Linear Algebra/Jordan Canonical Form/Solutions

Solutions Problem 1 Do the check for Example 2.3. Answer We are required to check that That calculation is easy. Problem 2 Each matrix is in Jordan form. State its characteristic polynomial and its minimal polynomial. <li> <li> <li> <li> <li> <li> <li> <li> <li> Answer <li> The characteristic polynomial is and the minimal polynomial is the same. <li> The characteristic polynomial is . The minimal polynomial is . <li> The characteristic polynomial is and the minimal polynomial is the same. <li> The characteristic polynomial is The minimal polynomial is the same. <li> The characteristic polynomial is . The minimal polynomial is . <li> The characteristic polynomial is and the minimal polynomial is the same. <li> The characteristic polynomial is and the minimal polynomial is the same. <li> The characteristic polynomial is and the minimal polynomial is . <li> The characteristic polynomial is and the minimal polynomial is the same. Problem 3 Find the Jordan form from the given data. <li> The matrix is with the single eigenvalue . The nullities of the powers are: has nullity two, has nullity three, has nullity four, and has nullity five. <li> The matrix is with two eigenvalues. For the eigenvalue the nullities are: has nullity two, and has nullity four. For the eigenvalue the nullities are: has nullity one. Answer <li> The transformation is nilpotent (that is, is the entire space) and it acts on a string basis via two strings, and . Consequently, can be represented in this canonical form. and therefore is similar to this this canonical form matrix. <li> The restriction of the transformation is nilpotent on the subspace , and the action on a string basis is given as . The restriction of the transformation is nilpotent on the subspace , having the action on a string basis of and . Consequently the Jordan form is this (note that the blocks are arranged with the least eigenvalue first). Problem 4 Find the change of basis matrices for each example. <li> Example 2.13 <li> Example 2.14 <li> Example 2.15 Answer For each, because many choices of basis are possible, many other answers are possible. Of course, the calculation to check if an answer gives that is in Jordan form is the arbiter of what's correct. <li> Here is the arrow diagram. x150px The matrix to move from the lower left to the upper left is this. The matrix to move from the upper right to the lower right is the inverse of . <li> We want this matrix and its inverse. <li> The concatenation of these bases for the generalized null spaces will do for the basis for the entire space. The change of basis matrices are this one and its inverse. Problem 5 Find the Jordan form and a Jordan basis for each matrix. <li> <li> <li> <li> <li> <li> <li> Answer The general procedure is to factor the characteristic polynomial to get the eigenvalues , , etc. Then, for each we find a string basis for the action of the transformation when restricted to , by computing the powers of the matrix and finding the associated null spaces, until these null spaces settle down (do not change), at which point we have the generalized null space. The dimensions of those null spaces (the nullities) tell us the action of on a string basis for the generalized null space, and so we can write the pattern of subdiagonal ones to have . From this matrix, the Jordan block associated with is immediate . Finally, after we have done this for each eigenvalue, we put them together into the canonical form. <li> The characteristic polynomial of this matrix is , so it has only the single eigenvalue . (Thus, this transformation is nilpotent: is the entire space). From the nullities we know that 's action on a string basis is . This is the canonical form matrix for the action of on and this is the Jordan form of the matrix. Note that if a matrix is nilpotent then its canonical form equals its Jordan form. We can find such a string basis using the techniques of the prior section. The first basis vector has been taken so that it is in the null space of but is not in the null space of . The second basis vector is the image of the first under . <li> The characteristic polynomial of this matrix is , so it is a single-eigenvalue matrix. (That is, the generalized null space of is the entire space.) We have and so the action of on an associated string basis is . Thus, the Jordan form of T is and choosing vectors from the above null spaces gives this string basis (many other choices are possible). <li> The characteristic polynomial has two roots and they are the eigenvalues and . We handle the two eigenvalues separately. For , the calculation of the powers of yields and the null space of is the same. Thus this set is the generalized null space . The nullities show that the action of the restriction of to the generalized null space on a string basis is . A similar calculation for gives these null spaces. (The null space of is the same, as it must be because the power of the term associated with in the characteristic polynomial is two, and so the restriction of to the generalized null space is nilpotent of index at most two— it takes at most two applications of for the null space to settle down.) The pattern of how the nullities rise tells us that the action of on an associated string basis for is . Putting the information for the two eigenvalues together gives the Jordan