7.7E: Exercises for Section 7.7

Exercise \(\PageIndex{1}\)

Let \( U \) and \( W \) be subspaces of \( \mathbb{R}^3 \), where
\[
U = \text{span} \left\{ \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix} \right\}, \quad
W = \text{span} \left\{ \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} \right\}.
\nonumber\]

1. Describe \( U + W \) geometrically. Find a basis for \( U + W \) and state its dimension.
2. Describe \( U \cap W \) geometrically. Find a basis for \( U \cap W \) and state its dimension.

Answer

1. Since \(W \subseteq U \), \( U + W = U\), with basis \(\text{span} \left\{ \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix} \right\}\). Hence, \( \text{dim}(U+W)=2\).
2. Since \(W \subseteq U \), \( U \cap W = W\), with basis \(\text{span} \left\{ \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} \right\}\). Hence, \( \text{dim}(U \cap W)=1\).

Exercise \(\PageIndex{2}\)

Let

\[
U = \left\{ \begin{bmatrix}x \\ y \\ z\end{bmatrix} \in \mathbb{R}^3 \mid x + y + z = 0 \right\}, \quad
W = \left\{ \begin{bmatrix}x \\ y \\ z\end{bmatrix} \in \mathbb{R}^3 \mid x = y \right\}. \nonumber
\]

1. Describe \( U + W \) geometrically. Find a basis for \( U + W \) and state its dimension.
2. Describe \( U \cap W \) geometrically. Find a basis for \( U \cap W \) and state its dimension.

Answer

1. The sum \( U + W \) is the entire space \( \mathbb{R}^3 \). Dimension: \( \dim(U + W) = 3 \). Basis for \( U + W \): \(\left\{\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}, \quad \begin{bmatrix}0 \\ 1 \\ -1\end{bmatrix}, \quad \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}\right\}\).
2. The intersection \( U \cap W \) is a line in \( \mathbb{R}^3 \) through the origin. Dimension: \( \dim(U \cap W) = 1 \). Basis for \( U \cap W \): \(\left\{\begin{bmatrix}1 \\ 1 \\ -2\end{bmatrix} \right\}\).

Exercise \(\PageIndex{3}\)

Let
\[
U = \text{span} \left\{ \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}, \begin{bmatrix}2 \\ 4 \\ 6\end{bmatrix} \right\}, \quad
W = \text{span} \left\{ \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix} \right\}. \nonumber
\]

1. Find a basis for \( U + W \) and its dimension.
2. Find a basis for \( U \cap W \) and its dimension.

Exercise \(\PageIndex{4}\)

Let
\[
U = \left\{ \begin{bmatrix}x \\ y \\ z \\ w\end{bmatrix} \in \mathbb{R}^4 \mid x + y = 0,\ z - w = 0 \right\}, \quad
W = \left\{ \begin{bmatrix}x \\ y \\ z \\ w\end{bmatrix} \in \mathbb{R}^4 \mid x - z = 0,\ y + w = 0 \right\}.
\]
Find bases and dimensions for \( U \), \( W \), \( U + W \), and \( U \cap W \)

Exercise \(\PageIndex{5}\)

Let \(U,W\) be subspaces of a vector space \(V\) and consider \(U +W\) defined as the set of all \(u+w\) where \(u ∈ U\) and \(w ∈ W\). Show that \(U +W\) is a subspace of \(V\).

Exercise \(\PageIndex{6}\)

Let \(M,N\) be subspaces of a vector space \(\mathbb{R}^2\). Then \(N ∪M\) consists of all vectors which are in either \(M\) or \(N\). Show that \(N ∪M\) is not necessarily a subspace of \(\mathbb{R}^2\) by giving an example where \(N ∪M\) fails to be a subspace.

Exercise \(\PageIndex{7}\)

Suppose \(V\) and \(W\) both have dimension equal to \(7\) and they are subspaces of \(\mathbb{R}^{10}\). What are the possibilities for the dimension of \(V ∩W\)?

