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Matrix transposition: continuing learning by doing

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

Figure 1. Matrix transpose

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

Exercise 2. For the matrix

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

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

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

Definition. If $latex A=A^T,$ then $latex A$ 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 $latex \det A$ and $latex \det (A^T)?$ Consider just a $latex 2\times 2$ matrix.

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

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

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

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

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

Transposing (1) and using (2) we have

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

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

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

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

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

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