Direct sum

In Algebra, the term of direct sum applies to several different situations

Direct sum of vectorial subspaces

Direct sum of two vectorial subspaces

detailed Article: additional Subspaces

Are $F\_1$ and $F\_2$ two subspaces vectorial of the vector space E . It is said that $F\_1$ and $F\_2$ are all in all direct if and only so for any element U of $F\_1\; +\; F\_2$, there exists a single couple $\backslash \; (u\_1;\; u\_2)$ of $F\_1\; \backslash \; times\; F\_2$ such as $u\; =\; u\_1\; +\; u\_2$.

One as says in this case as the sum $F\_1\; +\; F\_2$ is direct.

In other words, the sum of two vectorial subspaces $F\_1$ and $F\_2$ is direct if the decomposition of any element of $F\_1\; +\; F\_2$ all in all of an element of $F\_1$ and an element of $F\_2$ is single.

The sum will then be noted: $F\_1\; \backslash oplus\; F\_2$.

One has the following usual characterizations:

• $F\_1$ and $F\_2$ is all in all direct if and only if, for all $u\_1$ of $F\_1$ and $u\_2$ of $F\_2$,

$u\_1\; +\; u\_2\; =\; 0\; \backslash \; Leftrightarrow\; u\_1\; =\; u\_2\; =\; 0$

• $F\_1$ and $F\_2$ is all in all direct if and only if

$F\_1\; \backslash \; course\; F\_2\; =\; \backslash \; \{0\; \backslash \}$

Case of finished dimension : when $F\_1$ and $F\_2$ are of finished size, the following assertions are equivalent:

1. the sum $F\_1\; +\; F\_2$ is direct.
2. $\backslash \; dim\; F\_1\; +\; \backslash \; dim\; F\_2\; =\; \backslash \; dim\; (F\_1\; +\; F\_2)$.
3. By juxtaposing (" réunissant") a base of $F\_1$ and a base of $F\_2$, one constitutes a base of $F\_1\; +\; F\_2$.

additional Subspaces : two subspaces $F\_1$ and $F\_2$ of E are known as additional when $E\; =\; F\_1\; \backslash \; oplus\; F\_2$. That means that for any element U of E , there exists a single couple $\backslash \; (u\_1;\; u\_2)$ of $F\_1\; \backslash \; times\; F\_2$ such as $\backslash \; U\; =\; u\_1\; +\; u\_2$.

Direct sum several vectorial subspaces

One can generalize the concept of direct sum to a finished family of vectorial subspaces of E .

It is said that a family $(F\_i)\; \_\; \{i=1\; \backslash \; cdots\; K\}$ of vectorial subspaces of E is all in all direct if and only if, for any element U of the sum $F\; =\; \backslash \; sum\_\; \{i=1\}\; ^k\; F\_i$, there exists a K - uplet single $(u\_1;\; u\_2;\; \backslash \; cdots;\; u\_k)$ of $F\_1\; \backslash \; times\; F\_2\; \backslash \; times\; \backslash \; cdots\; \backslash \; times\; F\_k$ such as $u\; =\; \backslash \; sum\_\; \{i=1\}\; ^k\; u\_i$.

One as says in this case as the sum F of the subspaces $(F\_i)\; \_\; \{i=1\; \backslash \; cdots\; K\}$ is direct.

In other words, the sum is direct if the decomposition of any element of $F\; =\; \backslash \; sum\_\; \{i=1\}\; ^k\; F\_i$ all in all of elements of the $F\_i\; \backslash ,$ is single.

To indicate a direct sum, one makes use of the notations $F\_1\; \backslash \; oplus\; F\_2\; \backslash \; oplus\; \backslash \; cdots\; \backslash \; oplus\; F\_k$ or $\backslash \; bigoplus\_\; \{I\; =\; 1\}\; ^kF\_i$.

