Tag Archives: jordan canonical form

Compute the matrix exponential of a given matrix

Compute the exponential of the following matrices.

  1. A = \begin{bmatrix} 2 & -2 & 14 \\ 0 & 3 & -7 \\ 0 & 0 & 2 \end{bmatrix}
  2. B = \begin{bmatrix} 9 & 4 & 5 \\ -4 & 0 & -3 \\ -6 & -4 & -2 \end{bmatrix}
  3. C = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}

Let P = \begin{bmatrix} -7 & -1 & -2 \\ 7 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}. Evidently, P^{-1}AP = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} is in Jordan canonical form. now P^{-1}\mathsf{exp}(A)P = \begin{bmatrix} e^2 & 0 & 0 \\ 0 & e^2 & 0 \\ 0 & 0 & e^3 \end{bmatrix} using this previous exercise. So \mathsf{exp}(A) = \begin{bmatrix} e^2 & 2e^2-2e^3 & 14e^3-14e^2 \\ 0 & e^3 & 7e^2 - 7e^3 \\ 0 & 0 & e^2 \end{bmatrix}. (WolframAlpha agrees.)

Now consider B; we found in this previous exercise an invertible matrix Q such that Q^{-1}BQ = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} is in Jordan canonical form. Again by this previous exercise we have Q^{-1}\mathsf{exp}(B)Q = \begin{bmatrix} e^2 & e^2 & 0 \\ 0 & e^2 & 0 \\ 0 & 0 & e^3 \end{bmatrix}, so that \mathsf{exp}(B) = \begin{bmatrix} 2e^2 + 3e^3 & 4e^2 & 3e^3-e^2 \\ -2e^3 & -e^2 & e^2-2e^3 \\ -2e^2-2e^3 & -4e^2 & e^2-2e^3 \end{bmatrix}. (WolframAlpha agrees.)

Finally, consider C. In this previous exercise, we exhibited an invertible matrix R such that R^{-1}CR = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \oplus \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} is in Jordan canonical form. Now R^{-1}\mathsf{exp}(C)R = \begin{bmatrix} e & e & 0 & 0 \\ 0 & e & 0 & 0 \\ 0 & 0 & e & e \\ 0 & 0 & 0 & e \end{bmatrix}, and so \mathsf{exp}(C) = eC. (WolframAlpha agrees.)

Compute the exponential of a matrix

Let D = \begin{bmatrix} 1 & 2 & -4 & 4 \\ 2 & -1 & 4 & -8 \\ 1 & 0 & 1 & -2 \\ 0 & 1 & -2 & 3 \end{bmatrix}. Compute \mathsf{exp}(D).


Let P = \begin{bmatrix} 0 & 1 & 2 & 0 \\ 2 & 0 & -2 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}. Evidently, P^{-1}DP = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} is in Jordan canonical form.

Now \mathsf{exp}(P^{-1}DP) = \mathsf{exp} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \oplus \mathsf{exp} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, using this previous exercise. By this previous exercise, we have \mathsf{exp} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} e & e \\ 0 & e \end{bmatrix}. So P^{-1}\mathsf{exp}(D)P = e P^{-1}DP, and we have \mathsf{exp}(D) = eD.

(WolframAlpha agrees.)

Matrices with square roots over fields of characteristic 2

Let F be a field of characteristic 2. Compute the Jordan canonical form of a Jordan block J of size n and eigenvalue \lambda over F. Characterize those matrices A over F which are squares; that is, characterize A such that A = B^2 for some matrix B.


Let J = [b_{i,j}] be the Jordan block with eigenvalue \lambda and size n. That is, b_{i,j} = \lambda if j = i, 1 if j = i+1, and 0 otherwise. Now J^2 = [\sum_k b_{i,k}b_{k,j}]; if k \neq i or k \neq i+1, then b_{i,k} = 0. Evidently then we have (J^2)_{i,j} = \lambda^2 if j = i, 1 if j = i+2, and 0 otherwise, Noting that 2 = 0. So J^2 - \lambda^2 I = \begin{bmatrix} 0 & I \\ 0_2 & 0 \end{bmatrix}, where 0_2 is the 2 \times 2 zero matrix and I is the (n-2) \times (n-2) identity matrix. Now let v = \begin{bmatrix} V_1 \\ V_2 \end{bmatrix}, where V_1 has dimension 2 \times 1. Now (J^2 - \lambda^2 I)v = \begin{bmatrix} V_2 \\ 0 \end{bmatrix}. That is, J^2 - \lambda^2 I ‘shifts’ the entries of v– so e_{i+2} \mapsto e_i and e_1, e_2 \mapsto 0. In particular, the kernel of J^2 - \lambda^2 I has dimension 2, so that by this previous exercise, the Jordan canonical form of J^2 has two blocks (both with eigenvalue \lambda^2.

