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Aug 18

Basis and dimension

category: Matrix algebra, MT2175 Further linear algebra, MT3042 Optimisation theory, Optimisation theory No Comments

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.

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