As in the case of 2 vectorial subspaces, one can characterize the direct sums by the unicity of the decomposition of the null vector:

the sum $F\; =\; \backslash \; sum\_\; \{i=1\}\; ^k\; F\_i$ is direct if and only if:

single the K - uplet $(u\_1;\; u\_2;\; \backslash \; cdots;\; u\_k)$ of $F\_1\; \backslash \; times\; F\_2\; \backslash \; times\; \backslash \; cdots\; \backslash \; times\; F\_k$ such as $\backslash \; sum\_\; \{i=1\}\; ^k\; u\_i\; =\; 0$ is that of which all the elements are null.

Remark : as soon as the family includes/understands at least 3 subspaces, it is not enough so that the sum is direct that their intersections two to two are reduced to $\backslash \; \backslash \; \{0\; \backslash \}$, i.e.:

$F\_i\; \backslash \; course\; F\_j\; =\; \backslash \; \{0\; \backslash \}$ for all I and all J , I different of J .

One will be convinced some by looking in $\backslash \; R^2$ the vectorial subspaces:

$F\_1=\; \backslash \; \{(X;\; 0),\; X\; \backslash \; in\; \backslash \; R\; \backslash \}$

$F\_2=\; \backslash \; \{(there;\; there),\; there\; \backslash \; in\; \backslash \; R\; \backslash \}$

$F\_3=\; \backslash \; \{(0;\; T),\; T\; \backslash \; in\; \backslash \; R\; \backslash \}$.

Their intersections two to two are reduced to {(0; 0)}, but their amount $\backslash \; F\; =\; F\_\; 1\; +\; F\_2\; +\; F\_3$ (equal to $\backslash \; \backslash \; R^2$) is not direct.

Indeed, the 3 vectors $u\_1=\; (1;\; 0),\; \backslash ,\; u\_2=\; (-\; 1;\; -1),\; \backslash ,\; u\_3=\; (0;\; 1)$ belongs respectively to $F\_1,\; \backslash ,\; F\_2,\; \backslash ,\; F\_3$; they are nonnull, and such as $\backslash \; u\_1\; +\; u\_2\; +\; u\_3=\; (0;\; 0)$: the decomposition of the null vector is not single.

On the other hand, it is shown that the subspaces of the family of the $\backslash \; (F\_\; \{I\})\; \_\; \{1\; \backslash \; geq\; I\; \backslash \; geq\; N\}$ are all in all direct in $\backslash \; E$ if and only if:

• $\backslash \; \backslash \; sum\_\; \{i=1\}\; ^n\; F\_i\; =\; E$
• $\backslash \; \backslash \; forall\; K\; \backslash \; in\; \backslash \; left\; \backslash \; \{1,\dots ,\; n-1\; \backslash \; right\; \backslash \},\; \backslash \; \backslash \; left\; (\backslash \; sum\_\; \{i=1\}\; ^\; \{K\}\; F\_\; \{I\}\; \backslash \; right)\; \backslash \; course\; F\_\; \{k+1\}\; =\; \backslash \; left\; \backslash \; \{0\_\; \{E\}\; \backslash \; right\; \backslash \}$

When the vectorial subspaces are of finished size, there is still the equivalence of the following assertions:

1. the $(F\_i)\; \_\; \{i=1\; \backslash \; cdots\; K\}$ are all in all direct.
2. $\backslash \; sum\_\; \{i=1\}\; ^k\; \backslash \; dim\; F\_i\; =\; \backslash \; dim\; \backslash \; left\; (\backslash \; sum\_\; \{i=1\}\; ^k\; F\_i\; \backslash \; right)$.
3. By juxtaposing a base $\backslash \; \backslash \; mathcal\; \{B\}\; \_1$ of $\backslash \; F\_1$,…, a base $\backslash \; \backslash \; mathcal\; \{B\}\; \_k$ of $\backslash \; F\_k$, one constitutes a base of the sum.

Example: are E a vector space on K of finished size, and F a endomorphism of E having exactly p eigenvalues (distinct) called $\backslash \; lambda\_1,\; \backslash ,\; \backslash \; dowries,\; \backslash ,\; \backslash \; lambda\_p$. One indicates by $\backslash \; \backslash \; mathrm\; \{Id\}$ the identical endomorphism of E .