Now J^2 - \lambda^2 I = (J + \lambda I)(J - \lambda I) = (J - \lambda I)^2, since F has characteristic 2. Note that J-\lambda I has order n, since (evidently) we have (J - \lambda I)e_{i+1} = e_i and (J - \lambda I)e_1 = 0. So J-\lambda I has order n. If n is even, then (J^2 - \lambda^2 I)^{n/2} = 0 while (J^2 - \lambda^2 I)^{n/2-1} \neq 0, and if n is odd, then (J^2-\lambda^2 I)^{(n+1)/2} = 0 while (J^2 - \lambda^2 I)^{(n+1)/2-1} \neq 0. So the minimal polynomial of J^2 is (x-\lambda^2)^{n/2} if n is even and (x-\lambda^2)^{(n+1)/2} if n is odd.

So the Jordan canonical form of J^2 has two Jordan blocks with eigenvalue \lambda^2. If n is even, these have size n/2,n/2, and if n is odd, they have size (n+1)/2, (n-1)/2.

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

  1. The eigenvalues of A are square in F
  2. For each eigenvalue \lambda of A, the Jordan blocks with eigenvalue \lambda 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 P^{-1}AP = \bigoplus H_i \oplus K_i is in Jordan canonical form, where H_i and K_i are Jordan blocks having the same eigenvalue \lambda_i and whose sizes differ by 0 or 1. By the first half of this exercise, H_i \oplus K_i is the Jordan canonical form of J_i^2, where J_i is a Jordan block with eigenvalue \sqrt{\lambda_i}. Now A is similar to the direct sum of these J_i^2, and so Q^{-1}AQ = J^2. Then A = (Q^{-1}JQ)^2 = B^2 is square.

Conversely, suppose A = B^2 is square, and say P^{-1}BP = J is in Jordan canonical form. So P^{-1}AP = J^2. Letting J_i denote the Jordan blocks of J, we have P^{-1}AP = \bigoplus J_i^2. The Jordan canonical form of J_i^2 has two blocks with eigenvalue \lambda_i^2 and whose sizes differ by 0 or 1, by the first half of this exercise. So the Jordan blocks of A all have eigenvalues which are square in F and can be paired so that the sizes in each pair differ by 0 or 1.

Matrices with square roots

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


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

  1. Every eigenvalue of A is square in F.
  2. The Jordan blocks of A 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 P^{-1}AP = \bigoplus J_i \oplus \bigoplus H_i \oplus \bigoplus K_i is in Jordan canonical form, where J_i are the Jordan blocks having nonzero eigenvalue \lambda_i and H_i and K_i are the blocks with eigenvalue 0 such that the sizes of H_i and K_i differ by 0 or 1. As we showed in this previous exercise, J_i = Q_iM_i^2Q_i^{-1} is the Jordan canonical form of the square of the Jordan block M_i with eigenvalue \sqrt{\lambda_i}, and H_i \oplus K_i = T_iN_i^2T_i^{-1} is the Jordan canonical form of the square of the Jordan block whose size is \mathsf{dim}\ H_i + \mathsf{dim}\ K_i with eigenvalue 0. So we have P^{-1}AP = C^2, and so A = (P^CP^{-1})^2 = B^2 is a square.

Conversely, suppose A = B^2 is a square, and say P^{-1}BP = J is in Jordan canonical form. Now PAP^{-1} = J^2. Now say Q^{-1}J^2Q = K is in Jordan canonical form; then Q^{-1}PAP^{-1}Q = K. By the previous exercise, the nonzero eigenvalues of K are square in F, and the Jordan blocks having eigenvalue 0 can be paired so that the sizes in each pair differ by 0 or 1.

On the square of a Jordan block

Let J be a Jordan block of size n and eigenvalue \lambda over a field K whose characteristic is not 2.

  1. Suppose \lambda \neq 0. Prove that the Jordan canonical form of J^2 is the Jordan block of size n with eigenvalue \lambda^2.
  2. Suppose \lambda = 0. Prove that the Jordan canonical form of J^2 has two blocks (with eigenvalue 0) of size n/2, n/2 if n is even and of size (n+1)/2, (n-1)/2 if n is odd.

