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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, B ∈ $$M_n(\Bbb R)$$ be such that A + B = AB. Then AB = BA. [/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" 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="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR 2018 Part A Problem 19[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" 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" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]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"]

[/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" _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" hover_enabled="0" _i="1" _address="0.1.0.2.1"]
We need to play around with symbols.
Try to make (AB-BA) containing equation so that we can get the idea how to show it 0.This is a symmetric operation so remember if you use only B to get something you can also use A to get that so we have to take help of the symmetry.
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]
• Playing with the equation and multiplying the equation once left and once right by B we get A.B+B.B=A.B.B and B.A + B.B = B.A.B
• Now subtract it to get AB-BA term,which gives (AB-BA)=(AB-BA)B => (AB-BA)(I-B)=0
• Similarly use A in place of B to get (I-A)(AB-BA)=0
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]
• Now if somehow we can show (I-A) or (I-B) is invertible then we are done.
• So we go back to the actual equation. Remember the identity (1-a)(1-b)=1-a-b+ab.
• Observe that (I-A)(I-B)=I follows from the actual equation.Hence the result follows as (I-A) and (I-B) are invertible and are inverses of each other.
• The answer is therefore True.
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]
U+T are linear operators on R^n and U+T=UT.Then Img(T)=Img(U) and Ker(T)=Ker(U)
• Prove that Ker(T) is a subspace of Ker(U).
• Take adjoint of the equation and observe that Ker(U*) is a subspace of Ker(T*).
• Now we know that Ker(U*) is orthogonal to Img(U) and same goes for T.
• Hence conclude by that Img(T) is a subspace of Img(T).
• Therefore prove that Img(T)=Img(U) and Ker(T)=Ker(U) by Rank Nullity Theorem.

# Watch the video

[/et_pb_text][et_pb_code _builder_version="3.26.4" _i="4" _address="0.1.0.4"]

# Similar Problems

[/et_pb_text][et_pb_post_slider include_categories="12" _builder_version="3.23.3" _i="9" _address="0.1.0.9"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3" _i="10" _address="0.1.0.10"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

# 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, B ∈ $$M_n(\Bbb R)$$ be such that A + B = AB. Then AB = BA. [/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" 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="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR 2018 Part A Problem 19[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" 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" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]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"]

[/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" _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" hover_enabled="0" _i="1" _address="0.1.0.2.1"]
We need to play around with symbols.
Try to make (AB-BA) containing equation so that we can get the idea how to show it 0.This is a symmetric operation so remember if you use only B to get something you can also use A to get that so we have to take help of the symmetry.
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]
• Playing with the equation and multiplying the equation once left and once right by B we get A.B+B.B=A.B.B and B.A + B.B = B.A.B
• Now subtract it to get AB-BA term,which gives (AB-BA)=(AB-BA)B => (AB-BA)(I-B)=0
• Similarly use A in place of B to get (I-A)(AB-BA)=0
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]
• Now if somehow we can show (I-A) or (I-B) is invertible then we are done.
• So we go back to the actual equation. Remember the identity (1-a)(1-b)=1-a-b+ab.
• Observe that (I-A)(I-B)=I follows from the actual equation.Hence the result follows as (I-A) and (I-B) are invertible and are inverses of each other.
• The answer is therefore True.
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]
U+T are linear operators on R^n and U+T=UT.Then Img(T)=Img(U) and Ker(T)=Ker(U)
• Prove that Ker(T) is a subspace of Ker(U).
• Take adjoint of the equation and observe that Ker(U*) is a subspace of Ker(T*).
• Now we know that Ker(U*) is orthogonal to Img(U) and same goes for T.
• Hence conclude by that Img(T) is a subspace of Img(T).
• Therefore prove that Img(T)=Img(U) and Ker(T)=Ker(U) by Rank Nullity Theorem.

# Watch the video

[/et_pb_text][et_pb_code _builder_version="3.26.4" _i="4" _address="0.1.0.4"]

# Similar Problems

[/et_pb_text][et_pb_post_slider include_categories="12" _builder_version="3.23.3" _i="9" _address="0.1.0.9"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3" _i="10" _address="0.1.0.10"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

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