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Diagonilazibility in triangular matrix: TIFR GS 2018 Part A Problem 20

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" _i="1" _address="0.0.0.1"]Let A ∈ \(M_n(\Bbb R)\) be upper triangular with all diagonal entries 1 such that \( A \neq I\). Then A is not diagonalizable. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27.3" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="3.27.3" hover_enabled="0" _i="0" _address="0.1.0.0.0"]

TIFR GS 2018 Part A Problem 20[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27.3" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Linear Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27.3" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Hard[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27.3" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="on"]Linear Algebra; Hoffman and Kunze[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

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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?
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  • 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?
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  • 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.
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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 2x2 matrices!
  • Prove that it can be diagonalized.
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Watch the video

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