For entire I such as 1 ≤ I ≤ p, $E\_i\; =\; \backslash \; mathrm\; \{ker\}\; (F\; -\; \backslash \; lambda\_i\; \backslash ,\; \backslash \; mathrm\; \{Id\})$ is the clean subspace of F associated with the eigenvalue $\backslash \; \backslash \; lambda\_i$.
The two following properties are traditional:

• the sum $\backslash \; sum\_\; \{i=1\}\; ^p\; E\_i$ is direct.
• $\backslash \; bigoplus\_\; \{I\; =\; 1\}\; ^p\; E\_i\; =\; E$si and only if F is diagonalisable.

When it is the case, one constitutes a base $\backslash \; \backslash \; mathcal\; \{B\}$ of E diagonalisant F by juxtaposing a base $\backslash \; \backslash \; mathcal\; \{B\}\; \_1$ of $\backslash \; E\_1$,…, a base $\backslash \; \backslash \; mathcal\; \{B\}\; \_p$ of $\backslash \; E\_p$.

Orthogonal direct sum

One indicates here by E a space préhilbertien real or complex (real or complex vector space provided with a scalar product). That is to say a family $(F\_i)\; \_\; \{i=1\; \backslash \; cdots\; K\}$ of vectorial subspaces of E . If they are two to two orthogonal, their amount is direct. It is then called orthogonal direct sum .

A very simple example is the space $F^\; \backslash \; perp$ made up of the orthogonal vectors to all the vectors of a vectorial subspace F : it is all in all direct with F .

Two spaces which are at the same time additional and orthogonal are known as additional orthogonal . A vectorial subspace F of E, even if it has the additional ones, does not have of them necessarily one which is orthogonal for him. Sufficient conditions are that space E is of finished size or that space F is closed (proof). This question is related to the possibility of carrying out a orthogonal Projection.

When the vectorial subspaces are of finished size, one with the equivalence of the following assertions:

1. the $(F\_i)\; \_\; \{i=1\; \backslash \; cdots\; K\}$ are all in all direct orthogonal.
2. By juxtaposing an orthogonal base $\backslash \; \backslash \; mathcal\; \{B\}\; \_1$ of $\backslash \; F\_1$,…, an orthogonal base $\backslash \; \backslash \; mathcal\; \{B\}\; \_k$ of $\backslash \; F\_k$, one constitutes an orthogonal base of the sum.

External direct sum and Cartesian product

When two subspaces $F\_1$, $F\_2$ of a vector space E are all in all direct, the following application is bijective:

$F\_1\; \backslash \; times\; F\_2\; \backslash \; to\; F\_1\; \backslash \; oplus\; F\_2,\; (u\_1;\; u\_2)\; \backslash \; mapsto\; u\_1\; +\; u\_2$

There exists in this case a single structure of vector space on the Cartesian product $F\_1\; \backslash \; times\; F\_2$ such as this application is an isomorphism of vector spaces; the internal law and the external law are respectively defined by the relations:

$\backslash \; (u\_1;\; u\_2)\; +\; (v\_1;\; v\_2)\; =\; (u\_1\; +\; v\_1;\; u\_2\; +\; v\_2)$ and $\backslash \; alpha\; \backslash ,\; (u\_1;\; u\_2)\; =\; (\backslash \; alpha\; \backslash ,\; u\_1;\; \backslash \; alpha\; \backslash ,\; u\_2)$,

where $u\_1$, $v\_1$ is in $F\_1$, $u\_2$, $v\_2$ is in $F\_2$, and $\backslash \; alpha$ is in $K$.

This encourages, if $E\_1$ and $E\_2$ are two unspecified vector spaces on the same body $K$, to then define their sum direct, known as external .

