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# 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

Linear Algebra
Hard
##### Suggested Book
Linear Algebra; Hoffman and Kunze

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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