30
Jul 19

## Direct sums of subspaces

The definition of an orthogonal sum $L=L_{1}\oplus L_{2}$ requires two things: 1) every element $l\in L$ can be decomposed as $l=l_{1}+l_{2}$ with $l_{i}\in L_{i},$ $i=1,2,$ and 2) every element of $L_{1}$ is orthogonal to every element of $L_{2}.$ Orthogonality of $L_{1}$ to $L_{2}$ implies $L_{1}\cap L_{2}=\{0\}$ which, in turn, guarantees uniqueness of the representation $l=l_{1}+l_{2}.$ If we drop the orthogonality requirement but retain 1) and $L_{1}\cap L_{2}=\{0\},$ we get the definition of a direct sum.

Definition. Let $L_{1},L_{2}$ be two subspaces such that $L_{1}\cap L_{2}=\{0\}.$ The set $L=\{l_{1}+l_{2}:$ $l_{i}\in L_{i},$ $i=1,2\}$ is called a direct sum of $L_{1},L_{2}$ and denoted $L=L_{1}\dotplus L_{2}$.  The condition $L_{1}\cap L_{2}=\{0\}$ provides uniqueness of the representation $l=l_{1}+l_{2}.$

Exercise 1. Let $L=L_{1}\dotplus L_{2}.$ If $l\in L$ is decomposed as $l=l_{1}+l_{2}$ with $l_{i}\in L_{i},$ $i=1,2,$ define $P_{1}l=l_{1}.$ Then $P_{1}$ is linear and satisfies $P_{1}^{2}=P_{1},$ $\text{Img}(P_{1})=L_{1},$ $N(P_{1})=L_{2}$.

Under conditions of Exercise 1, $P_{1}$ is an oblique projector of $L$ onto $L_{1}$ parallel to $L_{2}.$

Exercise 2. Prove that dimension additivity extends to direct sums: if $L=L_{1}\dotplus L_{2},$ then $\dim L=\dim L_{1}+\dim L_{2}.$

Exercise 3. Let $L_{1},L_{2}$ be two subspaces of $C^{n}$ and suppose $n=\dim L_{1}+\dim L_{2}.$ Then to have $C^{n}=L_{1}+L_{2}$ it is sufficient to check that $L_{1}\cap L_{2}=\{0\},$ in which case $L=L_{1}\dotplus L_{2}$

Proof. Denote $L=L_{1}\dotplus L_{2}.$ By Exercise 2, $\dim L=\dim L_{1}+\dim L_{2}=n.$ If $C^{n}\setminus L$ is not empty, then we can complete a basis in $L$ with a nonzero vector from $C^{n}\setminus L$ to see that $\dim C^{n}\geq n+1$ which is impossible.

1
Aug 18

## Basis and dimension

### Basis and dimension

Definition 1. We say that vectors $x^{(1)},...,x^{(m)}$ form a basis in a subspace $L$ if 1) it is spanned by $x^{(1)},...,x^{(m)}$ and 2) these vectors are linearly independent. The number of vectors in the basis is called a dimension of $L$ and the notation is $\dim L.$

An orthogonal basis is a special type of a basis, when in addition to the above conditions 1)-2) the basis vectors are orthonormal. For the dimension definition to be correct, the number of vectors in any basis should be the same. We prove correctness in a separate post.

Exercise 1. In $R^n$ the unit vectors are linearly independent. Prove this fact 1) directly and 2) using the properties of an orthonormal system.

Direct proof. Any $x\in R^n$ can be represented as

(1) $x=x_1e_1+...+x_ne_n.$

If the right side is zero, then $x=0$ and all $x_i$ are zero.

Proof using orthonormality.  If the right side in (1) is zero, then $x_i=x\cdot e_i=0$ for all $i.$

Exercise 2. $\dim R^{n}=n.$

Proof. (1) shows that $R^n$ is spanned by $e_1,...,e_n.$ Besides, they are linearly independent by Exercise 1.

Definition 2. Let $L_1,\ L_2$ be two subspaces such that any element of one is orthogonal to any element of the other. Then the set $\{x_1+x_2:x_1\in L_1,\ x_2\in L_2\}$ is called an orthogonal sum of $L_1,\ L_2$ and denoted $L=L_1\oplus L_2.$

Exercise 3. If a vector $x$ belongs to both terms in the orthogonal sum of two subspaces $L=L_1\oplus L_2$, then it is zero. This means that $L_1\cap L_2=\{0\}.$

Proof. This is because any element of $L_1$ is orthogonal to any element of $L_2,$ so $x$ is orthogonal to itself, $0=x\cdot x=\|x\|^2$ and $x=0.$

Exercise 4 (dimension additivity) Let $L=L_1\oplus L_2$ be an orthogonal sum of two subspaces. Then $\dim L=\dim L_1+\dim L_2.$

Proof. Let $l_i=\dim L_i,$ $i=1,2.$ By definition, $L_1$ is spanned by some linearly independent vectors $y^{(1)},...,y^{(l_{1})}$ and $L_2$ is spanned by some linearly independent vectors $z^{(1)},...,z^{(l_{2})}.$ Any $x\in L$ can be decomposed as $x=y+z,$ $y\in L_1,$ $z\in L_2.$ Since $y,z$ can be further decomposed as $y=\sum_{i=1}^{l_1}a_iy^{(i)},$ $z=\sum_{i=1}^{l_2}b_iz^{(i)},$ the system $y^{(1)},...,y^{(l_1)}, z^{(1)},...,z^{(l_2)}$ spans $L.$

Moreover, this system is linearly independent. If

$\sum_{i=1}^{l_1}a_iy^{(i)}+\sum_{i=1}^{l_2}b_iz^{(i)}=0,$

then

$L_1\ni\sum_{i=1}^{l_1}a_iy^{(i)}=-\sum_{i=1}^{l_2}b_iz^{(i)}\in L_2.$

By Exercise 3 then $\sum_{i=1}^{l_1}a_iy^{(i)}=0,$ $\sum_{i=1}^{l_2}b_iz^{(i)}=0.$ By linear independence of the vectors in the two systems all coefficients $a_i,b_i$ must be zero.

The conclusion is that $\dim L=l_1+l_2.$