Understand the problem

True or false: Let $A$ be a $3 \times 3$ real symmetric matix s.t $A^6=I$. Then $A^2=I$
Source of the problem
TIFR 2018 Part A, Problem 6
Linear Algebra
Medium
Suggested Book
Linear Algebra, Hoffman and Kunze

Do you really need a hint? Try it first!

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.
If matrix $A$ then there exists $Q$ with $Q'Q=I$ such that $A = Q'BQ$,where $B$ is a diagonal matrix with diagonal entries being the eigenvalues of A which are real numbers.
Here assume that the eigen values are $a,b,c$.
We will use this spectral decomposition of A.
Observe that $A^6 = I \Rightarrow Q'(B^6)Q = I$.(Check!) $Q'(B^6)Q=I \Rightarrow Q[Q'(B^6)Q]Q'= QQ'= I \Rightarrow B^6=I$ . [as Q is orthogonal] $B^6$ is a diagonal matrix with diagonal entries real numbers raised to the power 6 i.e $a^6,b^6$ and $c^6$.
Now hint 3 implies $a^6=1,b^6=1$ and $c^6=1$. $a^2=1,b^2=1$ and $c^2=1$ as $a,b$ and $c$ are real numbers. This means $B^2=I$. $A^2 = Q'(B^2)Q = Q'Q = I$.
Hence the given statement is TRUE.

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