Decide whether each of the following is a subspace ofM22. If so, provide a basis. If | StudySoup

Textbook Solutions for Linear Algebra: A Geometric Approach

Chapter 3.6 Problem 3

Question

Decide whether each of the following is a subspace of \(\mathcal{M}_{2 \times 2}\). If so, provide a basis. If not, give a reason.

a. \(\left\{A \in \mathcal{M}_{2 \times 2}:\left[\begin{array}{l}1 \\ 2\end{array}\right] \in \mathbf{N}(A)\right\}\)

b. \(\left\{A \in \mathcal{M}_{2 \times 2}:\left[\begin{array}{l}1 \\ 2\end{array}\right] \in \mathbf{C}(A)\right\}\)

c. \(\left\{A \in \mathcal{M}_{2 \times 2}: \operatorname{rank}(A)=1\right\}\)

d. \(\left\{A \in \mathcal{M}_{2 \times 2}: \operatorname{rank}(A) \leq 1\right\}\)

e. \(\left\{A \in \mathcal{M}_{2 \times 2}: A \text { is in echelon form }\right\}\)

f. \(\left\{A \in \mathcal{M}_{2 \times 2}: A\left[\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right] A\right\}\)

g. \(\left\{A \in \mathcal{M}_{2 \times 2}: A\left[\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right] A^{\top}\right\}\)

Solution

Step 1 of 7

a)

Given \(\left\{ {A \in {{\rm M}_{2 \times 2}}:\left[ {\begin{array}{*{20}}1\\2\end{array}} \right] \in {\rm N}\left( A \right)} \right\}\)

Let \(A,B \in W\).where \(\left[ {\begin{array}{*{20}}1\\2\end{array}} \right] \in {\rm N}\left( A \right)\) and \(\left[ {\begin{array}{*{20}}1\\2\end{array}} \right] \in {\rm N}\left( B \right)\).

Hence \(A\left[ {\begin{array}{*{20}}1\\2\end{array}} \right] = B\left[ {\begin{array}{*{20}}1\\2\end{array}} \right] = 0\)

Let, \(\alpha ,\beta  \in F\)

\(\left( {\alpha A + \beta B} \right)\left[ {\begin{array}{*{20}}1\\2\end{array}} \right] = \alpha  \cdot A\left[ {\begin{array}{*{20}}1\\2\end{array}} \right] + \beta  \cdot B\left[ {\begin{array}{*{20}}1\\2\end{array}} \right]\\ = 0 + 0\\ = 0\)

Therefore, \(\left[ {\begin{array}{*{20}}1\\2\end{array}} \right] \in N\left( {\alpha A + \beta B} \right)\)

Hence, \(\alpha A + \beta B \in W\)

Therefore, it can be stated that \(\left\{ {A \in {{\rm M}_{2 \times 2}}:\left[ {\begin{array}{*{20}}1\\2\end{array}} \right] \in {\rm N}\left( A \right)} \right\}\) is a subspace of \({{\rm M}_{2 \times 2}}\).

The basis for \(W\) is obtained below:

Let, \(\left\{ {A = \left[ {\begin{array}{*{20}}a&b\\c&d\end{array}} \right]} \right\}\)

Now

\(A\left[ {\begin{array}{*{20}}1\\2\end{array}} \right] = 0\)

\(\left[ {\begin{array}{*{20}}a&b\\c&d\end{array}} \right]\left[ {\begin{array}{*{20}}1\\2\end{array}} \right] = 0\)

\(\left[ {\begin{array}{*{20}}{a + 2b}\\{c + 2d}\end{array}} \right] = 0\)

From the above we can write:

\(a + 2b = 0\)

\(a =  - 2b\)

\(c + 2d = 0\)

\(c =  - 2d\)

The matrix \(A\) can be rewritten as

\(A = \left[ {\begin{array}{*{20}}{ - 2b}&b\\{ - 2d}&d\end{array}} \right]\)

Therefore \(W = \left\{ {\left[ {\begin{array}{*{20}}{ - 2b}&b\\{ - 2d}&d\end{array}} \right]:b,d \in F} \right\}\)

Basis: therefore \(W = \left\{ {\left[ {\begin{array}{*{20}}{ - 2}&1\\0&0\end{array}} \right],\left[ {\begin{array}{*{20}}0&0\\{ - 2}&1\end{array}} \right]} \right\}\)

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full solution

Title Linear Algebra: A Geometric Approach 2 
Author Ted Shifrin, Malcolm Adams
ISBN 9781429215213

Decide whether each of the following is a subspace ofM22. If so, provide a basis. If

Chapter 3.6 textbook questions

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