Problem 5. Show that the eigenvalues of a unitary matrix...

Problem 5. Show that the eigenvalues of a unitary matrix have unit modulus and that the corresponding eigenvectors are orthogonal. (Hint: Compare with the methodology for Hermitian matrices.) Problem 6. Riley and Hobson problem #1.12 Given matrix $A = egin{pmatrix} 1 & alpha & 0 \ eta & 1 & 0 \ 0 & 0 & 1 end{pmatrix}$ where $alpha$ and $eta$ are non-zero complex numbers, find its eigenvalues and eigenvectors. Find the respective conditions for (a) the eigenvalues to be real (b) the eigenvectors to be orthogonal. (c) Show that the conditions are jointly satisfied if and only if A is Hermitian.

Transcript

00:01 In this question we have to show that first of all the eigenvalues of the unit sorry unit tree matrix have unit modulus and the corresponding eigenvector they are orthogonal are orthogonal so this is what we have to show that first of all we consider let a be the unitary matrix.

00:43 First of all it must be a square matrix.

00:47 It satisfies the condition that a times conjugate transpose of a that is denoted by a times a theta let's say this is nothing but the conjugate transpose of the matrix a.

01:08 So a times the transpose equals to transpose times a and this would always be equals to the identity matrix correct now let us talk about the eigenvalues how they have unit modulus so we consider that let lambda be the eigenvalue of the matrix a then what we will get is a v equals to lambda times of v now if we take this is our equation one taking the conjugate transpose of this equation we will get we conjugate v transpose times a conjugate equals to lambda conjugate times again v conjugate we are going to get this is our equation number two if you multiply these two we get we conjugate a conjugate times a v equals to lambda conjugate v conjugate lambda v we can group the terms together this is going to be b conjugate v because a conjugate a is just identity correct and here lambda is lambda times lambda conjugate will give us mod of lambda square similarly v times we conjugate will give us mod of v square that is mod here also we get modulus of v square.

02:43 These two will get cancelled.

02:45 We have modulus of lambda square equals to 1.

02:51 So this will imply the modulus of lambda is just equal to 1.

02:57 This will imply that the eigenvalues have unit modulus...

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