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[parent] direct sum of matrices (Definition)

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

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

$\displaystyle \begin{pmatrix} A & O \ O & B \end{pmatrix}= \begin{pmatrix} 3&... ... 0&0 \ 2&5 & 0&0 \ 0&0 & 1&2 \ 0&0 & 4&2 \ 0&0 & -7&8 \ \end{pmatrix}$
Remark. It is not hard to see that the $ \oplus$ operation on matrices is associative:
$\displaystyle (A\oplus B)\oplus C = A \oplus (B\oplus C),$
because both sides lead to
$\displaystyle \begin{pmatrix} A & O & O \ O & B & O \ O & O & C \end{pmatrix}$
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
$\displaystyle T_{A\oplus B}: k^{n+q}\to k^{m+p}$
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$,

$\displaystyle (T_1\oplus T_2)(v_1,v_2):=(T_1(v_1),T_2(v_2))$
where $ v_i\in V_i$. Based on this definition, it is not hard to see that
$\displaystyle T_{A\oplus B}=T_A \oplus T_B$
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

$\displaystyle [T_1\oplus T_2]_{\beta}=[T_1]_{\beta_1}\oplus [T_2]_{\beta_2}.$



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Cross-references: linear orders, ordered basis, finite dimensional, subspace, restriction, standard ordered basis, standard ordered bases, field, linear transformations, vector spaces, sides, associative, operation, right, zero matrices, represent, mean, direct sum, matrix
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This is version 5 of direct sum of matrices, born on 2007-11-04, modified 2007-11-04.
Object id is 10030, canonical name is DirectSumOfMatrices.
Accessed 1012 times total.

Classification:
AMS MSC15-01 (Linear and multilinear algebra; matrix theory :: Instructional exposition )

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Adding to Your Entry by pjohara15 on 2008-01-19 19:37:50
Dear Chi Woo:

Thank you so much. I am working on two(2) proofs related to linear operators and matrices that appear directly related to what your addition is stating.

The challenge will then be to articulate this connection.
It may be some time before I actually make the addition. If you like, I can notify you beforehand.

PJO
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Adding to Your Entry by pjohara15 on 2008-01-19 10:10:02
Hello -

I am wondering if you would mind my adding a short exposition to you post concerning Sum and Direct Sum.

I have proved some fundamental results and would like to post them and get some input.

Patrick J. O'Hara
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