direct sum of matrices

Direct sum of matrices

Let $A$ be an $m\times n$ matrix and $B$ be a $p\times q$ matrix. By the direct sum of $A$ and $B$ , written $A\oplus B$ , we mean the $(m+p)\times(n+q)$ matrix of the form

\begin{pmatrix}A&O\\ O&B\end{pmatrix}

where the $O$ ’s represent zero matrices. The $O$ on the top right is an $m\times q$ matrix, while the $O$ on the bottom left is $n\times p$ .

For example, if $A=\begin{pmatrix}3&-1\\ 2&5\end{pmatrix}$ and $B=\begin{pmatrix}1&2\\ 4&0\\ -7&8\end{pmatrix}$ , then

\begin{pmatrix}A&O\\ O&B\end{pmatrix}=\begin{pmatrix}3&-1&0&0\\ 2&5&0&0\\ 0&0&1&2\\ 0&0&4&2\\ 0&0&-7&8\\ \end{pmatrix}

Remark. It is not hard to see that the $\oplus$ operation on matrices is associative:

(A\oplus B)\oplus C=A\oplus(B\oplus C),

because both sides lead to

\begin{pmatrix}A&O&O\\ O&B&O\\ O&O&C\end{pmatrix}

In fact, we can inductively define the direct sum of $n$ matrices unambiguously.

Direct sums of linear transformations

The direct sum of matrices is closely related to the direct sum of vector spaces and linear transformations. Let $A$ and $B$ be as above, over some field $k$ . We may view $A$ and $B$ as linear transformations $T_{A}:k^{n}\to k^{m}$ and $T_{B}:k^{q}\to k^{p}$ using the standard ordered bases. Then $A\oplus B$ may be viewed as the linear transformation

T_{A\oplus B}:k^{n+q}\to k^{m+p}

using the standard ordered basis, such that

•

the restriction of $T_{A\oplus B}$ to the subspace $k^{n}$ (embedded in $k^{n+q}$ ) is $T_{A}$ , and
•

the restriction of $T_{A\oplus B}$ to $k^{q}$ is $T_{B}$ .

The above suggests that we can define direct sums on linear transformations. Let $T_{1}:V_{1}\to W_{1}$ and $T_{2}:V_{2}\to W_{2}$ be linear transformations, where $V_{i}$ and $W_{j}$ are finite dimensional vector spaces over some field $k$ such that $V_{1}\cap V_{2}=0$ . Then define $T_{1}\oplus T_{2}:V_{1}\oplus V_{2}\to W_{1}\oplus W_{2}$ such that for any $v\in V_{1}\oplus V_{2}$ ,

(T_{1}\oplus T_{2})(v_{1},v_{2}):=(T_{1}(v_{1}),T_{2}(v_{2}))

where $v_{i}\in V_{i}$ . Based on this definition, it is not hard to see that

T_{A\oplus B}=T_{A}\oplus T_{B}

for any matrices $A$ and $B$ .

More generally, if $\beta_{i}$ is an ordered basis for $V_{i}$ , then $\beta:=\beta_{1}\cup\beta_{2}$ extending the linear orders on $\beta_{i}$ , such that if $v_{i}\in\beta_{1}$ and $v_{j}\in\beta_{2}$ , then $v_{i}<v_{j}$ is an ordered basis for $V_{1}\oplus V_{2}$ , and

[T_{1}\oplus T_{2}]_{\beta}=[T_{1}]_{\beta_{1}}\oplus[T_{2}]_{\beta_{2}}.

Title	direct sum of matrices
Canonical name	DirectSumOfMatrices
Date of creation	2013-03-22 17:36:48
Last modified on	2013-03-22 17:36:48
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	8
Author	CWoo (3771)
Entry type	Definition
Classification	msc 15-01
Related topic	DirectSum