First suppose \lambda \neq 0.

Lemma: Let e_i denote the ith standard basis element (i.e. e_i = [\delta_{i,1}\ \delta_{i,2}\ \ldots\ \delta_{i,n}]^\mathsf{T}. If 1 \leq i < n-2, then (J^2-\lambda^2I)e_{i+2} = 2\lambda e_{i+1} + e_i, (J^2-\lambda^2I)e_{2} = 2\lambda e_1, and (J^2-\lambda^2I)e_1 = 0. Proof: We have J^2-\lambda^2I = (J+\lambda I)(J-\lambda I). Evidently, (J-\lambda I)e_{i+2} = e_{i+1}. Now J+\lambda I = [b_{j,k}], where b_{j,k} = 2\lambda if j = k and 1 if k = j+1 and 0 otherwise. So (J+\lambda I)e_{i+1} = [\sum_k b_{j,k} \delta_{k,i+1}] = [b_{j,i+1}] = 2\lambda e_{i+1} + e_i if i > 1, and similarly (J^2-\lambda^2 I)(e_2) = 2\lambda e_1. \square

In particular, note that (J^2 - \lambda^2 I)^{n-1} e_n = 2^{n-1}\lambda^{n-1} e_1 is nonzero (here we use the noncharacteristictwoness of K), while (J^2 - \lambda^2 I)^n = 0. So the minimal polynomial of J^2 is (x-\lambda^2)^n. Thus the Jordan canonical form of J^2 has a single Jordan block, of size n, with eigenvalue \lambda^2.

Now suppose \lambda = 0. Evidently, we have Je_{i+1} = e_i and Je_1 = 0. Now J^n = 0 and J^{n-1} \neq 0. If n is even, we have (J^2)^{n/2} = 0 and (J^2)^{n/2-1} = J^{n-2} \neq 0, so that the minimal polynomial of J^2 is x^{n/2}. If n is odd, we have (J^2)^{(n+1)/2} = 0 while (J^2)^{(n+1)/2-1} = J^{n-1} \neq 0, so the minimal polynomial of J^2 is x^{(n+1)/2}.

Next, we claim that \mathsf{ker}\ J^2 has dimension 2. To this end, note that J^2e_{i+2} = e_i is nonzero, while J^2e_2 = J^2e_1 = 0. By this previous exercise, J^2 has 2 Jordan blocks; thus the blocks of J^2 both have eigenvalue 0 and are of size n/2,n/2 if n is even, and (n+1)/2, (n-1)/2 if n is odd.

A relation between rank and the number of Jordan blocks of a finitely generated module over a PID

Let F be a field, V and n-dimensional F-vector space, and T a linear transformation on V. Suppose the eigenvalues of T are contained in F, and let \lambda be one such eigenvalue. Let k be a natural number, and let r_k = \mathsf{dim}_F\ (T-\lambda I)^k V. Prove that r_{k-1} - 2r_k + r_{k+1} is the number of Jordan blocks of T with eigenvalue \lambda and having size k.


We will use the notation established in this previous exercise. To wit, \lambda_i are the eigenvalues of T, and we have V = \bigoplus_i \bigoplus_j F[x]/(x-\lambda_i)^{a_{i,j}}, where a_{i,j} is the ‘size’ (that is, dimension over F) of its corresponding Jordan block. For brevity we say V_i = \bigoplus_j F[x]/(x-\lambda_i)^{a_{i,j}}.

We showed in that previous exercise that \mathsf{dim}_F\ (T-\lambda_t I)^kV = \sum_{j,a_{t,j} > k} (a_{t,j} - k) + \mathsf{dim}_F\ \bigoplus_{i,i\neq t} V_i. Thus we have the following.

r_{k-1} - 2r_k + r_{k+1}  =  \sum_{j,a_{t,j} > k-1} (a_{t,j} - k + 1) - 2 \sum_{j,a_{t,j} > k} (a_{t,j} - k) + \sum_{j,a_{t,j} > k+1} (a_{t,j} - k - 1)
 =  \sum_{j,a_{t,j} > k+1} (a_{t,j} - k+1 - 2(a_{t,j}-k) + a_{t,j}-k-1) + \sum_{j,a_{t,j} = k+1} (a_{t,j} - k + 1 - 2(a_{t,j}-k) + \sum_{j,a_{t,j} = k} (a_{t,j} - k + 1)
 =  \sum_{j,a_{t,j} = k} 1

This final sum is precisely the number of Jordan blocks with eigenvalue \lambda_t and having size k, as desired.

