Question

Let \( B_1 = \left\{(2,1,1,1),(1,1,1,1),(1,1,2,1)\right\} \hspace{2mm} \)and \( \hspace{2mm}B_2 =\left\{(2,1,2,2)\right\}\hspace{2mm} \)be two subsets of\( \hspace{2mm} \mathbb{R}^4, E_1 \) be a subspace spanned by\( B_1, E_2 \)be a subspace spanned by \( B_2 \), and L be a linear operator on \( \mathbb{R}^4 \) defined by

\( L(v)=(-w +4x-y+z,-w+3x,-w+2x+y,-w+2x+z)\hspace{2mm},v=(w,x,y,z) \)

(a) Show that \( B_1 \) is a basis for \( E_1 \) and \( B_2 \) is a basis for \( E_2 \)

(b) Show that \( E_1 \) and \( E_2 \) are L-invariant. Find the matrices \( A_{1} =[L_{E_1}]_{B_1} \) and \( A_2=[L_{E_2}]_{B_2} \)

 

Answer 1

Solution

(a) Show that \( B_1 \) is a basis for \( E_1 \) and \( B_2 \) is a basis for 

\( \bullet \hspace{2mm}For\hspace{2mm}E_2 \)

\( \implies E_1=span\left\{\begin{pmatrix}2\\ 1\\ 1\\ 1\end{pmatrix},\begin{pmatrix}1\\ 1\\ 1\\ 1\end{pmatrix},\begin{pmatrix}1\\ 1\\ 2\\ 1\end{pmatrix}\right\} \)

Check \( \begin{pmatrix}2&1&1&1\\ 1&1&1&1\\ 1&1&2&1\end{pmatrix}\sim \begin{pmatrix}1&0&0&0\\ 1&1&1&1\\ 0&0&1&0\end{pmatrix} \implies L.I \)

Therefore.  B is a basis for \( E_1 \) and \( B_2 \) is basis for \( E_2 \) is obvious.

(b) Show that \( E_1 \) and \( E_2 . \) are L-invariant. Find the matrices \( A_{1} =[L_{E_1}]_{B_1} \) and \( A_2=[L_{E_2}]_{B_2} \)

we have  \( E_1=span\left\{\begin{pmatrix}2\\ 1\\ 1\\ 1\end{pmatrix},\begin{pmatrix}1\\ 1\\ 1\\ 1\end{pmatrix},\begin{pmatrix}1\\ 1\\ 2\\ 1\end{pmatrix}\right\} \)

Let \( x\in E_1\implies x=\begin{pmatrix}2\alpha \:_1+\alpha \:_2+\alpha \:_3\\ \alpha \:_1+\alpha \:_2+\alpha \:_3\\ \alpha \:_1+\alpha \:_2+2\alpha \:_3\\ \alpha _1+\alpha _2+\alpha _3\end{pmatrix} \)

\( L(x_1)=\bigg(x_1,x_2,x_3,x_4\bigg) \)

where.  \( x_1=2\alpha_1+3\alpha_2-2\alpha_3 \hspace{2mm},x_2=\alpha_1+2\alpha_2+2\alpha_3\hspace{2mm},x_3=\alpha_2+\alpha_3\hspace{2mm} \)

\( ,x_4=\alpha_1+2\alpha_2+2\alpha_3 \)

Then \( L(E_1)=span\left\{\begin{pmatrix}2\\ 1\\ 0\\ 1\end{pmatrix},\begin{pmatrix}3\\ 2\\ 1\\ 2\end{pmatrix},\begin{pmatrix}-2\\ 2\\ 1\\ 2\end{pmatrix}\right\} \)

we will show that\( L(E_1)\subseteq E_1 \)

Let  \( x=\begin{pmatrix}u\\ v\\ w\end{pmatrix}\in L\left(E_1\right)\implies \begin{pmatrix}u\\ v\\ w\end{pmatrix}=\alpha _1\begin{pmatrix}2\\ 1\\ 0\\ 1\end{pmatrix}+\alpha _2\begin{pmatrix}3\\ 2\\ 1\\ 2\end{pmatrix}+\alpha _3\begin{pmatrix}-2\\ 2\\ 1\\ 2\end{pmatrix} \)

\( \bullet \) if  \( x\in E_1 \hspace{2mm}so \hspace{2mm} x=\beta _1\:\begin{pmatrix}2\\ 1\\ 1\\ 1\end{pmatrix}+\:\beta _2\:\begin{pmatrix}1\\ 1\\ 1\\ 1\end{pmatrix}\:+\beta _3\begin{pmatrix}1\\ 1\\ 2\\ 1\end{pmatrix} \)

we will show that there exists  \( \beta _1,\:\beta _2\:,\beta _3 \)

we get  \( \implies\begin{cases} 2\alpha_1+3\alpha_2-2\alpha_3=2\beta_1+\beta_2+\beta_3 & \quad \\ \alpha_1+2\alpha_2+2\alpha_3=\beta_1+\beta_2+\beta_3 & \quad \\ \alpha_2+\alpha_3=\beta_1+\beta_2+2\beta_3 & \quad \end{cases} \)

since there are three equations satisfy by  \( \beta_1,\beta_2,\beta_3 \)   so there exists \( \hspace{2mm}\beta_1,\beta_2,\beta_3\hspace{2mm} \) which satisfy \( \hspace{2mm}x\in E_1\hspace{3mm} \)since \( \hspace{2mm}x\in L(E_1)\hspace{2mm} \)Then \( L(E_1)\subseteq E_1 \hspace{2mm}Thus E_1 \) is stable by L.

