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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"]

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" _i="1" _address="0.1.0.2.1" hover_enabled="0"]
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?
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" _i="2" _address="0.1.0.2.2" hover_enabled="0"]
• 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?
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3" hover_enabled="0"]
• 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.
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]
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.

# 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"]

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" _i="1" _address="0.1.0.2.1" hover_enabled="0"]
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?
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" _i="2" _address="0.1.0.2.2" hover_enabled="0"]
• 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?
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3" hover_enabled="0"]
• 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.
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]
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.

# Similar Problems

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