External direct sum of two K - vector spaces

The external direct sum of two $K$-espaces vectorial $E\_1$ and $E\_2$ is the Cartesian product $E\_1\; \backslash \; times\; E\_2$ on which one defines

• an addition:

$\backslash \; (u\_1;\; u\_2)\; +\; (v\_1;\; v\_2)\; =\; (u\_1\; +\; v\_1;\; u\_2\; +\; v\_2)$

• an external multiplication by the elements of $K$:

$\backslash \; alpha\; \backslash ,\; (u\_1;\; u\_2)\; =\; (\backslash \; alpha\; \backslash ,\; u\_1;\; \backslash \; alpha\; \backslash ,\; u\_2)$ (where $\backslash \; alpha\; \backslash \; in\; K$) Provided with these two laws of composition, the unit $E\_1\; \backslash \; times\; E\_2$ is a vector space on $K$.

Consequently, $\backslash \; tilde\; \{E\_1\}\; =\; E\_1\; \backslash \; times\; \backslash \; \{0\; \backslash \}$ and $\backslash \; tilde\; \{E\_2\}\; =\; \backslash \; \{0\; \backslash \}\; \backslash \; times\; E\_2$ are two subspaces of $E\_1\; \backslash \; times\; E\_2$, respectively isomorphs with $E\_1$ and $E\_2$ (one has " plongé" $E\_1$, $E\_2$ in the Cartesian product); the relation $E\_1\; \backslash \; times\; E\_2\; =\; \backslash \; tilde\; \{E\_1\}\; \backslash \; oplus\; \backslash \; tilde\; \{E\_2\}$ justifies the name of external direct sum .

When $E\_1$ and $E\_2$ are of finished size, it is the same of their external direct sum, and:

$\backslash \; dim\; (E\_1\; \backslash \; times\; E\_2)\; =\; \backslash \; dim\; E\_1\; +\; \backslash \; dim\; E\_2$

(because $E\_1\; \backslash \; times\; E\_2$ is direct sum of the two subspaces $\backslash \; tilde\; \{E\_1\}$ and $\backslash \; tilde\; \{E\_2\}$, which has even dimension that $\backslash \; E\_1$, $\backslash \; E\_2$ respectively).

External direct sum of several K - vector spaces

One defines in the same way the external direct sum $\backslash \; E\_1\; \backslash \; times\; \backslash \; cdots\; \backslash \; times\; E\_k$ of K vector spaces $E\_1,\; \backslash \; dowries,\; E\_k$ on the same body $K$.

When $E\_1,\; \backslash \; dowries,\; E\_k$ are of finished size, it is the same of their external direct sum, and:

$\backslash \; dim\; (E\_1\; \backslash \; times\; \backslash \; cdots\; \backslash \; times\; E\_k)\; =\; \backslash \; dim\; E\_1\; +\; \backslash \; cdots\; +\; \backslash \; dim\; E\_k$.

Notice in connection with other algebraic structures

One defines in a similar way the external direct sum of a finished number of additive groups, or rings, or of has - modules on the same ring has .

For example, if $A\_1$ and $A\_2$ are two rings, one defines on $A\_1\; \backslash \; times\; A\_2$ two laws of composition internal:

• an addition:

$\backslash \; (a\_1;\; a\_2)\; +\; (b\_1;\; b\_2)\; =\; (a\_1\; +\; b\_1;\; a\_2\; +\; b\_2)$

• a multiplication:

$\backslash \; (a\_1;\; a\_2)\; \backslash \; cdot\; (b\_1;\; b\_2)\; =\; (a\_1\; b\_1;\; a\_2\; b\_2)$

Provided with these two laws of composition, the unit $A\_1\; \backslash \; times\; A\_2$ is a ring. It will be noted that even if $A\_1$ and $A\_2$ are just, their Cartesian product is not it: $a\_1$, $a\_2$ being two nonnull elements of $A\_1$, $A\_2$ respectively, one a: $\backslash \; (a\_1;\; 0)\; \backslash \; cdot\; (0;\; a\_2)\; =\; (0;\; 0)$.

Internal bonds

• vector Space

Category: linear algebra

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