ a linear operator such that \begin{equation} \det\left[T-I\lambda\right]_{B_{c}}=(2-\lambda)^4\qquad \text{and}\qquad T((0,0,0,1)) = (1,0,0,1)\end{equation} Is T diagonalizable?

I've tried to use the following reasoning: Since the algebraic multiplicity of the eigenvalue $2$ is $4$, we know that $T$ is diagonalizable if (and only if) the dimension of the eigenspace associated with the eigenvalue $2$ is $4$. But we also know that there's a vector in $\mathbb{R}^4$ that don't belong to $S_{2}$ (because $(0,0,0,1)$ is not an eigenvector). Can I use this argument to show that $T$ is not diagonalizable? Thanks in advance

Answer 1

An endomorphism φ of a finite dimensional vector space over a field F is diagonalizable if and only if its minimal polynomial factors completely over F into distinct linear factors. The fact that there is only one factor X − λ for every eigenvalue

from https://en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)

I like to say that a matrix diagonalizes as some $P^{-1}AP = D$ if and only if the minimal polynomial is squarefree. They phrase that as "factors...into distinct linear factors"

If the minimal polynomial really were $\lambda - 2,$ we would have $A-2I = 0,$ so that $A=2I,$ and everything would be an eigenvector. Since they give a vector that is not an eigenvector, that does it. The minimal polynomial is not linear, it is one of $(\lambda-2)^2$ or $(\lambda-2)^3$ or $(\lambda-2)^4.$ Note that i am using the reverse order polynomial $\det(\lambda I - A).$ Oh, and $A$ is the square matrix defined by $T$

Answer 2

Since $T((0,0,0,1)) \neq 2 \cdot (0,0,0,1)$ we know that $(0,0,0,1)$ isn't in the eigenspace $\operatorname{Ker}(T-2I)$. The dimension of the eigenspace is therefore strictly less than the dimension of $\mathbb{R}^4$.

Answer 3

If diagonalized, the diagonal matrix would simply be $$D=2I_4$$ As the result your matrix is $$B^{-1}DB=2I_4$$

That contradicts $$A(1,0,0,0)^T=(1,0,0,1)^T$$