Compute the Jordan canonical form of a given matrix over a finite field

Let p be a prime. Compute the Jordan canonical form over \mathbb{F}_p of the n \times n matrix whose diagonal entries are 0 and whose off-diagonal entries are 1. (n \geq 2)


Let A = [1 - \delta_{i,j}]. The computation we carried out here shows that (A - (n-1)I)(A+I) = 0 over \mathbb{F}_p (indeed, over any ring with 1). So the minimal polynomial of A divides (x-(n-1))(x+1). Note that A-(n-1)I and A+I are nonzero (any off-diagonal entry is 1), so the minimal polynomial of A is precisely m(x) = (x-(n-1))(x+1).

Let V = \mathbb{F}_p^n be an \mathbb{F}_p[x]-module via A as usual. Now the invariant factors m_k(x) of V divide m(x), and we have V = \bigoplus_k \mathbb{F}[x]/(m_k(x)).

We claim that (A+I)V has dimension 1 over \mathbb{F}_p. Indeed, if v = [v_1\ \ldots\ v_n]^\mathsf{T}, then (A+I)v = (\sum v_i)[1\ \ldots\ 1]^\mathsf{T}.

Using Lemmas 2 and 3 from this previous exercise, we have (x+1)V = \bigoplus_k (x+1)\mathbb{F}_p[x]/(m_k(x)) \cong_{\mathbb{F}_p[x]} \bigoplus_{(x+1)|m_k} \mathbb{F}_p[x]/(m_k/(x+1)) \oplus \bigoplus_{(x+1,m_k) = 1} \mathbb{F}_p[x]/(m_k(x)). Note that any mapping which realizes an isomorphism as \mathbb{F}_p[x]-modules is necessarily also an \mathbb{F}_p-vector space isomorphism; since (A+I)V has dimension 1, the right hand side of this isomorphism also has dimension 1 over \mathbb{F}_p. Now one of the left summands is \mathbb{F}_p[x]/(x-(n-1)), corresponding to the minimal polynomial. Thus the remaining summands must be trivial; that is, there are no invariant factors which are not divisible by x+1, and every invariant factor other than the minimal polynomial is exactly x+1.

So the elementary divisors of A are x+1 (n-1 times) and x-(n-1). The corresponding Jordan canonical form is J = [b_{i,j}] where b_{i,j} = n-1 if i=j=n, -1 if i=j \neq n, and 0 otherwise.

Compute the Jordan canonical form of a given matrix over QQ

Compute the Jordan canonical form over \mathbb{Q} of the n \times n matrix whose diagonal entries are 0 and whose off-diagonal entries are 1.


Let A = [1 - \delta_{i,j}]. Note the following.

(A - (n-1)I)(A + I)  =  [1 - \delta_{i,j} - \delta_{i,j}(n-1)][1 - \delta_{i,j} + \delta_{i,j}]
 =  [1-\delta_{i,j}n][1]
 =  [\sum_k (1 - \delta_{i,k}n)1]
 =  [\sum_k 1 - \sum_k \delta_{i,k}n]
 =  [n-n]
 =  [0]

So the minimal polynomial of A divides (x-(n-1))(x+1). Since A \neq -I and A \neq (n-1)I, in fact this is the minimal polynomial of A.

Suppose now that v = [v_1\ \ldots\ v_n] is an eigenvector with eigenvalue n-1. Evidently then for each i, we have \sum_{k \neq i} v_k = (n-1)v_i, and thus \sum_k v_k = nv_i. Since n \neq 0, we have v_i = v_j for all i,j, so that v = v_1[1\ \ldots\ 1]^\mathsf{T}. In particular, the eigenspace with eigenvalue n-1 has dimension 1. By the lemma we proved in this previous exercise, n-1 has multiplicity 1 in the characteristic polynomial of A. So the elementary divisors of A are x+1 n-1 times and x-(n-1). The Jordan canonical form of A is thus J = [b_{i,j}], where b_{i,j} = n-1 if i=j=n, -1 if i = j \neq n, and 0 otherwise.

Compute the Jordan canonical form of a given matrix over a finite field

Let p be a prime. Compute the Jordan canonical form over the finite field \mathbb{F}_p of the n \times n matrix whose every entry is 1. (n \geq 2.)


Let A = [1]. Now the computation in this previous exercise holds over \mathbb{F}_p (indeed, over any ring with 1), and so we have A(A-nI) = 0.

