9
Sep 18

Applications of the diagonal representation III

Applications of the diagonal representation III

6. Absolute value of a matrix

Exercise 1. For a square matrix A the matrix A^TA is non-negative.

Proof. x^TA^TAx=(Ax)^TAx=\|Ax\|^2\geq 0 for any x.

For a complex number c the absolute value is defined by |c|=(\bar{c}c)^{1/2}. Since transposition of matrices is similar to conjugation of complex numbers, this leads us to the following definition.

Definition 1. By Exercise 1 from the previous post and Exercise 1, the matrix A^TA has non-negative eigenvalues. Hence we can define the absolute value of A as a square root of A^TA,  |A|=(A^TA)^{1/2}.

7. Polar form

A complex nonzero number c in polar form is c=\rho e^{i\theta} where \rho >0 is the absolute value of c and \theta is a real angle, so that |e^{i\theta}|=1. The matrix analog of this form obtains when the condition \rho >0 is replaced by \det A\neq 0, the absolute value of A from Definition 1 is used and an orthogonal matrix plays the role of e^{i\theta}.

Exercise 2. For a symmetric matrix A, its determinant equals the product of its eigenvalues.

Proof. \det A=\det(Udiag[\lambda_1,...,\lambda_n]U^{-1})=(\det U)(\det diag[\lambda_1,...,\lambda_n])(\det(U^{-1}))

=(\det U)^2\det diag[\lambda_1,...,\lambda_n]=\lambda_1...\lambda_n.

Exercise 3. Let \det A\neq 0. Put U=A|A|^{-1}. Then U is orthogonal and the polar form of A is A=U|A|.

Proof. Let \lambda_1,...,\lambda_n be the eigenvalues of A^TA. They are real and non-negative. In fact they are all positive because by Exercise 2 their product equals \det(A^TA)=(\det A)^2.

Hence, |A|^{-1} exists and it is symmetric. U also exists and is orthogonal: U^TU=(|A|^{-1})^TA^TA|A|^{-1}=|A|^{-1}|A|^2|A|^{-1}=I. Finally, from the definition of U, A=U|A|.

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