form of the transformation . We can take elements of the nullspaces to get an appropriate basis. <li> The characteristic polynomial is . For the eigenvalue , calculation of the powers of yields this. The null space of is the same, and so this is the generalized null space . Thus the action of the restriction of to on an associated string basis is . For , computing the powers of yields and so the action of on a string basis for is . Therefore the Jordan form is and a suitable basis is this. <li> The characteristic polynomial of this matrix is . This matrix has only a single eigenvalue, . By finding the powers of we have and so the action of on an associated string basis is . The Jordan form is this and one choice of basis is this. <li> The characteristic polynomial has only a single root, so the matrix has only a single eigenvalue . Finding the powers of and calculating the null spaces shows that the action of the nilpotent map on a string basis is and . Therefore the Jordan form is and an appropriate basis (a string basis associated with ) is this. <li> The characteristic polynomial is a bit large for by-hand calculation, but just manageable . This is a single-eigenvalue map, so the transformation is nilpotent. The null spaces and the nullities show that the action of on a string basis is and . The Jordan form is and finding a suitable string basis is routine. Problem 6 Find all possible Jordan forms of a transformation with characteristic polynomial . Answer There are two eigenvalues, and . The restriction of to could have either of these actions on an associated string basis. The restriction of to could have either of these actions on an associated string basis. In combination, that makes four possible Jordan forms, the two first actions, the second and first, the first and second, and the two second actions. Problem 7 Find all possible Jordan forms of a transformation with characteristic polynomial . Answer The restriction of to can have only the action . The restriction of to could have any of these three actions on an associated string basis. Taken together there are three possible Jordan forms, the one arising from the first action by (along with the only action from ), the one arising from the second action, and the one arising from the third action. Problem 8 Find all possible Jordan forms of a transformation with characteristic polynomial and minimal polynomial . Answer The action of on a string basis for must be . Because of the power of in the minimal polynomial, a string basis for has length two and so the action of on must be of this form. Therefore there is only one Jordan form that is possible. Problem 9 Find all possible Jordan forms of a transformation with characteristic polynomial and minimal polynomial . Answer There are two possible Jordan forms. The action of on a string basis for must be . There are two actions for on a string basis for that are possible with this characteristic polynomial and minimal polynomial. The resulting Jordan form matrics are these. Problem 10 Diagonalize these. <li> <li> Answer <li> The characteristic polynomial is . For we have (of course, the null space of is the same). For , (and the null space of is the same). We can take this basis to get the diagonalization. <li> The characteristic polynomial is . For , and the null space of is the same. For and the null space of is the same. We can take this basis to get a diagonalization. Problem 11 Find the Jordan matrix representing the differentiation operator on . Answer The transformation is nilpotent. Its action on is . Its Jordan form is its canonical form as a nilpotent matrix. Problem 12 Decide if these two are similar. Answer Yes. Each has the characteristic polynomial . Calculations of the powers of and gives these two. (Of course, for each the null space of the square is the entire space.) The way that the nullities rise shows that each is similar to this Jordan form matrix and they are therefore similar to each other. Problem 13 Find the Jordan form of this matrix. Also give a Jordan basis. Answer Its characteristic polynomial is which has complex roots . Because the roots are distinct, the matrix is diagonalizable and its Jordan form is that diagonal matrix. To find an associated basis we compute the null spaces. For instance, and so we get a description of the null space of by solving this linear system. (To change the relation so that the leading variable is expressed in terms of the free variable , we can multiply both sides by .) As a result, one such basis is this. Problem 14 How many similarity classes are there for matrices whose only eigenvalues are and ? Answer We can count the possible classes by counting the possible canonical representatives, that is, the possible Jordan form matrices. The characteristic polynomial must be either or . In the case there are two possible actions of on a string basis for . There are two associated Jordan form matrices. Similarly there are two Jordan form matrices that could arise out of . So in total there are four possible Jordan forms. Problem 15 Prove that a matrix is diagonalizable if and only if its minimal polynomial has only linear factors. Answer Jordan form is unique. A diagonal matrix is in Jordan form. Thus the Jordan form of a diagonalizable matrix is its diagonalization. If the minimal polynomial has factors to some power higher than one