If n is not divisible by p, then n is a unit in \mathbb{F}_p. If v = [v_1\ \ldots\ v_n]^\mathsf{T} is an eigenvector with eigenvalue n, then evidently (as we saw in the exercise referenced above) v = v_1[1\ \ldots\ 1]^\mathsf{T}. So the eigenspace of n has dimension 1, the characteristic polynomial of A is x^{n-1}(x-n), and the Jordan canonical form of A is J = [b_{i,j}] where b_{n,n} = n and b_{i,j} = 1 otherwise.

Suppose p divides n. Since A \neq 0, the minimal polynomial of A is now x^2. We claim that the remaining invariant factors of A are x. There is probably a slick linear-algebraic way to argue this, but I don’t see it. Instead I will show explicitly that A is similar to a matrix in Jordan canonical form. (If anyone can offer a simplification of this proof I’d be grateful!)

If n = 2, then the minimal polynomial of A is the only invariant factor, so the Jordan form of A is \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}. Suppose n \geq 2.

Let U = [1\ \ldots\ 1]^\mathsf{T} have dimension (n-1) \times 1, and let W be the (n-1) \times (n-1) matrix whose every entry is 1. Note that UU^\mathsf{T} = W and U^\mathsf{T}U = [n-1]. Let P = \begin{bmatrix} 1 & 0 \\ U & I \end{bmatrix} and Q = \begin{bmatrix} 1 & 0 \\ -U & I \end{bmatrix}. Evidently QP = I, so that Q = P^{-1}. Moreover, we have A = \begin{bmatrix} 1 & U^\mathsf{T} \\ U & W \end{bmatrix}, and P^{-1}AP = B = \begin{bmatrix} n & U^\mathsf{T} \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & U^\mathsf{T} \\ 0 & 0 \end{bmatrix}.

Now let V = [1\ \ldots\ 1] have dimension 1 \times (n-2). Let R = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -V \\ 0 & 0 & I \end{bmatrix} and S = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & V \\ 0 & 0 & I \end{bmatrix}. Evidently, RS = I, so that S = R^{-1}. Now B = \begin{bmatrix} 0 & 1 & V \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, and we have R^{-1}BR = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. If we squint just right, this matrix is in Jordan canonical form, and is similar to A.

Compute the Jordan canonical form of a given matrix over QQ

Compute the Jordan canonical form over \mathbb{Q} of the n \times n matrix whose every entry is equal to 1.


Let A = [1] be an n \times n matrix. Note the following.

A(A-nI)  =  [1][1-\delta_{i,j}n]
 =  [\sum_k 1_{i,k}(1_{k,j} - \delta_{k,j}n)]
 =  [\sum_k 1 - \sum_k \delta_{k,j} n]
 =  [n-n]
 =  [0]

So we have A(A-nI) = 0, and thus the minimal polynomial of A divides x(x-n). Since A \neq 0 and A \neq nI, the minimal polynomial of A is x(x-n).

Next we prove a lemma.

Lemma: Let A be an n \times n matrix over a field F containing the eigenvalues of A, and let \lambda be an eigenvalue of A with eigenspace W \subseteq F^n. If the minimal polynomial of A over F is a product of distinct linear factors, then the dimension of W (over F) is precisely the multiplicity of \lambda among the roots of the characteristic polynomial of A. Proof: Let V = F^n. As an F[x] module via A, we have V = \bigoplus_{i=1}^n F[x]/(x-\lambda_i), where \lambda_i are the eigenvalues of A. The eigenspace of \lambda consists precisely of those direct factors with \lambda_i = \lambda, each of which has dimension 1 over F. \square

For our given matrix A, we claim that the eigenspace W of the eigenvalue n has dimension 1 over F. To see this, let v = [v_1\ \ldots\ v_n]^\mathsf{T} and suppose Av = nv. Evidently, every entry of Av is \sum_k v_k, and so we have nv_i = nv_j for all i,j. Since n \neq 0, we have v_i = v_j for all i,j, and so v = v_1[1\ \ldots\ 1]^\mathsf{T}. So W has a basis containing one element, and so has dimension 1 over F as desired.

By the lemma, n has multiplicity 1 among the roots of the characteristic polynomial of A. Since the remaining roots are 0, the invariant factors of A are x n-2 times and x(x-n). The corresponding Jordan canonical form matrix is J = [b_{i,j}], with b_{n,n} = n and b_{i,j} = 0 otherwise.