Question #106348

Let λ ∈ R be an eigenvalue of an orthogonal matrix A. Show that λ = ±1.

(Hint: consider the norm of Av, where v is an eigenvector of A associated with the

eigenvalue λ.)


Also, find diagonal orthogonal matrices B, C such that 1 is an eigenvalue of B

and −1 is an eigenvalue of C.

Expert's answer

A square matrix AA is said to be orthogonal if AA′=IAA'=I =A′A=A'A

where A′=A'= Transpose of AA and I=I= Identity matrix.

Suppose λ∈R\lambda \in R be a eigenvalue of AA .

Then there exists a non zero eigenvector XX such that

AX=λXAX=\lambda X ......(1)......(1)

Taking transpose of both sides of the above equality, we get

(AX)′=(λX)′(AX)'=(\lambda X)'

  ⟹  \implies X′A′=X′λX'A'=X' \lambda

  ⟹  \implies X′A′=λX′X'A'=\lambda X'

Multiplying both sides by AXAX we get,

X′A′AX=λX′AXX'A'AX=\lambda X'AX

  ⟹  \impliesX′X=λX′λXX'X=\lambda X' \lambda X [from equation (1)]

  ⟹  X′X=λ2X′X\implies X'X={ \lambda }^2 X'X

  ⟹  (1−λ2)X′X=0\implies (1- { \lambda}^2)X'X=0

Since ,X≠0  ⟹  X′X≠0.X \neq0 \implies X'X\neq0.

Therefore,(1−λ2)=0(1-\lambda^2)=0

  ⟹  λ2=1\implies \lambda^2=1

Hence λ=1,−1\lambda =1,-1 .

(Proved)(Proved) .

Let B=B= (1001)\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} which is a diagonal matrix and BB′=B′B=IBB'=B'B=I

Hence ,BB is an othogonal matrix ,whose eigen values are +1,+1.+1, +1.

Let C=(−100−1)C=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} which is a diagonal matrix and CC′=C′C=ICC'=C'C=I

Hence ,CC is an orthogonal matrix, whose eigen values are −1,−1.-1,-1. .




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