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Chapter 4: Problem 40

Prove Theorem 4.17 : Let \(W\) be a subspace of an \(n\) -dimensional vector space \(V\). Then \(\operatorname{dim} W \leq n\). In particular, if \(\operatorname{dim} W=n,\) then \(W=V\).

Short Answer

Expert verified
In summary, to prove Theorem 4.17, we showed that any basis of subspace \(W\) can be extended to form a basis of \(V\), implying \(\operatorname{dim} W \leq n\). If \(\operatorname{dim} W=n\), the basis of \(W\) is already a basis of \(V\), so \(W\) spans all of \(V\) and \(W=V\).

Step by step solution

01

Show that the dimension of \(W\) cannot be more than the dimension of \(V\).

Let \(W\) be a subspace of an \(n\)-dimensional vector space \(V\). Since \(W\) is a subspace of \(V\), it must contain vectors from \(V\). Therefore, any basis of \(W\) is a set of linearly independent vectors and can be extended to form a basis of \(V\). The number of vectors in the basis of \(W\) (i.e., the dimension of \(W\)) cannot be more than the number of vectors in the basis of \(V\) (i.e., the dimension of \(V\), which is \(n\)). This is because adding more linearly independent vectors to a basis of \(V\) will result in a linearly dependent set, which contradicts the definition of a basis. Hence, the dimension of \(W\) is less than or equal to \(n\).
02

Show that if the dimension of \(W\) is equal to the dimension of \(V\), then \(W\) must be equal to \(V\).

Suppose the dimension of \(W\) is equal to the dimension of \(V\). In this case, the basis of \(W\) will have the same number of vectors as the basis of \(V\). Since the basis of \(W\) is a set of linearly independent vectors, and it has the same number of vectors as the basis of \(V\), we can extend it to form a basis of \(V\). However, since the number of basis vectors is the same in both \(W\) and \(V\), the extension of the basis of \(W\) to form a basis of \(V\) is trivial 鈥� it means the basis of \(W\) is already a basis of \(V\). As a result, \(W\) spans all of \(V\), and we can conclude that \(W\) is equal to \(V\). So, we have shown that if \(W\) is a subspace of an \(n\)-dimensional vector space \(V\), the dimension of \(W\) is less than or equal to \(n\). Furthermore, if the dimension of \(W\) is equal to \(n\), then \(W\) must be equal to \(V\). Thus, we have proven Theorem 4.17.

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Most popular questions from this chapter

Let \(V\) be the set of ordered pairs \((a, b)\) of real numbers. Show that \(V\) is not a vector space over \(\mathbf{R}\) with addition and scalar multiplication defined by (i) \((a, b)+(c, d)=(a+d, b+c)\) and \(k(a, b)=(k a, k b)\), (ii) \((a, b)+(c, d)=(a+c, b+d)\) and \(k(a, b)=(a, b)\), (iii) \((a, b)+(c, d)=(0,0)\) and \(k(a, b)=(k a, k b)\), (iv) \(\quad(a, b)+(c, d)=(a c, b d)\) and \(k(a, b)=(k a, k b)\).

For \(M \in M_{n \times n}(C)\), let \(\bar{M}\) be the matrix such that \((\bar{M})_{i j}=\overline{M_{i j}}\) for all \(i, j\), where $\overline{M_{i j}}\( is the complex conjugate of \)M_{i j}$. (a) Prove that \(\operatorname{det}(\bar{M})=\overline{\operatorname{det}(M)}\). (b) A matrix \(Q \in \mathrm{M}_{n \times n}(C)\) is called unitary if $Q Q^{*}=I\(, where \)Q^{*}=\overline{Q^{t}}\(. Prove that if \)Q$ is a unitary matrix, then \(|\operatorname{det}(Q)|=1\).

Suppose \(u\) and \(v\) belong to a vector space \(V\). Simplify each of the following expressions: (a) \(E_{1}=4(5 u-6 v)+2(3 u+v)\), (c) \(\quad E_{3}=6(3 u+2 v)+5 u-7 v\), (b) \(E_{2}=5(2 u-3 v)+4(7 v+8)\), (d) \(E_{4}=3(5 u+2 / v)\).

For \(k=1,2, \ldots, 5,\) find the number \(n_{k}\) of linearly independent subsets consisting of \(k\) columns for each of the following matrices: (a) \(A=\left[\begin{array}{lllll}1 & 1 & 0 & 2 & 3 \\ 1 & 2 & 0 & 2 & 5 \\ 1 & 3 & 0 & 2 & 7\end{array}\right]\) (b) \(B=\left[\begin{array}{lllll}1 & 2 & 1 & 0 & 2 \\ 1 & 2 & 3 & 0 & 4 \\ 1 & 1 & 5 & 0 & 6\end{array}\right]\)

A matrix \(Q \in \mathrm{M}_{n \times n}(R)\) is called orthogonal if $Q Q^{t}=I\(. Prove that if \)Q\( is orthogonal, then \)\operatorname{det}(Q)=\pm 1$.
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