Let . Compute .

Let . Evidently, is in Jordan canonical form.

Now , using this previous exercise. By this previous exercise, we have . So , and we have .

(WolframAlpha agrees.)

unnecessary lemmas. very sloppy. handwriting needs improvement.

Let . Compute .

Let . Evidently, is in Jordan canonical form.

Now , using this previous exercise. By this previous exercise, we have . So , and we have .

(WolframAlpha agrees.)

Let be an matrix over with 1 on the first superdiagonal and 0 elsewhere. Compute . Next, suppose is an Jordan block with eigenvalue , and compute .

First, given a parameter , we define a matrix as follows: , where if and otherwise. (That is, is the matrix with 1 on the th superdiagonal and 0 elsewhere, and .)

Lemma: for . Proof: We proceed by induction. Certainly the result holds for and . Now suppose it holds for some . We have . Consider entry : . If , then . So this sum is . Now if , then the whole sum is 0; otherwise, it is 1. So in fact .

Now . As we saw in this previous exercise, powers of past the th are all zero; so in fact we have . Note that the indices of nonzero entries of the are mutually exclusive. So is the matrix whose entry is if and 0 otherwise.

Now let be the Jordan block with eigenvalue . Note that ; using this previous exercise, we have . (Where is as above.)

Let and be commuting matrices. Prove that , where is a formal power series over a field with characteristic zero.

We will work in the ring of formal power series over with coefficients which are formal power series over . (Note that and commute.) Note that if is a formal power series and a polynomial, then is the formal power series obtained by ‘collecting like terms’. Remember the binomial theorem.

= | ||

= | ||

= | ||

= | ||

= | ||

= | ||

= | . |

In particular, if and commute, then .

Let be a power series over with radius of convergence . Let be an matrix over , and let be a nonsingular matrix. Prove the following.

- If converges, then converges, and .
- If and converges, then and converge and .
- If is a diagonal matrix with diagonal entries , then converges, converges for each , and is diagonal with diagonal entries .

Suppose converges. Then (by definition) the sequence of matrices converges entrywise. Let , , and . Now . That is, the entry of is . Since each sequence converges, this sum converges as well. In particular, converges (again by definition). Now since for each , the corresponding sequences for each are equal for each term, and so have the same limit. Thus .

Now suppose . We have . Since converges in each entry, each of and converge in each entry. So and converge. Again, because for each the corresponding sequences and are the same, they converge to the same limit, and thus .

Finally, suppose is diagonal. Then in fact we have , and so by the previous part, . In particular, converges, and is diagonal with diagonal entries as desired.

Let be a formal power series over with radius of convergence . Let be the matrix norm introduced in this previous exercise.

Given a matrix over , we can construct a sequence of matrices by taking the th partial sum of . That is, . This gives us sequences where is the entry of . Suppose converges to for each , and let . In this situation, we say that converges to , and that . (In other words, converges precisely when each entrywise sequence converges.)

- Prove that if , then converges.
- Deduce that for all matrices , the following power series converge: , , and .

[Disclaimer: My facility with even simple analytical concepts is laughably bad, but I’m going to give this a shot. Please let me know what’s wrong with this solution.]

We begin with a lemma.

Lemma: For all , , where the subscripts denote taking the entry. Proof: By the definition of matrix multiplication, we have , where and . (I’m abusing the notation a bit here; is an element of , which we think of as a function on .) Now by the triangle inequality. Note that is the sum of all possible -fold products of (absolute values of) entries of , and that we have bounded above by a sum of some distinct -fold products of (absolute values of) entries of . In particular, since the missing terms are all positive. Thus .

Let us define the formal power series by . What we have shown (I think) is that , using the triangle inequality.

Now recall, by the Cauchy-Hadamard theorem, that the radius of convergence of satisfies . In particular, has the same radius of convergence as . So in fact the sequence converges. Now the sequence is bounded and monotone increasing, and so must also converge. That is, is absolutely convergent, and so is convergent. Thus converges.

Now we will address the convergence of the series . Recall that the radius of convergence of satisfies . Now if is even and if is odd. So if is even and if is odd.

Brief aside: Suppose . Certainly , so that . Now , so that . In particular, if , then .

By our brief aside, we have that if is odd, and if is even. So (skipping redundant terms) we have satisfies . We know from the Brief Aside that this sequence is monotone decreasing. Now let . Consider now the sequence , , , et cetera. only finitely many terms of this sequence are less than or equal to 1. So there must exist some such that the product is greater than 1. Now , so that . Thus the sequence converges to 0, and so the radius of convergence of is .

