Raising the bar basis Archives

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.