Matrix additive-multiplicative :TIFR 2018 Part A Problem 19

[et_pb_section fb_built="1" _builder_version="3.22.4" fb_built="1" _i="0" _address="0"][et_pb_row _builder_version="3.25" _i="0" _address="0.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.0.0"][et_pb_text _builder_version="3.22.4" 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_padding="20px|20px|20px|20px" _i="0" _address="0.0.0.0"]

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"]

Start with hints

[/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.

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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.22.4" 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="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]

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"]

Start with hints

[/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.

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Watch the video

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Connected Program at Cheenta

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Matrix additive-multiplicative :TIFR 2018 Part A Problem 19

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