By a similar argument, the radii of convergence for and are also .

Given an matrix over , define , where bars denote the complex modulus. Prove the following for all and all .

Say and .

We have by the triangle inequality. Rearranging, we have as desired.

Now latex \sum_{i,j} \sum_k |a_{i,k}|b_{k,j}|$ by the triangle inequality. Now since all the new terms are positive, and rearranging, we have .

Finally, we have .

Let be a field of characteristic 2. Compute the Jordan canonical form of a Jordan block of size and eigenvalue over . Characterize those matrices over which are squares; that is, characterize such that for some matrix .

Let be the Jordan block with eigenvalue and size . That is, if , if , and 0 otherwise. Now ; if or , then . Evidently then we have if , if , and 0 otherwise, Noting that . So , where is the zero matrix and is the identity matrix. Now let , where has dimension . Now . That is, ‘shifts’ the entries of – so and . In particular, the kernel of has dimension 2, so that by this previous exercise, the Jordan canonical form of has two blocks (both with eigenvalue .

Now , since has characteristic 2. Note that has order , since (evidently) we have and . So has order . If is even, then while , and if is odd, then while . So the minimal polynomial of is if is even and if is odd.

So the Jordan canonical form of has two Jordan blocks with eigenvalue . If is even, these have size , and if is odd, they have size .

Now let be an arbitrary matrix over (with eigenvalues in ). We claim that is a square if and only if the following hold.

- The eigenvalues of are square in
- For each eigenvalue of , the Jordan blocks with eigenvalue can be paired up so that the sizes of the blocks in each pair differ by 0 or 1.

To see the ‘if’ part, suppose is in Jordan canonical form, where and are Jordan blocks having the same eigenvalue and whose sizes differ by 0 or 1. By the first half of this exercise, is the Jordan canonical form of , where is a Jordan block with eigenvalue . Now is similar to the direct sum of these , and so . Then is square.

Conversely, suppose is square, and say is in Jordan canonical form. So . Letting denote the Jordan blocks of , we have . The Jordan canonical form of has two blocks with eigenvalue and whose sizes differ by 0 or 1, by the first half of this exercise. So the Jordan blocks of all have eigenvalues which are square in and can be paired so that the sizes in each pair differ by 0 or 1.

Let be a field whose characteristic is not 2. Characterize those matrices over which have square roots. That is, characterize such that for some matrix .

We claim that has a square root if and only if the following hold.

- Every eigenvalue of is square in .
- The Jordan blocks of having eigenvalue 0 can be paired up in such a way that the sizes of the blocks in each pair differ by 0 or 1.

First we tackle the ‘if’ part. Suppose is in Jordan canonical form, where are the Jordan blocks having nonzero eigenvalue and and are the blocks with eigenvalue 0 such that the sizes of and differ by 0 or 1. As we showed in this previous exercise, is the Jordan canonical form of the square of the Jordan block with eigenvalue , and is the Jordan canonical form of the square of the Jordan block whose size is with eigenvalue 0. So we have , and so is a square.

Conversely, suppose is a square, and say is in Jordan canonical form. Now . Now say is in Jordan canonical form; then . By the previous exercise, the nonzero eigenvalues of are square in , and the Jordan blocks having eigenvalue 0 can be paired so that the sizes in each pair differ by 0 or 1.

Let be a Jordan block of size and eigenvalue over a field whose characteristic is not 2.

- Suppose . Prove that the Jordan canonical form of is the Jordan block of size with eigenvalue .
- Suppose . Prove that the Jordan canonical form of has two blocks (with eigenvalue 0) of size if is even and of size if is odd.

First suppose .

Lemma: Let denote the th standard basis element (i.e. . If , then , , and . Proof: We have . Evidently, . Now , where if and if and 0 otherwise. So if , and similarly .

In particular, note that is nonzero (here we use the noncharacteristictwoness of ), while . So the minimal polynomial of is . Thus the Jordan canonical form of has a single Jordan block, of size , with eigenvalue .

Now suppose . Evidently, we have and . Now and . If is even, we have and , so that the minimal polynomial of is . If is odd, we have while , so the minimal polynomial of is .

Next, we claim that has dimension 2. To this end, note that is nonzero, while . By this previous exercise, has 2 Jordan blocks; thus the blocks of both have eigenvalue 0 and are of size if is even, and if is odd.