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.)
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Textbook Solutions for Linear Algebra: A Geometric Approach
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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