Raising the bar Analytical function of a matrix Archives

Sep 18

Applications of the diagonal representation I

Applications of the diagonal representation I

When $A$ is symmetric, we can use the representation

(1) $A=UA_UU^T$ where $A_U=diag[\lambda_1,...,\lambda_n].$

1. Matrix positivity in terms of eigenvalues

Exercise 1. Let $A$ be a symmetric matrix. It is positive (non-negative) if and only if its eigenvalues are positive (non-negative).

Proof. By definition, we have to consider the quadratic form $x^TAx=x^TUA_UU^Tx=(U^Tx)^TA_UU^Tx.$ Denoting $y=U^Tx,$ we know that an orthogonal matrix preserves the norm: $\|y\|=\|x\|.$ This allows us to obtain a lower bound for the quadratic form

$x^TAx=y^TA_Uy=\sum\lambda_iy_i^2\geq\min_i\lambda_i\sum y_i^2=\min_i\lambda_i\|x\|^2.$

This implies the statement.

2. Analytical functions of a matrix

Definition 1. For a square matrix $A$ all non-negative integer powers $A^m$ are defined. This allows us to define an analytical function of a matrix $f(A)=\sum_{m=0}^\infty\frac{f^{(m)}(0)}{m!}A^m$ whenever a function $f$ has the Taylor decomposition $f(t)=\sum_{m=0}^\infty\frac{f^{(m)}(0)}{m!}t^m,$ $t\in R.$

Example 1. The exponential function $f(t)=e^t$ has the decomposition $e^t=\sum_{m=0}^\infty\frac{1}{m!}t^m.$ Hence, $e^{At}=\sum_{m=0}^\infty\frac{1}{m!}A^mt^m.$ Differentiating this gives

$\frac{de^{At}}{dt}=\sum_{m=1}^\infty\frac{1}{m!}A^mmt^{m-1}$ ( the constant term disappears)

$=A\sum_{m=1}^\infty\frac{1}{(m-1)!}A^{m-1}t^{m-1}=Ae^{At}.$

This means that $e^{At}$ solves the differential equation $x^\prime(t)=Ax(t).$ To satisfy the initial condition $x(t_0)=x_0,$ instead of $e^{At}$ we can consider $x(t)=e^{A(t-t_0)}x_0.$ This matrix function solves the initial value problem

(2) $x^\prime(t)=Ax(t),$ $x(t_{0})=x_0.$

Calculating all powers of a matrix can be a time-consuming business. The process is facilitated by the knowledge of the diagonal representation.

Exercise 2. If $A$ is symmetric, then $A^m=UA_U^mU^{-1}$ for all non-negative integer $m.$ Hence, $f(A)=U\left(\sum_{m=0}^\infty\frac{f^{(m)}(0)}{m!}A_U^m\right)U^{-1}=Uf(A_U)U^{-1}.$

Proof. The equation

(3) $A^2=UA_UU^{-1}UA_UU^{-1}=UA_UA_UU^{-1}=UA_U^2U^{-1}=Udiag[\lambda_1^2,...,\lambda_n^2]U^{-1}$

shows that to square $A$ it is enough to square $A_{U}.$ In a similar fashion we can find all non-negative integer powers of $A.$

Example 2. $e^{At}=U\left(\sum_{m=0}^\infty\frac{1}{m!}diag[\lambda_{1}^m,...,\lambda_n^m]t^m\right)U^{-1}.$

3. Linear differential equation

Example 1 about the exponential matrix function involves a bit of guessing. With the diagonal representation at hand, we can obtain the same result in a more logical way.

One-dimensional case. $\frac{dx}{dt}=ax$ is equivalent to $\frac{dx}{x}=adt.$ Upon integration this gives

$\log x(t)-\log x(t_0)=\int_{t_0}^td(\log x)=\int_{t_0}^tadt=a(t-t_0)$

(4) $x(t)=e^{a(t-t_0)}x_0.$

General case. In case of a $2\times 2$ matrix the first equation of (2) is $\frac{dx_1}{dt}=a_{11}x_1+a_{12}x_2.$ The fact that the x's on the right are mixed up makes the direct solution difficult. The idea is to split the system and separate the x's.

Premultiplying (2) by $U^{-1}$ we have $\frac{d(U^{-1}x)}{dt}=U^{-1}Ax=U^{-1}AUU^{-1}x$ or, denoting $U^{-1}x=y$ and using (1), $\frac{dy}{dt}=A_Uy.$ The last system is a collection of one-dimensional equations $\frac{dy_i}{dt}=\lambda_iy_i,$ $i=1,...,n.$ Let $y_0=U^{-1}x_0$ be the initial vector. From (4) $y_i=e^{\lambda_i(t-t_0)}y_{0i}.$ In matrix form this amounts to $y=e^{A_U(t-t_0)}y_0.$ Hence, as in Exercise 2, $x=Uy=Ue^{A_U(t-t_0)}U^{-1}Uy_0=e^{A(t-t_0)}x_0.$