Raising the bar

Applications of the diagonal representation III

6. Absolute value of a matrix

We know that for any matrix $latex A$, the matrix $latex A^TA$ is symmetric and non-negative.

For a complex number $latex c$ the absolute value is defined by $latex |c|=(\bar{c}c)^{1/2}$. 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, the matrix $latex A^TA$ has non-negative eigenvalues. Hence we can define the absolute value of $latex A$ as a square root of $latex A^TA,$  $latex |A|=(A^TA)^{1/2}.$

7. Polar form

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

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

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

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

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

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

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

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