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[/et_pb_text][et_pb_text _builder_version="3.26.6" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]True or false: Let $latex A$ be a $latex 3 \times 3$ real symmetric matix s.t $latex A^6=I$. Then $latex A^2=I$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.26.8" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.26.8"]TIFR 2018 Part A, Problem 6[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.26.8" open="off"]Linear Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.26.8" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.26.8" 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"]

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.26.8" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.26.8"]First investigate the properties of Real Symmetric Matrices.
The most important property of a real symmetric matrix A is encoded in “Spectral Decomposition” of A.
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.26.8"]
If matrix $latex A$ then there exists $latex Q$ with $latex Q'Q=I$ such that $latex A = Q'BQ$,where $latex B$ is a diagonal matrix with diagonal entries being the eigenvalues of A which are real numbers.
Here assume that the eigen values are $latex a,b,c$.
We will use this spectral decomposition of A.
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.26.8"]
Observe that $latex A^6 = I \Rightarrow Q'(B^6)Q = I$.(Check!)
$latex Q'(B^6)Q=I \Rightarrow Q[Q'(B^6)Q]Q'= QQ'= I \Rightarrow B^6=I$ . [as Q is orthogonal]
$latex B^6$ is a diagonal matrix with diagonal entries real numbers raised to the power 6 i.e $latex a^6,b^6$ and $latex c^6$.
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.26.8"]
Now hint 3 implies $latex a^6=1,b^6=1$ and $latex c^6=1$.
$latex a^2=1,b^2=1$ and $latex c^2=1$ as $latex a,b$ and $latex c$ are real numbers. This means $latex B^2=I$.
$latex A^2 = Q'(B^2)Q = Q'Q = I$.
Hence the given statement is TRUE.

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