then the Jordan form has subdiagonal 's, and so is not diagonal. Problem 16 Give an example of a linear transformation on a vector space that has no non-trivial invariant subspaces. Answer One example is the transformation of that sends to . Problem 17 Show that a subspace is invariant if and only if it is invariant. Answer Apply Lemma 2.7 twice; the subspace is invariant if and only if it is invariant, which in turn holds if and only if it is invariant. Problem 18 Prove or disprove: two matrices are similar if and only if they have the same characteristic and minimal polynomials. Answer False; these two matrices each have and . Problem 19trace The trace of a square matrix is the sum of its diagonal entries. <li> Find the formula for the characteristic polynomial of a matrix. <li> Show that trace is invariant under similarity, and so we can sensibly speak of the "trace of a map". (Hint: see the prior item.) <li> Is trace invariant under matrix equivalence? <li> Show that the trace of a map is the sum of its eigenvalues (counting multiplicities). <li> Show that the trace of a nilpotent map is zero. Does the converse hold? Answer <li> The characteristic polynomial is this. Note that the determinant appears as the constant term. <li> Recall that the characteristic polynomial is invariant under similarity. Use the permutation expansion formula to show that the trace is the negative of the coefficient of . <li> No, there are matrices and that are equivalent (for some nonsingular and ) but that have different traces. An easy example is this. Even easier examples using matrices are possible. <li> Put the matrix in Jordan form. By the first item, the trace is unchanged. <li> The first part is easy; use the third item. The converse does not hold: this matrix has a trace of zero but is not nilpotent. Problem 20 To use Definition 2.6 to check whether a subspace is invariant, we seemingly have to check all of the infinitely many vectors in a (nontrivial) subspace to see if they satisfy the condition. Prove that a subspace is invariant if and only if its subbasis has the property that for all of its elements, is in the subspace. Answer Suppose that is a basis for a subspace of some vector space. Implication one way is clear; if is invariant then in particular, if then . For the other implication, let and note that is in as any subspace is closed under linear combinations. Problem 21 Is invariance preserved under intersection? Under union? Complementation? Sums of subspaces? Answer Yes, the intersection of invariant subspaces is invariant. Assume that and are invariant. If then by the invariance of and by the invariance of . Of course, the union of two subspaces need not be a subspace (remember that the - and -axes are subspaces of the plane but the union of the two axes fails to be closed under vector addition, for instance it does not contain .) However, the union of invariant subsets is an invariant subset; if then or so or . No, the complement of an invariant subspace need not be invariant. Consider the subspace of under the zero transformation. Yes, the sum of two invariant subspaces is invariant. The check is easy. Problem 22 Give a way to order the Jordan blocks if some of the eigenvalues are complex numbers. That is, suggest a reasonable ordering for the complex numbers. Answer One such ordering is the dictionary ordering. Order by the real component first, then by the coefficient of . For instance, but . Problem 23 Let be the vector space over the reals of degree polynomials. Show that if then is an invariant subspace of under the differentiation operator. In , does any of , ..., have an invariant complement? Answer The first half is easy— the derivative of any real polynomial is a real polynomial of lower degree. The answer to the second half is "no"; any complement of must include a polynomial of degree , and the derivative of that polynomial is in . Problem 24even In , the vector space (over the reals) of degree polynomials, and are the even and the odd polynomials; is even while is odd. Show that they are subspaces. Are they complementary? Are they invariant under the differentiation transformation? Answer For the first half, show that each is a subspace and then observe that any polynomial can be uniquely written as the sum of even-powered and odd-powered terms (the zero polynomial is both). The answer to the second half is "no": is even while is odd. Problem 25 Lemma 2.8 says that if and are invariant complements then has a representation in the given block form (with respect to the same ending as starting basis, of course). Does the implication reverse? Answer Yes. If has the given block form, take to be the first vectors of , where is the upper left submatrix. Take to be the remaining vectors in . Let and be the spans of and . Clearly and are complementary. To see is invariant ( works the same way), represent any with respect to , note the last components are zeroes, and multiply by the given block matrix. The final components of the result are zeroes, so that result is again in . Problem 26square root A matrix is the square root of another if . Show that any nonsingular matrix has a square root. Answer Put the matrix in Jordan form. By non-singularity, there are no zero eigenvalues on the diagonal. Ape this example: to construct a square root. Show that it holds up under similarity: if then .