16
Jun 18

Matrix transposition: continuing learning by doing

category: Matrix algebra, MT2175 Further linear algebra No Comments

Matrix transposition: continuing learning by doing

Definition. For any matrix A, its transposed A^T is obtained by putting rows of A into columns of A^T.

Figure 1. Matrix transpose

Figure 1. Matrix transpose

Exercise 1. See Figure 1 borrowed from Wikipedia. Based on that illustration, what is the relationship between A and (A^T)^T?

Exercise 2. For the matrix

A=\left(\begin{array}{cc}1&2\\3&4\\5&6 \end{array}\right)

which of the products AA^T and A^TA exists? Find the one(s) that exist. Repeat the same for

A=\left(\begin{array}{c}1\\3\\5\end{array}\right).

Definition. If A=A^T, then A is called symmetric.

Exercise 3. Can a non-square matrix be symmetric? The above definition is in matrix terms. What does it mean in terms of matrix elements?

Exercise 4. What is the relationship between \det A and \det (A^T)? Consider just a 2\times 2 matrix.

Exercise 5. What happens if you apply transposition to a product AB?

Solution. Partitioning A  and B into rows and columns, respectively, we find the elements of the product C as dot products of the rows of A by the columns of B:

(1) C=\left(\begin{array}{ccc}A_1\cdot B^1&...&A_1\cdot B^k \\...&...&... \\A_n\cdot B^1&...&A_n\cdot B^k\end{array}\right)=\left(\begin{array}{ccc}A_1B^1&...&A_1B^k \\...&...&... \\ A_nB^1&...&A_nB^k\end{array}\right).

In the second expression the dots are omitted because the A_j are rows and the B^i are columns, so that the matrix product definition can be applied to write A_j\cdot B^i=A_jB^i. Further, (B^i)^T are rows and (A_j)^T are columns, so we can write

(2) A_jB^i=(B^i)^T(A_j)^T.

Transposing (1) and using (2) we have

C^T=\left(\begin{array}{ccc}A_1B^1&...&A_nB^1 \\...&...&... \\A_1B^k&...&A_nB^k\end{array} \right)=\left(\begin{array}{ccc}(B^1)^T(A_1)^T&...&(B^1)^T(A_n)^T \\...&...&... \\(B^k)^T(A_1)^T&...&(B^k)^T(A_n)^T\end{array}\right)

=\left(\begin{array}{c}(B_1)^T \\... \\(B_k)^T\end{array}\right)\left(\begin{array}{ccc}(A^1)^T&...&(A^n)^T\end{array}\right)=B^TA^T.

We have proved that (AB)^T=B^TA^T.

Exercise 6. What is the relationship between transposition and inversion? More precisely, if A is invertible, then what can you say about (A^T)^{-1}?

Solution. By definition, AA^{-1}=I. Transposing this gives (A^{-1})^TA^T=I^T=I. This shows that (A^T)^{-1}=(A^{-1})^T.

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