This lecture present some useful properties of matrix powers.
Table of contents
We can take non-negative integer powers of a square matrix by repeatedly multiplying by itself.
If is a positive integer and is a matrix, then
We adopt the convention thatwhere is the identity matrix.
Let be a matrix and let be the space of all column vectors.
Remember that the null space of is the subspace
In other words, the null space is formed by all the vectors of that are mapped into the zero vector by the linear transformation defined by .
Here is the first interesting result: the larger a matrix power, the bigger the null space.
Proposition Let be a matrix. Then, the null spaces of the powers of satisfyfor any non-negative integer .
Since and is the only vector mapped into the zero vector by the identity transformation, we haveFor any non-negative integer , let . Then, We can multiply both sides by , so as to getwhich implies that . Thus, all the members of are also members of . In other words, is included in .
As we take increasingly greater matrix powers, if at a certain point the growth of null spaces stops, then it stops forever.
Proposition Let be a matrix. If there exist a positive integer such thatthenfor any integer .
We only need to prove that implies because then we can generate a chain of equalities so as to obtain for any . We already know thatWe need to prove inclusion in the other direction. Suppose . Then,or, equivalently,Thus, we have and, by the hypothesis that , we also have The latter fact means thatorIn other words,Thus, all the members of are also members of . Hence, is included in , which is what we needed to prove.
The null spaces do not grow indefinitely as we take larger powers, but they eventually stabilize.
Proposition Let be a matrix. Then,
The proof is by contradiction. Suppose thatBy the previous proposition this implies thatfor all integers (otherwise we would have ). Therefore,where 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 must be at least equal to . This is impossible because is a subspace of the space of vectors (hence can have at most dimensions). Thus, the initial hypothesis has led us to a contradiction. As a consequence, we must have
By combining the previous two propositions, we obtainfor any integer .
Note that the last proposition does not imply that the null spaces stabilize exactly at , as it may well happen thatfor . However, the proposition guarantees that the stabilization point is not beyond .
Let be the space of all column vectors. Remember that the column space of a matrix , which coincides with the range of the linear transformation defined by the matrix, is
By the rank-nullity theorem, for any matrix power , we havewhere and denote the dimensions of the column space and the null space respectively.
We have demonstrated above that increases as we increase the power , up to . As a consequence, must decrease as we increase , up to .
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 a matrix power, the smaller its column space.
Proposition Let be the space of all column vectors. Let be a matrix. Then, the column spaces of the powers of satisfyfor any non-negative integer .
Since , for any . Therefore,For any non-negative integer , let . Then, by the definition of column space, there exists such that which can be written equivalently aswhich implies that . Thus, all the members of are also members of . In other words, includes .
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 be a matrix. If there exist a positive integer such thatthenfor any integer .
We only need to prove that implies because then we can generate a chain of equalities so as to obtain for any . We already know thatWe need to prove inclusion in the other direction. Let be the space of all column vectors. Take any and define By definition, . The hypothesis that implies that . Thus, there exists such thatorBy putting (1) and (2) together, we getThus, we have just proved that, for any , there exists such that (3) holds. Now, suppose . Then, there exists such thator By (3), there exists , such thatIn other words, . Thus, all the members of are also members of . In other words,
The column spaces do not grow indefinitely as we take larger powers, but they eventually stabilize.
Proposition Let be a matrix. Then,
The proof is by contradiction. Suppose thatBy the previous proposition this implies thatfor all integers (otherwise we would have ). Therefore,where is the space of all column vectors and 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 is , the dimension of must be less that . 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
By combining the previous two propositions, we obtainfor any integer .
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 , as it may well happen thatfor . However, the proposition guarantees that the stabilization point is not beyond .
By putting together the rank-nullity theorem and the propositions above, we see that, when we increase the power by one unit, then there are two mutually exclusive possibilities:
the dimension of the null space increases by at least one unit and the dimension of the row space decreases by the same amount;
the dimensions of and remain unchanged (and they never change again when we further increase ).
In particular, the column and null spaces must stabilize at the same point: there exists a unique integer such thatandfor any integer .
Please cite as:
Taboga, Marco (2017). "Matrix power", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/matrix-power.
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