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1. Eigen Values of an orthogonal matrix will always be +1 and the modulus will also be always |1|? I have a doubt on this point .

Let λ be A eigenvalue and  Ax=λx.

(1) (x^t)Ax = λ(x^t)x > 0. Because (x^t)x > 0, then λ > 0.

(2) (|λ|^2)(x^t)x = {(Ax)^t}Ax = (x^t)(A^t)Ax = (x^t)x. So |λ|=1. Then λ=−1 or 1.

Note: A in (2) may have a complex eigenvalue with absolute value 1.

@Bharat Bhushan plz write the power properly….

@samarpita modulus will always be $1$ but eigen values need not be always $\pm 1.$ As mentioned in the above answer, It can be complex eigen values. If your orthogonal matrix is real-symmetric then you can say eigen values are $\pm 1$ and modulus is $1$ because for real-symmetric matrices, eigen values are real.

For a square matrix $A,$ orthogonality means $AA^T = A^T A = I$

According to Eigen-Value Equation:  $Ax = \lambda x$ where $x$ is an eigen vector for corresponding eigen value $\lambda.$

Pre-multiplying both sides with $A^T$  gives

$A^T Ax = \lambda A^T x \implies Ix = \lambda A^T x \implies x = \lambda A^T x \implies \frac{1}{\lambda} x = A^T x.$ It means matrix $A^T$ has eigen value $\frac{1}{\lambda}$ and same eigen vector $x.$

Since matrices $A$ and $A^T$ have the same eigen values (can be proved easily)

So, $\lambda = \frac{1}{\lambda} \implies \lambda – \frac{1}{\lambda} = 0 \implies \lambda^2 = 1.$

Since $x^2 = |x|^2,$ So, $\lambda^2 = 1 \implies |\lambda|^2 = 1$

If $\lambda$ is real then $\lambda = \pm 1$

otherwise $\lambda$ will be a complex number. So, $|\lambda|^2 = 1 \implies |\lambda|= 1$

$|\lambda| = 1$ shows that $\lambda$ will be on the unit circle because considering $\lambda = a+ib,$ $|\lambda|^2 = 1$ shows that $a^2 + b^2 = 1$ which is the equation of a circle and point $(a,b)$ lie on a circle. You can also write $\lambda = e^{i\theta} = \cos \theta + i \sin \theta$

So, whether $\lambda$ is real or complex, it will always lie on unit circle(circle with radius $1$) (only real eigen value is $\pm 1$) and its absolute value/modulus will always be $1$.

@ankitgupta.1729 got it sir

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