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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?

BAYMAX
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3 Answers3

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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 $N\ne 0$.

jgon
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  • $N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices? – BAYMAX Dec 20 '17 at 05:43
  • Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)\mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes. – jgon Dec 20 '17 at 05:45
  • Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$? – BAYMAX Dec 20 '17 at 05:47
  • Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $\begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ \end{pmatrix}$, which squares to 0. – jgon Dec 20 '17 at 05:48
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Since the matrix is upper triangular with all diagonal entries $1$, its characteristic equation is $(1-\lambda)^n=0$. This give an eigenvalue of $1$ with algebraic multiplicity $n$. Also, finding the corresponding eigenspace through the null space of $A-\lambda I$ tells us that it is generated by a single vector $[1,0,...,0]^T$. Thus its geometric multiplicity is $1$, different from $n$ and hence it is not diagonalizable.

ZSMJ
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  • The null space of $A - \lambda I$ is $\mathrm{Span} (1, 0, \dots, 0)$ if all the entries just above the diagonal ($A_{i, i + 1}$ for $i = 1, \dots, n - 1$) are nonzero, but is not generally true: consider $A = \begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}$. – jskattt797 Oct 16 '21 at 20:07
  • I think G.M. of $1 < n$ = A.M of $1$... is enough reason here. – user_1729 Oct 12 '23 at 08:02
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The eigenvalues of an upper triangular matrix are the diagonal entries, so $A \neq I_n$ has only one eigenvalue, $1$. An $n \times n$ matrix is diagonalizable if and only if there is a basis $b_1, b_2, \dots, b_n$ of $\mathbb{R}^n$ consisting of eigenvectors of $A$. But no such basis can exist, as it would cause $A e_j = A \left( \sum_{i = 1}^n \alpha_i b_i \right) = \sum_i \alpha_i Ab_i = \sum_i \alpha_i b_i = e_j$ for each standard basis vector $e_j \in \mathbb{R}^n$, producing the contradiction $$A = \begin{bmatrix} | & | & \dots & | \\ Ae_1 & Ae_2 & \dots & Ae_n \\ | & | & \dots & | \end{bmatrix} = \begin{bmatrix} | & | & \dots & | \\ e_1 & e_2 & \dots & e_n \\ | & | & \dots & | \end{bmatrix} = I_n$$

jskattt797
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