# Dimension of vector space problem

1. Sep 15, 2008

### neelakash

1. The problem statement, all variables and given/known data

Suppose V1 (dim. n1) and V29dim. n2) are two vector subspaces such that any element in V1 is orthogonal to any element in V2.Show that the dimensionality of V1+V2 is n1+n2

2. Relevant equations
3. The attempt at a solution

The subspace V1 is spanned by n1 linearly indipendent (mutually orthogonal) vectors.

The subspace V2 is spanned by n2 linearly indipendent (mutually orthogonal) vectors.

Also, the space V1+V2 is spanned by (n1+n2) mutuallyorthogonal vectors.Since, any element in V1 is perpendicular to any element in V2.Thus,the space will be spanned by (n1+n2) linearly independent vectors.

Clearly, the dimension will be n1+n2.

Please tell me if there is a fault in my argument.

Last edited: Sep 15, 2008
2. Sep 15, 2008

Let $$\mathcal{B}_1$$ be the basis for $$V_1$$ and $$\mathcal{B}_2$$ the basis for $$V_2$$

Oops - I hit post when I meant to hit preview.

1. Show that the set $$\mathcal{B} = \mathcal{B}_1 \cup \mathcal{B}_2$$ is linearly independent (since you can assume that all basis vectors in the original two basis sets are orthonormal, this should be rather easy). Note that $$\mathcal{B}$$ contains exactly $$n_1 + n_2$$ vectors
2. Pick an arbitrary $$\vec{y} \in V_1 + V_2$$ and show that $$\vec{y}$$ can be written as a linear combination of vectors in $$\mathcal{B}$$ (this should be easy too)
3. Show that if you have any set of $$n_1 + n_2 + 1$$ vectors in $$V_1 + V_2$$ they must be linearly dependent .

Step 1 shows the obvious set to be linearly independent
Step 2 shows that the set spans $$V_1 + V_2$$
Step 3 shows that the set is a maximally independent set of vectors, and these 3 results taken together get you to your answer.

3. Sep 15, 2008

### HallsofIvy

Staff Emeritus
This is not a sentence!

You need to explain (perhaps by completing the sentence fragment) how you know that all n1+ n2 vectors are linearly independent. Yes, you can choose the n1 vectors in V1 to be mutually orthogonal. Yes, you can choose the n2 vectors in V2 to be mutually orthogonal. Now you need to state clearly that all n1 vectors in the basis for V1 are orthogonal to all the n2 vectors in the basis for V2 because of the condition "any element in V1 is orthogonal to any element in V2".

4. Sep 16, 2008

### neelakash

Let me formulate the problem this way:

Vector space $$\ V_1$$ contains $$\ n_1$$ linearly indipendent vectors.Hence the basis set is:

$$\ B_1 = {|\ u_1>,|\ u_2>, |\ u_3>...|\ u_i>,...|\ u_k>,...|\ u_n1>}$$

Similarly,vector space $$\ V_2$$ contains $$\ n_2$$ linearly indipendent vectors. Hence, the basis set is:

$$\ B_2 = {|\ v_1>,|\ v_2>, |\ v_3>...|\ v_p>,...|\ v_r>,...|\ v_n2>}$$

Given:
$$<\ u_i|\ u_k>=0$$............................A [for all i,j,k in the set [1,2,...,n1]]

$$<\ v_p|\ v_r>=0$$............................B [for all p,q,r in the set [1,2,...,n2]]

Additionally, $$<\ u_i|\ v_p>=0$$...........C [for all i,j,k,...p,q,r]

Let us consider the set:

$$\ B'={|\ u_1>,|\ u_2>,...|\ u_n1,|\ v_1>,|\ v_2>,...|\ v_n2>}$$

containing $$\ n_1+\ n_2$$ linearly indipendent vectors.

The basis set $$\ B$$ for the vector space $$\ V$$will be a subset of this.The number of elements in the set $$\ B$$ will be less than or equal to $$\ n_1+\ n_2$$ depending on whether all the $$\ n_1$$ vectors in $$\ B_1$$ are orthogonal to all $$\ n_2$$ vectors in $$\ B_2$$.

Since,we have C satisfied by all $$\ u_i$$ and $$\ v_p$$,

$$\ B=\ B_1 \ U \ B_2$$ has all distinct elements because of A,B and C.So, there are total $$\ n_1+\ n_2$$ elements.All of them are linearly indipendent of each other. And they are therefore, a good choice of basis of the vector space V.

Hence, the vector space V is spanned by $$\ n_1+\ n_2$$ vectors.No more,no less.Insertion of any non-trivial extra element in $$\ B'$$ will result in $$\ n_1+\n_2+\1$$ vectors and the set of vectors will be lineraly dependent.

With our choice, $$\ B=\ B'$$.Any vector in vector space V can be written as a linear combination of those $$\ n_1+\ n_2$$ vectors in $$\ B$$.So, the dimension of the vector space $$\ V$$ will be $$\ n_1+\ n_2$$