# 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 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} 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 DirectSumOfMatrices 2013-03-22 17:36:48 2013-03-22 17:36:48 CWoo (3771) CWoo (3771) 8 CWoo (3771) Definition msc 15-01 DirectSum