Higher Mathematics 1st Paper · Classes 11-12 · Chapter 1 · 1.1

Transpose, Symmetric and Other Special Matrices.

What happens when you tip a matrix on its side? The rows stand up and become columns. And some matrices, when squared, come back as themselves, some collapse to zero, some turn into I. This note is about those moves and their names.

1. The transpose and its two rules

Swap the rows of a matrix A for its columns and you get a new matrix, the transpose of A, written A', Aᵗ or Aᵀ. Take A = [[3, 5, 7], [2, 4, 9]]; then A' = [[3, 2], [5, 4], [7, 9]]. If A has order m × n, then A' has order n × m: here a 2 × 3 matrix turns into a 3 × 2 one.

There are two rules: (A + B)' = A' + B' and (AB)' = B'A'. A sum transposes term by term, but a product reverses its order. Writing A'B' is wrong; it must be B'A'.

The book checks the rule in exercise 1(A) 7(b). Take A = [[1, 0, −2], [2, −1, 3]] and B = [[−3, 1], [2, 4], [1, −2]]. Multiplying gives AB = [[−5, 5], [−5, −8]], so (AB)' = [[−5, −5], [5, −8]]. Work out B'A' separately and the very same matrix appears. Hence (AB)' = B'A'.

2. Symmetric and skew symmetric

A square matrix A = (a_ij) is symmetric if A' = A, that is a_ij = a_ji. For example [[1, 2, −3], [2, −7, 4], [−3, 4, 1]]. The entries on the two sides of the principal diagonal match in pairs.

If A' = −A, that is a_ij = −a_ji, then A is skew symmetric. On the diagonal i = j, so a_ii = −a_ii, which forces every skew-symmetric matrix to have zeros on its principal diagonal. For example [[0, 3, 4], [−3, 0, 7], [−4, −7, 0]].

In the book's Task 3, B = [[1, −3, 5], [−3, 2, 7], [5, 7, 4]] has B' = B, so B is symmetric. C = [[0, −8, 5], [8, 0, −4], [−5, 4, 0]] has C' = −C, so C is skew symmetric. Below, flip six matrices, those two included. One transpose tells you which is which.

3. Four names, four conditions

A square matrix A is idempotent if A² = A. For instance A = [[3, −6], [1, −2]] satisfies A² = A. If Aⁿ = 0 for some positive integer n, A is nilpotent, and the smallest such n is its index. For A = [[4, 4], [−4, −4]], A² = 0, so the index is 2.

If A² = I, A is involutory. For instance B = [[1, 3], [0, −1]] satisfies B² = I. Since A·A = I, an involutory matrix is its own inverse: A⁻¹ = A. If AA' = A'A = I, A is orthogonal, for example [[cos x, sin x], [−sin x, cos x]]. That last name is the only one of the four that needs the transpose.

Now drop each of six conditions into the box of its name. Symmetric and skew symmetric join in as well.

4. Trace and sub-matrix

The sum of the entries on the principal diagonal of a square matrix is its trace. The trace of [[5, −2, −6], [−3, 2, 9], [2, 0, 3]] is 5 + 2 + 3 = 10. And a matrix formed by deleting any number of rows and columns of a matrix is a sub-matrix of the original.

5. Square it and compare

Every proof in exercise 7 is the same single move: square the matrix, then compare the result with A, 0 or I.

For A = [[2, −2, −4], [−1, 3, 4], [1, −2, −3]], A² = A, so A is idempotent (exercise 1(A) 7(c)).

For C = [[−5, −8, 0], [3, 5, 0], [1, 2, −1]], C² = I, so C is involutory. For D = (1/3)[[−1, 2, 2], [2, −1, 2], [2, 2, −1]], DD' = I, so D is orthogonal (7(e), 7(f)). There is a failing case too: N = [[√3, 1], [−1, √3]] has NN' = [[4, 0], [0, 4]], which is not I₂, so N is not orthogonal (Example 5(a)).

Try it on five 2 × 2 matrices. Show A², then choose the name.

For a nilpotent matrix, squaring once is not always enough. B = [[1, 1, 3], [5, 2, 6], [−2, −1, −3]] has B² = [[0, 0, 0], [3, 3, 9], [−1, −1, −3]], which is not the zero matrix. But B³ = 0. So B is nilpotent of index 3 (exercise 7(d)). Raise the power one step at a time.

Everything so far is in the book. Here is one more move that comes up in exams but is missing from it.

6. Check yourself

Three questions in the exam's style.

Have a go at the second.

The last one. Think about what you have to square.

The whole note on seven cards.

Source: Higher Mathematics 1st Paper, Classes 11-12 (NCTB), Chapter 1 “Matrices and Determinants”, section 1.1.