# True or False, diagonalization problem

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 $$$$\det\left[T-I\lambda\right]_{B_{c}}=(2-\lambda)^4\qquad \text{and}\qquad T((0,0,0,1)) = (1,0,0,1)$$$$ 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

• Yes, that's fully correct. – egreg Dec 31 '18 at 23:44

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 $$T$$
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 $$\mathbb{R}^4$$.
If diagonalized, the diagonal matrix would simply be $$D=2I_4$$ As the result your matrix is $$B^{-1}DB=2I_4$$
That contradicts $$A(1,0,0,0)^T=(1,0,0,1)^T$$