How do you prove that every Jordan block is similar to its transpose?

How can you prove that every Jordan block is similar to its transpose?

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  • 8 years ago
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    Let J(n,k) denote the nxn Jordan block having the single eigenvalue k. (So J(n,k) is an nxn matrix with k's down the diagonal, 1's immediately above the diagonal, and 0s everywhere else.) Let S_n denote the nxn matrix obtained by writing the rows of the nxn identity matrix in the reverse order. (So S_n has a 1 in the row 1, column n entry, a 1 in the row 2, column n - 1 entry, ...., and a 1 in the row n, column 1 entry, and zeros everywhere else). A short matrix calculation shows that S_n^(-1) = S_n and that

    S_n J(n,k) S_n = J(n,k)^T.

    So, this shows that J(n,k) is similar to J(n,k)^T by explicitly exhibiting an invertible matrix that implements the similarity.

    If you don't like taking the matrix calculation on faith (or simply checking it for n = 2, 3, etc and noticing the pattern), one way to prove the matrix calculation is as follows. Note that

    (1) Two nxn matrices M and N are equal if and only if M e_j = N e_j for all 1 <= j <= n, where e_j denotes the nx1 column vector with a 1 in the nth entry and zeros elsewhere.

    (2) For any 1 <= j <= n one has S_n e_j = e_{n - j + 1}.

    (3) For any 1 < j <= n one has J(n,k) e_j = k e_j + e_{j-1}, and J(n,k) e_1 = k e_1.

    (4) For any 1 <= j < n one has J(n,k)^T e_j = k e_j + e_{j+1} and J(n,k)^T e_n = k e_n.

    For any 1 < j < n you can then see

    S_n J(n,k) S_n e_j = S_n J(n,k) e_{n - j + 1} [by (2)]

    = S_n [k e_{n - j + 1} + e_{n - j}) [by (3)]

    = k S_n e_{n - j + 1} + S e_{n - j}

    = k e_{n - (n - j + 1) + 1} + e_{n - (n - j) + 1} [by (2)]

    = k e_j + e_{j + 1}

    = J(n,k)^T e_j [by (4)]

    and likewise

    S_n J(n,k) S_n e_1 = S_n J(n,k) e_n [by (2)]

    = S_n (k e_n + e_{n - 1}) [by (3)]

    = k S_n e_n + S_n e_{n - 1}

    = k e_{n - n + 1} + e_{n - (n - 1) + 1} [by (2)]

    = k e_1 + e_2

    = J(n,k)^T e_1 [by (4)]

    and

    S_n J(n,k) S_n e_n = S_n J(n,k) e_1 [by (2)]

    = S_n (k e_1) [by (3)]

    = k S_n e_1

    = k e_{n - 1 + 1} [by (2)]

    = k e_n

    = J(n,k)^T e_1 [by (4)].

    By (1), this establishes that S_n J(n,k) S_n = J(n,k)^T.

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