linear algebra - Diagonalizability of $A in M_{n}(Bbb{R})$ (Upper traingular matrix) with all diagonal entries 1 and $A neq I$?

If $A \in M_{n}(\Bbb{R})$ is an Upper triangular matrix with diagonal entries $1$ such that $A \neq I$, then what can we say about the diagonalizability of $A$ ?

I know that if the matrix has distinct eigenvalues or the set of eigenvectors are linearly independent then the matrix is diagonalizable.

And that the eigenvalues of the triangular matrices are given by the diagonal elements like here and here, but they work nocely if we had distinct elements on the main diagonal.

But in my case I have same value 1 on the main diagonal, how can I approach about the diagonalizability of the matrix?

Answer

It isn't diagonalizable. Your matrix is of the form $I+N$ where $I$ is the identity matrix and $N$ is strictly upper triangular. You can check that being strictly upper triangular means that $N^n=0$, since if the $e_i$ are the basis vectors, and $E_m$ is the subspace spanned by $e_1,\ldots,e_m$, with $E_0=0$ then you can check that $NE_m \subseteq E_{m-1}$, so $N^nE_n = E_0=0$. However since $A\ne I$, $N\ne 0$. Now suppose $M$ diagonalizes $A$, so $MAM^{-1} = D$ for $D$ a diagonal matrix. Then $M(I+N)M^{-1}=MIM^{-1}+MNM^{-1} = D$, but $MIM^{-1}=I$, so we in fact have $MNM^{-1}=D-I$, so in fact $N$ must be diagonalizable. However we still have $(MNM^{-1})^n=0$, so if $D-I$ has elements $\lambda_1,\ldots,\lambda_n$ on the diagonal, then $\lambda_i^n=0$, which implies $\lambda_i=0$, hence $D-I=0$. Thus we would have that $D=I$, and $N=0$, contradicting the assumption that $A\ne I$, so $N\ne 0$.