Question

Show that if \(A\) is diagonalizable, with all eigenvalues less than 1 in magnitude, then \({A^k}\) tends to the zero matrix as \(k \to \infty \). (Hint: Consider \({A^k}x\) where \(x\) represents any one of the columns of \(I\).)

Short answer

Since all eigenvalues are in magnitude less than one, \({\lambda ^k} \to 0\) as \(k\) increases, so \(D\) becomes zero matrix.

Step-by-step solution

Definition of zero matrix and Diagonal Matrix

In mathematics, particularly linear algebra, a zero matrix or null matrix is a matrix all of whose entries are zero.

It also serves as the additive identity of the additive group of matrices, and is denoted by the symbol or followed by subscripts corresponding to the dimension of the matrix as the context sees fit.

A diagonal matrix is matrix where all nondiagonal entries are zero.

Showing if \(A\) is diagonalizable, with all eigenvalues less than 1 in magnitude then \({A^k}\) tends to Zero Matrix

Since\(A\)is diagonalizable,\(A = PD{P^{ - 1}}\)with some invertible matrix\(P\)and diagonal matrix\(D\)with eigenvalues on its diagonal.

Then we have,

\(\begin{aligned}{c}{A^2} &= AA\\ &= PD{P^{ - 1}}PD{P^{ - 1}}\\ &= P{D^2}{P^{ - 1}}\end{aligned}\)

\(\begin{aligned}{c}{A^3} &= {A^2}A\\ &= P{D^2}{P^{ - 1}}PD{P^{ - 1}}\\ &= P{D^3}{P^{ - 1}}\end{aligned}\)

This implies that \({A^k} = P{D^k}{P^{ - 1}}\).

When raising the diagonal matrix to power\(k\), it just raises every element on the diagonal to power\(k\).

Since all eigenvalues are in magnitude less than one, \({\lambda ^k} \to 0\) as \(k\) increases, so \(D\) becomes zero matrix.