3

Let’s say there are two matrices $A$ and $B$ where $$A=\begin{bmatrix}a&b\\b&a\end{bmatrix}$$ $$B=\begin{bmatrix}c&d\\d&c\end{bmatrix}$$

$A$ and $B$ together make up a third matrix $C$ where $$C=\begin{bmatrix}A&B\\B&A\end{bmatrix}=\begin{bmatrix}\begin{bmatrix}a&b\\b&a\end{bmatrix}&\begin{bmatrix}c&d\\d&c\end{bmatrix}\\\begin{bmatrix}c&d\\d&c\end{bmatrix}&\begin{bmatrix}a&b\\b&a\end{bmatrix}\end{bmatrix}$$

My question is of syntax. Specifically, did I define $C$ using $A$ and $B$ properly? It seems ambiguous to me in the sense of $C$ being a matrix of matrices and not of $A$ and $B$ ‘s elements; when I in fact want $C$ to be a matrix of the elements and not of the matrices.

Edit: it looks like I have described a block matrix.

UpTide
  • 169

1 Answers1

4

Let $R$ be a ring.

Let $M=\mathcal M_{2\times 2}(R)$ be the ring of $2\times 2$ matrices over $R$.

Let $A,B$ be in $M$. Then we can associate and write both follwing matrices. (The person writing the matrices has to give the one or the other sense.)

  • The matrix $$\begin{bmatrix}A & B \\ B & A\end{bmatrix}\in\mathcal M_{2\times 2}(M)\ .$$

  • The block matrix $$\left[\begin{array}{c|c}A & B \\\hline B & A\end{array}\right]\in\mathcal M_{4\times 4}(R)\ .$$

(There are obvious ring homomorphisms between the two spaces of matrices. Using this, the "multiplication of block matrices" is possible.)


Later edit:

I decided to insert some explicit examples after the discussion in the comments. In my oppinion, the usage of block matrix multiplication is underestimated, it should be a standard tool in the school. Here, to have an easy game of inserting examples, i will use sage. Here is my dialog with sage.

sage: A = matrix( ZZ, 2, 2, [1,5,4,7] )
sage: B = matrix( ZZ, 2, 2, [1,8,4,9] )
sage: C = matrix( ZZ, 2, 2, [0,1,8,2] )
sage: D = matrix( ZZ, 2, 2, [0,7,7,2] )
sage: M = block_matrix( 2, 2, [A,B,C,D] )
sage: M
[1 5|1 8]
[4 7|4 9]
[---+---]
[0 1|0 7]
[8 2|7 2]
sage: A, B, C, D
(
[1 5]  [1 8]  [0 1]  [0 7]
[4 7], [4 9], [8 2], [7 2]
)

Now to the above $2\times 2$ block matrix named $M$ we associate a plain matrix named $X$.

sage: X = M.matrix_from_rows_and_columns( [0,1,2,3], [0,1,2,3] )
sage: X
[1 5 1 8]
[4 7 4 9]
[0 1 0 7]
[8 2 7 2]

The question in the comment has now the following answer.

The $(1,1)$ entry in $M$ is the $2\times 2$ matrix $A$.

The $(1,2)$ entry in $M$ is the $2\times 2$ matrix $B$.

The $(1,1)$ entry in $X$ is the number $1$.

The $(1,2)$ entry in $X$ is the number $5$.

We want now to see the utility of block matrices. For this note that we can multiply the blocks "as if" they were numbers. (Well, here, they are indeed "numbers" in a ring, the ring of $2\times 2$ matrices over $\Bbb Z$. But also more general patterns of block matrices are allowed / make sense. Let us understand but the present situation.)

We have formally $$M = \left[\begin{array}{c|c}A & B \\\hline C & D\end{array}\right]\in\mathcal M_{2\times 2}\ .$$ Then we can for instance compute $$ M^2 = \left[\begin{array}{c|c}A & B \\\hline C & D\end{array}\right] \left[\begin{array}{c|c}A & B \\\hline C & D\end{array}\right] = \left[\begin{array}{c|c}AA+BC & AC+BD \\\hline CA+DC & CB+DD\end{array}\right] \ . $$ In our example:

sage: M
[1 5|1 8]
[4 7|4 9]
[---+---]
[0 1|0 7]
[8 2|7 2]
sage: M^2
[ 85  57  77  76]
[104  91  95 141]
[ 60  21  53  23]
[ 32  65  30 135]
sage: A*A + B*C, A*C + B*D
(
[ 85  57]  [ 96  34]
[104  91], [119  64]
)
sage: C*A + D*C, C*B + D*D
(
[60 21]  [ 53  23]
[32 65], [ 30 135]
)
sage: 

Sage has a rather mathematically oriented thinking, so i hope the above can be easily digested.

dan_fulea
  • 32,856
  • So for the matrix, not block, would $n$ in $M_n$ be equal to 4? If it isn’t then why not? For the block matrix is $M_n$’s $n$ equal to 4? If it isn’t why not? – UpTide Aug 03 '18 at 13:03
  • @UpTide A "block matrix" is a special kind of matrix, it comes with the information of its entries and the information of two partitions, one for the rows, one for the columns, the two partitions determine the blocks of the "block matrix". (A "block" does not make sense alone. So it is hard to understand the question in the above comment "So for the matrix, not block, would $n$ in $M_n$ be equal to 4?" There are no $n$, $M_n$ defined in the context.) For the question "If it isn’t then why not?" i could not find a suitable comment. – dan_fulea Aug 03 '18 at 16:52
  • I think I would understand better with an example. If we say that your matrix $A$ can only contain $0$ as an element, and we say matrix $B$ can only contain $1$ as an element; then we reference the element $M_{1,1}$, what will the element be equal to? $0$ or $1$? Can we reference element $M_{3,3}$? What would it be? I'm not trying to be dense, I just truly don't understand. – UpTide Aug 04 '18 at 01:00