Chapter 3: Problem 10
Let \(V \subset \mathbb{R}^{n}\) be a subspace. Show that \(V \cap V^{\perp}=\\{0\\}\).
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Chapter 3: Problem 10
Let \(V \subset \mathbb{R}^{n}\) be a subspace. Show that \(V \cap V^{\perp}=\\{0\\}\).
These are the key concepts you need to understand to accurately answer the question.
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\({ }^{*}\) a. Construct a \(3 \times 3\) matrix \(A\) with \(\mathbf{C}(A) \subset \mathbf{N}(A)\). b. Construct a \(3 \times 3\) matrix \(A\) with \(\mathbf{N}(A) \subset \mathbf{C}(A)\). c. Do you think there can be a \(3 \times 3\) matrix \(A\) with \(\mathbf{N}(A)=\mathbf{C}(A)\) ? Why or why not? d. Construct a \(4 \times 4\) matrix \(A\) with \(\mathbf{C}(A)=\mathbf{N}(A)\).
Prove that if \(U\) and \(V\) are subspaces of \(\mathbb{R}^{n}\) and \(W\) is a subspace of \(\mathbb{R}^{n}\) containing all the vectors of \(U\) and all the vectors of \(V\) (that is, \(U \subset W\) and \(V \subset W\) ), then \(U+V \subset W\). This means that \(U+V\) is the smallest subspace containing both \(U\) and \(V\).
Find a basis for the intersection of the subspaces $$ V=\operatorname{Span}((1,0,1,1),(2,1,1,2)) \quad \text { and } \quad W=\operatorname{Span}((0,1,1,0),(2,0,1,2)) \subset \mathbb{R}^{4} . $$
Use the definition of a vector space \(V\) to prove the following: a. \(0 \mathbf{u}=\mathbf{0}\) for every \(\mathbf{u} \in V\). b. \(-\mathbf{u}=(-1) \mathbf{u}\) for every \(\mathbf{u} \in V\). (Hint: The distributive property 7 is all important.)
In each case, construct a matrix with the requisite properties or explain why no such matrix exists. "a. The column space has basis \(\left[\begin{array}{l}1 \\ 0 \\\ 1\end{array}\right]\), and the nullspace contains \(\left[\begin{array}{l}1 \\\ 2 \\ 0\end{array}\right]\). b. The nullspace contains \(\left[\begin{array}{l}1 \\ 0 \\\ 1\end{array}\right],\left[\begin{array}{r}-1 \\ 2 \\ 1\end{array}\right]\), and the row space contains \(\left[\begin{array}{r}1 \\ 1 \\\ -1\end{array}\right]\). *c. The column space has basis \(\left[\begin{array}{l}1 \\\ 0 \\ 1\end{array}\right],\left[\begin{array}{l}0 \\ 1 \\\ 1\end{array}\right]\), and the row space has basis \(\left[\begin{array}{l}1 \\\ 1 \\ 1\end{array}\right],\left[\begin{array}{l}2 \\ 0 \\\ 1\end{array}\right]\). d. The column space and the nullspace both have basis \(\left[\begin{array}{l}1 \\\ 0\end{array}\right]\). e. The column space and the nullspace both have basis \(\left[\begin{array}{l}1 \\\ 0 \\ 0\end{array}\right]\).
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