1 votes 1 votes 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 . Linear Algebra linear-algebra eigen-value matrix + – samarpita asked May 9, 2022 samarpita 9.7k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes 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. Ref 1: https://math.stackexchange.com/questions/653133/eigenvalues-in-orthogonal-matrices#653143 Ref 2: https://www.youtube.com/watch?v=PFDu9oVAE-g Ref 3: https://math.mit.edu/~gs/linearalgebra/linearalgebra5_6-1.pdf Ref 4: https://math.mit.edu/~gs/linearalgebra/ila0601.pdf Ref 5: http://scipp.ucsc.edu/~haber/ph116A/Rotation2.pdf Bharat Bhushan answered May 9, 2022 • selected May 10, 2022 by samarpita Bharat Bhushan comment Share Follow See all 4 Comments See all 4 4 Comments reply samarpita commented May 9, 2022 reply Follow Share @Bharat Bhushan plz write the power properly…. 1 votes 1 votes samarpita commented May 9, 2022 reply Follow Share @Sachin Mittal 1 @ankitgupta.1729 @Shaik Masthan @Deepak Poonia sir can you plz confirm this 1 votes 1 votes ankitgupta.1729 commented May 9, 2022 reply Follow Share @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 $1$. 4 votes 4 votes samarpita commented May 9, 2022 reply Follow Share @ankitgupta.1729 got it sir 1 votes 1 votes Please log in or register to add a comment.