I was reading the definition of block matrix from Wikipedia but I can't understand it. The definition is:

A block matrix or a partitioned matrix is a matrix that is interpreted as having been broken into sections called blocks or submatrices.

What is the meaning of interpreted? I mean based on that definition every matrix can be viewed as a block matrix.

  • $\begingroup$ Yes, every matrix can be viewed as a block matrix with little $1\times 1$ blocks, but for what purpose? $\endgroup$ May 11, 2022 at 21:40

1 Answer 1


An $(m \times n)$-matrix $A$ represents a linear transformation $T:V \to W$ with respect to a basis $\{v_1, \dots, v_n\} \subseteq V$ and $\{w_1, \dots, w_m\} \subseteq W$ by explicitly describing how to express each $T(v_j)$ as a linear combination of vectors $w_i$ in column $j$: $$ T(v_j) = \sum_{i=1}^m a_{ij}\, w_i $$ However, sometimes it makes sense to partition the basis into chunks. Maybe pairs of basis vectors that span little $2 \times 2$ subspaces, maybe larger collections of basis vectors. (The blocks of the partitions don't have to be the same size.) If your linear transformation respects the partitions of each basis, then the transformation is essentially a matrix of little rectangular matrix blocks.

There are lots of situations where this comes up, which you undoubtedly read about in the article. Some that come to mind immediately:

  1. Transformation $T:V \to W$ with subspaces $Y \subseteq V$ and $Z \subseteq W$ that satisfies $T(Y) \subseteq Z$, hence descends to a quotient $\tilde{T}: V/Y \to W/Z$. In a basis that respects these subspaces, the matrix will be block upper triangular with blocks $A$ and $B$ on the diagonal representing $T\vert_Y: Y \to Z$ and $\tilde{T}:V/Y \to W/Z$, respectively. See this question and its answers for more.
  2. As a special case of the previous example if $V \equiv V_1 \oplus V_2$ and $W = W_1 \oplus W_2$ are decompositions of the vector spaces as direct sums, and $T(V_1) \subseteq W_1$ and $T(V_2) \subseteq W_2$, then the map splits as $T \equiv T_1 \oplus T_2$, where the summands are restrictions of $T$ to appropriate subspaces, and the matrix for $T$ in a basis that respects this decomposition will be block diagonal. See this question and its answers for more.
  3. Converting a linear transformation between complex vector spaces, represented by a matrix $A \in \mathbb{C}^{m \times n}$, to a linear transformation between real vector spaces of double dimensions, represented by by a matrix $A_{\mathbb{R}} \in \mathbb{R}^{2m \times 2n}$, where each complex entry $$ c_{ij} = a_{ij} + \sqrt{-1}\, b_{ij} $$ in $A$ is replaced by the $2 \times 2$ block in $A_{\mathbb{R}}$ $$ \begin{bmatrix} a_{ij} & -b_{ij} \\ b_{ij} & \phantom{-}a_{ij} \end{bmatrix}. $$
  4. By permuting the bases (conjugating by the appropriate permutation matrices), the previous example can me modified to sort all the real parts from the imaginary parts, converting a matrix with $m \times n$ blocks of size $2 \times 2$ into a matrix with $2 \times 2$ blocks of size $m \times n$. See this question and its answers for more. Here's what the permutation looks like in the $2n = 2 \cdot 5 = 10$ case in $2$-line notation: $$ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 1 & 6 & 2 & 7 & 3 & 8 & 4 & 9 & 5 & 10 \end{pmatrix}. $$ This is the perfect riffle shuffle.

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