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Chapter 3: Problem 18

Show that if \(U\) and \(V\) are subspaces of \(\mathbb{R}^{n}\) and \(U \cap V=\\{0\\},\) then \\[ \operatorname{dim}(U+V)=\operatorname{dim} U+\operatorname{dim} V \\]

Short Answer

Expert verified
Taking the basis of U as \(B_{U}=\{u_{1}, u_{2}, \ldots, u_{k}\}\) and V as \(B_{V}=\{v_{1}, v_{2}, \ldots, v_{l}\}\), we can show that the sum of U and V, U+V, has a basis \(B_{U\oplus V} = B_{U} \cup B_{V}\), formed from linearly independent vectors. Since the intersection is only the zero vector, it follows that the dimension of U+V equals the sum of their individual dimensions, i.e., \(\operatorname{dim}(U+V)=\operatorname{dim}(U)+\operatorname{dim}(V)\).

Step by step solution

01

Write definitions of the sum and dimension of the subspaces U and V

The sum of two subspaces U and V, denoted by U+V, is the set of all vectors in \(\mathbb{R}^{n}\) that can be written as the sum of a vector in U and a vector in V. The dimension of a subspace W, denoted by \(\operatorname{dim}(W)\), is the number of vectors in a basis for W.
02

Relate the dimension of the sum of the subspaces U and V

To find the dimension of U+V, we'll first write down the basis for U+V as a union of the bases of U and V. Let \(B_{U}=\{u_{1}, u_{2}, \ldots, u_{k}\}\) be a basis for U and \(B_{V}=\{v_{1}, v_{2}, \ldots, v_{l}\}\) be a basis for V. Since the intersection of U and V is only the zero vector, the vectors in basis \(B_{U}\) and \(B_{V}\) are linearly independent. Therefore, the set \(B_{U\oplus V} = B_{U} \cup B_{V}\) forms a basis for U+V. Notice that \(\operatorname{dim}(U+V) = k+l = \operatorname{dim}(U) + \operatorname{dim}(V)\).
03

Relate the dimension of U+V to the dimensions of U and V

Now that we have expressed the dimension of U+V in terms of the dimensions of U and V, we can use this to prove the given statement. Since we have shown that the set \(B_{U\oplus V} = B_{U} \cup B_{V}\) forms a basis for U+V, and we know that the union of two linearly independent sets is also linearly independent, it follows that the dimension of the sum of the subspaces U and V is equal to the sum of the dimensions of the individual subspaces. Thus, we have proven that if U and V are subspaces of \(\mathbb{R}^{n}\) with their intersection being the zero vector, then \(\operatorname{dim}(U+V)=\operatorname{dim}(U)+\operatorname{dim}(V)\).

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Most popular questions from this chapter

In Exercise 2 of Section \(3,\) indicate whether the given vectors form a basis for \(\mathbb{R}^{3}\).

Which of the sets that follow are spanning sets for \(P_{3} ?\) Justify your answers. (a) \(\left\\{1, x^{2}, x^{2}-2\right\\}\) (b) \(\left\\{2, x^{2}, x, 2 x+3\right\\}\) (c) \(\left\\{x+2, x+1, x^{2}-1\right\\}\) (d) \(\left\\{x+2, x^{2}-1\right\\}\)

Let \(S\) denote the set of all infinite sequences of real numbers with scalar multiplication and addition defined by $$\begin{aligned} \alpha\left\\{a_{n}\right\\} &=\left\\{\alpha a_{n}\right\\} \\ \left\\{a_{n}\right\\}+\left\\{b_{n}\right\\} &=\left\\{a_{n}+b_{n}\right\\} \end{aligned}$$ Show that \(S\) is a vector space.

Prove: If \(A\) is an \(m \times n\) matrix and the column vectors of \(A\) span \(\mathbb{R}^{m}\), then \(A\) has a right inverse. [Hint: Let e jenote the \(j\) th column of \(I_{m},\) and solve \(A \mathbf{x}=\mathbf{e}_{j}\) for \(j=1, \ldots, m ].\)

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