\( \bullet \hspace{2mm} \)show\( \hspace{2mm}E_2 \) is L-invariant

\( \implies E_2=\left\{\begin{pmatrix}2\\ 1\\ 2\\ 2\end{pmatrix}\right\}\iff L\bigg(2t,t,2t,2t\bigg)=\bigg(2t,t,2t,2t\bigg) \)

then for \( x\in E_2 : L(x)=x \)

so \( L(E_1)\subseteq E_2 \implies E_2 \hspace{2mm} \)is stable by L.

\( \bullet \hspace{2mm}Find \hspace{2mm} \)\( A_1=[L_{E_1}]_{B_1} \hspace{2mm} \)and\( \hspace{2mm} B=[L_{E_1}]_{B_1} \)

Let  \( x\in \mathbb{R}^4\implies [x]_{B_1}=\bigg(\alpha_1,\alpha_2,\alpha_3\bigg) \)

where \( [x]_{B_1}=\alpha _1\begin{pmatrix}2\\ 1\\ 1\\ 1\end{pmatrix}\:+\alpha _2\begin{pmatrix}1\\ 1\\ 1\\ 1\end{pmatrix}\:+\alpha _3\begin{pmatrix}1\\ 1\\ 2\\ 1\end{pmatrix}=\begin{pmatrix}a\\ b\\ c\\ d\end{pmatrix} \)

\( \implies \begin{pmatrix}2&1&1\\ 1&1&1\\ 1&1&2\\ 1&1&1\end{pmatrix}\begin{pmatrix}\alpha _1\\ \alpha _2\\ \alpha _3\end{pmatrix}=\begin{pmatrix}a\\ b\\ c\\ d\end{pmatrix} \)

\( \iff \left( \begin{array}{ccc|c} 2 & 1 & 1 & a\\ 1 & 1 & 1 & b\\ 1 & 1 & 2 & c\\ 1 & 1 & 1 & d \end{array} \right)\sim \left( \begin{array}{ccc|c} 2 & 1 & 1 & a\\ -1 & 0 & 0 & b-a\\ 0 & 1 & 2 & c+b-a\\ 0 & 0 & 0 & d-b \end{array} \right) \)

\( \sim \left( \begin{array}{ccc|c} 0 & 1 & 1 & a+2b-2a\\ -1 & 0 & 0 & b-a\\ 0 & 1 & 2 & c+b-a\\ 0 & 0 & 0 & d-b \end{array} \right) \sim \left( \begin{array}{ccc|c} 0 & 1 & 1 & -a+2b\\ 1 & 0 & 0 & -b+a\\ 0 & 0 & 1 & c+b-2b\\ 0 & 0 & 0 & d-b \end{array} \right) \)

since  \( L(E_1)\subseteq E_1\implies d-b=0 \)

\( \implies \begin{cases} \alpha_1=a-b & \quad \\ \alpha_3=-b+c & \quad \\ \alpha_2=2b-a-c+b=3b-a-c & \quad \end{cases} \)

\( [x]_{B_1}=\begin{pmatrix}a-b\\ 3b-a-c\\ c-b\end{pmatrix}\hspace{2mm}, L\begin{pmatrix}2\\ 1\\ 1\\ 1\end{pmatrix}=\left(2,1,1,1\right) \)

\( L\begin{pmatrix}1\\ 1\\ 1\\ 1\end{pmatrix}=\left(3,2,2,2\right), L\begin{pmatrix}1\\ 1\\ 2\\ 1\end{pmatrix}=\left(2,2,3,2\right) \)

\( \implies [L_{E_1}]_{B_1}=\begin{pmatrix}1&1&0\\ 0&1&1\\ 0&0&1\end{pmatrix}\hspace{3mm} Let \hspace{2mm}x\in \mathbb{R}^4\implies [x]_{B_1}=\alpha_1 \)

\( \iff \begin{pmatrix}a\\ b\\ c\\ d\end{pmatrix}=\alpha _1\begin{pmatrix}2\\ 1\\ 2\\ 2\end{pmatrix}\implies [L_{E_2}]_{B_2}=1 \) then\( A_2=(1) \)

Answer.

Therefore.

a). B is a basis for \( E_1 \) and \( B_2 \) is basis for \( E_2 \) is obvious.

b). \( L(E_1)\subseteq E_2 \implies E_2 \) is stable by L

\( \)