Exercise \(\PageIndex{8}\)

Suppose \(V\) has dimension \(p\) and \(W\) has dimension \(q\) and they are each contained in a subspace, \(U\) which has dimension equal to \(n\) where \(n > \text{max}(p,q)\). What are the possibilities for the dimension of \(V ∩W\)?

Answer

Let \(\{x_1,\cdots ,x_k\}\) be a basis for \(V∩W\). Then there is a basis for \(V\) and \(W\) which are respectively \[\{x_1, \cdots ,x_k, y_{k+1},\cdots ,y_p\},\:\{x_1,\cdots ,x_k, z_{k+1},\cdots z_q\}\nonumber\] It follows that you must have \(k+p-k+q-k\leq n\) and so you must have \[p+q-n\leq k\nonumber\]

Exercise \(\PageIndex{9}\)

Suppose \(A\) is an \(m\times n\) matrix and \(B\) is an \(n\times p\) matrix. Show that \[\text{dim}(\text{ker}(AB))\leq\text{dim}(\text{ker}(A))+\text{dim}(\text{ker}(B)).\nonumber\] Consider the subspace, \(B(\mathbb{R}^p )∩\text{ker}(A)\) and suppose a basis for this subspace is \(\{\vec{w}_1,\cdots ,\vec{w}_k\}\). Now suppose \(\{\vec{u}_1,\cdots ,\vec{u}_r\}\) is a basis for \(\text{ker}(B)\). Let \(\{\vec{z}_1,\cdots ,\vec{z}_k\}\) be such that \(B\vec{z}_1 =\vec{w}_i\) and argue that \[\text{ker}(AB)⊆ span\{\vec{u}_1,\cdots ,\vec{u}_r,\vec{z}_1,\cdots ,\vec{z}_k\}.\nonumber\]

Answer

Here is how you do this. Suppose \(AB\vec{x} =\vec{0}\). Then \(B\vec{x} ∈ \text{ker}(A) ∩ B(\mathbb{R}^p)\) and so \(B\vec{x} =\sum\limits_{i=1}^k B\vec{z}_i\) showing that \[\vec{x}-\sum\limits_{i=1}^k\vec{z}_i\in\text{ker}(B)\nonumber\] Consider \(B(\mathbb{R}^p )∩\text{ker}(A)\) and let a basis be \(\{\vec{w}_1,\cdots ,\vec{w}_k\}\). Then each \(\vec{w}_i\) is of the form \(B\vec{z}_i =\vec{w}_i\). Therefore, \(\{\vec{z}_1,\cdots ,\vec{z}_k\}\) is linearly independent and \(AB\vec{z}_i = 0\). Now let \(\{\vec{u}_1,\cdots ,\vec{u}_r\}\) be a basis for \(\text{ker}(B)\). If \(AB\vec{x} =\vec{0}\), then \(B\vec{x} ∈ \text{ker}(A)∩B(\mathbb{R}^p)\) and so \(B\vec{x} =\sum\limits_{i=1}^k c_iB\vec{z}_1\) which implies \[\vec{x}-\sum\limits_{i=1}^k c_i\vec{z}_i\in\text{ker}(B)\nonumber\] and so it is of the form \[\vec{x}-\sum\limits_{i=1}^kc_i\vec{z}_i=\sum\limits_{j=1}^r d_j\vec{u}_j\nonumber\] It follows that if \(AB\vec{x} =\vec{0}\) so that \(\vec{x} ∈ \text{ker}(AB)\), then \[\vec{x}\in span (\vec{z}_1,\cdots ,\vec{z}_k,\vec{u}_1, \cdots ,\vec{u}_r ).\nonumber\] Therefore, \[\begin{aligned}\text{dim}(\text{ker}(AB))&\leq k+r=\text{dim}(B(\mathbb{R}^p)∩\text{ker}(A))+\text{dim}(\text{ker}(B)) \\ &\leq\text{dim}(\text{ker}(A))+\text{dim}(\text{ker}(B))\end{aligned}\]