# Understand the problem

Let A ∈ \(M_n(\Bbb R)\) be upper triangular with all diagonal entries 1 such that \( A \neq I\). Then A is not diagonalizable.

##### Source of the problem

TIFR GS 2018 Part A Problem 20

##### Topic

Linear Algebra

##### Difficulty Level

Hard

##### Suggested Book

Linear Algebra; Hoffman and Kunze

# Start with hints

Do you really need a hint? Try it first!

Let us start by assuming A is diagonalizable. Let’s see what it means to say A is diagonalizable.

- A is diagonalizable if there exists a basis of \(\Bbb R^n\) consisting of eigenvalues of A.
- Now as A is upper triangular the diagonal elements of A are the eigenvalues of A. So the only eigenvalue of A is 1.
- Now what we can say about a linear operator A with a basis of eigenvectors and all the eigenvectors have eigenvalue 1?

- Now write any vector as a linear combination of eigenvectors of A.
- Prove every vector under A is an eigenvector corresponding to eigenvalue 1.
- Doesn’t it seem geometrically that A is identity i.e. keeps every vector the same?

- Yes, indeed. Prove it mathematically by taking the Euclidean Basis vectors \(e_1,e_2,..,e_n\)
- Hence the only such matrix is I, so no non-identity upper triangular matrix fits this category.
- So the answer is True.

Food for Thought:

- What would have happened if the diagonal entries are not all equal?
- If it is hard to see try to play with 2×2 matrices!
- Prove that it can be diagonalized.

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