Search for probability and statistics terms on Statlect
StatLect
Index > Matrix algebra

Matrix power

by , PhD

This lecture present some useful properties of matrix powers.

Table of Contents

Integer matrix powers

We can take non-negative integer powers of a square matrix A by repeatedly multiplying A by itself.

If $m$ is a positive integer and A is a $K imes K$ matrix, then[eq1]

We adopt the convention that[eq2]where I is the $K imes K$ identity matrix.

Null space

Let A be a $K imes K$ matrix and let $S$ be the space of all Kx1 column vectors.

Remember that the null space of A is the subspace[eq3]

In other words, the null space is formed by all the vectors of $S$ that are mapped into the zero vector by the linear transformation defined by A.

The larger the power the bigger the null space

Here is the first interesting result: the larger a matrix power, the bigger the null space.

Proposition Let A be a $K imes K$ matrix. Then, the null spaces of the powers of A satisfy[eq4]for any non-negative integer $m$.

Proof

Since $A^{0}=I$ and 0 is the only vector mapped into the zero vector by the identity transformation, we have[eq5]For any non-negative integer $m$, let [eq6]. Then, [eq7]We can multiply both sides by A, so as to get[eq8]which implies that [eq9]. Thus, all the members of [eq10] are also members of [eq11]. In other words, [eq10] is included in [eq13].

If null spaces stop growing then they never resume

As we take increasingly greater matrix powers, if at a certain point the growth of null spaces stops, then it stops forever.

Proposition Let A be a $K imes K$ matrix. If there exist a positive integer $m$ such that[eq14]then[eq15]for any integer $jgeq 1$.

Proof

We only need to prove that [eq16] implies [eq17] because then we can generate a chain of equalities so as to obtain [eq18] for any $j$. We already know that[eq19]We need to prove inclusion in the other direction. Suppose [eq20]. Then,[eq21]or, equivalently,[eq22]Thus, we have [eq23]and, by the hypothesis that [eq24], we also have [eq25]The latter fact means that[eq26]or[eq27]In other words,[eq28]Thus, all the members of [eq29] are also members of [eq30]. Hence, [eq31] is included in [eq30], which is what we needed to prove.

Stabilization of null spaces

The null spaces do not grow indefinitely as we take larger powers, but they eventually stabilize.

Proposition Let A be a $K imes K$ matrix. Then,[eq33]

Proof

The proof is by contradiction. Suppose that[eq34]By the previous proposition this implies that[eq35]for all integers $mleq K-1$ (otherwise we would have [eq36]). Therefore,[eq37]where $subset $ denotes strict inclusion (i.e., each null space has at least one more member than the null space immediately preceding it). Since null spaces are subspaces, having at least one more member is the same as having at least an additional dimension. As a consequence, the dimension of [eq38] must be at least equal to $K+1$. This is impossible because [eq39] is a subspace of the space of Kx1 vectors (hence [eq40] can have at most K dimensions). Thus, the initial hypothesis has led us to a contradiction. As a consequence, we must have [eq41]

By combining the previous two propositions, we obtain[eq42]for any integer $jgeq 1$.

Note that the last proposition does not imply that the null spaces stabilize exactly at [eq43], as it may well happen that[eq44]for $m<K-1$. However, the proposition guarantees that the stabilization point is not beyond [eq45].

Column space

Let $S$ be the space of all Kx1 column vectors. Remember that the column space of a $K imes K$ matrix A, which coincides with the range of the linear transformation defined by the matrix, is[eq46]

By the rank-nullity theorem, for any matrix power $A^{m}$, we have[eq47]where [eq48] and [eq49] denote the dimensions of the column space and the null space respectively.

We have demonstrated above that [eq50] increases as we increase the power $m$, up to [eq51]. As a consequence, [eq52] must decrease as we increase $m$, up to [eq53].

In other words, the behavior of column spaces is opposite to that of null spaces. The next sections discuss this point in more detail.

The larger the power the smaller the column space

The larger a matrix power, the smaller its column space.

Proposition Let $S$ be the space of all Kx1 column vectors. Let A be a $K imes K$ matrix. Then, the column spaces of the powers of A satisfy[eq54]for any non-negative integer $m$.

Proof

Since $A^{0}=I$, $s=A^{0}s$ for any $sin S$. Therefore,[eq55]For any non-negative integer $m$, let [eq56]. Then, by the definition of column space, there exists $tin S$ such that [eq57]which can be written equivalently as[eq58]which implies that [eq59]. Thus, all the members of [eq60] are also members of [eq61]. In other words, [eq62] includes [eq60].

If column spaces stop shrinking then they never resume

As we take increasingly greater matrix powers, if at a certain point the column spaces stop shrinking, then they will never shrink again.

Proposition Let A be a $K imes K$ matrix. If there exist a positive integer $m$ such that[eq64]then[eq65]for any integer $jgeq 1$.

Proof

We only need to prove that [eq66] implies [eq67] because then we can generate a chain of equalities so as to obtain [eq68] for any $j$. We already know that[eq69]We need to prove inclusion in the other direction. Let $S$ be the space of all Kx1 column vectors. Take any $sin S$ and define [eq70]By definition, [eq71]. The hypothesis that [eq72] implies that [eq73]. Thus, there exists $tin S$ such that[eq74]or[eq75]By putting (1) and (2) together, we get[eq76]Thus, we have just proved that, for any $sin S$, there exists $tin S$ such that (3) holds. Now, suppose [eq73]. Then, there exists $sin S$ such that[eq78]or [eq79]By (3), there exists $tin S$, such that[eq80]In other words, [eq81]. Thus, all the members of [eq82] are also members of [eq83]. In other words, [eq84]

Stabilization of column spaces

The column spaces do not grow indefinitely as we take larger powers, but they eventually stabilize.

Proposition Let A be a $K imes K$ matrix. Then,[eq85]

Proof

The proof is by contradiction. Suppose that[eq86]By the previous proposition this implies that[eq87]for all integers $mleq K-1$ (otherwise we would have [eq88]). Therefore,[eq89]where $S$ is the space of all Kx1 column vectors and $supset $ denotes strict inclusion. In other words, each column space has at least one less member than the column space immediately preceding it. Since column spaces are subspaces, having at least one less member is the same as having at least one less dimension. Since the dimension of $S$ is K, the dimension of [eq90] must be less than $-1$. This is impossible because the dimension cannot be a negative number. Thus, the initial hypothesis has led us to a contradiction. As a consequence, we must have [eq91]

By combining the previous two propositions, we obtain[eq92]for any integer $jgeq 1$.

A comment similar to those made for null spaces is in order: the last proposition does not imply that the columns spaces stabilize exactly at [eq93], as it may well happen that[eq94]for $m<K-1$. However, the proposition guarantees that the stabilization point is not beyond [eq95].

Null and column spaces stabilize at the same point

By putting together the rank-nullity theorem [eq96] and the propositions above, we see that, when we increase the power $m$ by one unit, then there are two mutually exclusive possibilities:

  1. the dimension of the null space [eq13] increases by at least one unit and the dimension of the row space [eq98] decreases by the same amount;

  2. the dimensions of [eq13] and [eq100] remain unchanged (and they never change again when we further increase $m$).

In particular, the column and null spaces must stabilize at the same point: there exists a unique integer $m$ such that[eq65]and[eq15]for any integer $jgeq 1$.

How to cite

Please cite as:

Taboga, Marco (2021). "Matrix power", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/matrix-power.

The books

Most of the learning materials found on this website are now available in a traditional textbook format.

Featured pages
Explore
Main sections
About
Glossary entries
Share
  • To enhance your privacy,
  • we removed the social buttons,
  • but don't forget to share.