- #1
mathmari
Gold Member
MHB
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Hey!
Let $B=\{b_1, \ldots , b_5\}$ be a basis of the real vector space $V$ and let $\Phi$ be an endomorphis of $V$ with
\begin{align*}\Phi (b_1)& =4b_1+2b_2 -2b_4-3b_5 \\ \Phi (b_2)& = -2b_3 +b_5 \\ \Phi (b_3)& = -4b_2+2b_3 -b_5 \\ \Phi (b_4)& =-2b_1 +3b_3+b_4-b_5 \\ \Phi (b_5)& = 3b_2 +2b_5\end{align*}
I have found the following transformation matrix of $\Phi$ in relation to $B$:
\begin{equation*}A_{\Phi}=\begin{pmatrix} \ 4& \ 0& \ 0& -2& \ 0\\ \ 2& \ 0& -4& \ 0& \ 3\\ \ 0& -2& \ 2& \ 3& \ 0 \\ -2& \ 0& \ 0& \ 1& \ 0 \\ -3& \ 1& -1& -1& \ 2\end{pmatrix}\end{equation*} Then I want to check if $\Phi$ is bijective. To check this do we use the transormation matrix? (Wondering) I want to show also that the set $C=\{c_1, c_2, c_3\}$, that consists of the vectors $c_1=b_2+b_3+b_5, \ c_2=-b_3+b_5, \ c_3=b_2+b_5$, is a basis of a vector subspace $U$ of $V$ with $\Phi (U)\subset U$.
To show that the set $C$ is a basis, we have to show that the vectors $c_1, c_2, c_3$ are linearly independent, right? (Wondering)
We have that \begin{equation*}x c_1+y c_2+z c_3=0 \Rightarrow x (b_2+b_3+b_5)+y (-b_3+b_5)+z (b_2+b_5)=0 \Rightarrow (x+y )b_2+(x -y )b_3+(x +y +z )b_5=0\end{equation*}
Since $b_2, b_3, b_5$ are elements of a basis we get $x+y =x -y =x +y +z =0$ and so $x=y=z=0$.
Therefore, $c_1, c_2, c_3$ are linearly independent and the set $C$ is a basis. To show that $\Phi (U)\subset U$ I have done the following:
Let $c\in U$. Simce $C$ is a basis, $c$ can be written as a linear combination of the $c_1, c_2, c_3$, $c=a_1c_1+a_2c_2+a_3c_3$.
To show that $\Phi (U)\subset U$, we have to show that $\Phi (c)\in U$.
We have that $\Phi (c)=a_1\Phi (c_1)+a_2\Phi (c_2)+a_3\Phi (c_3)$.
Calculating the $\Phi (c_1), \Phi (c_2), \Phi (c_3)$ we get:
\begin{align*}\Phi (c_1) = & \Phi (b_2+b_3+b_5)=\ldots =-b_2+2b_5 \\ \Phi (c_2) = & \Phi (-b_3+b_5)=\ldots =7b_2-2b_3+3b_5 \\ \Phi (c_3) = & \Phi (b_2+b_5)=\ldots =3b_2-2b_3+3b_5 \end{align*}From the definitions of $c_1,c_2,c_3$ we get that $c_1+c_2-c_3=b_5, 2c_3-(c_1+c_2)=b_2, b_3=c_1-c_3$.
The $c_1, c_2, c_3$ are elements of the vector subspace $U$. So, each linear combination will also be in $U$.
Since we can write the $b_2, b_3, b_5$ as a linear combination of the $c_1, c_2, c_3$, we get that $b_2, b_3, b_5\in U$.
So, every linear combination of the $b_2, b_3, b_5$ is also in $U$.
Therefore $\Phi (c_1), \Phi (c_2), \Phi (c_3)\in U$.
And from that we get that every linear combination of them is also in $U$, and so $\Phi (c)=a_1\Phi (c_1)+a_2\Phi (c_2)+\Phi (c_3)\in U$. Is everything correct? Could I improve something? (Wondering)
Let $B=\{b_1, \ldots , b_5\}$ be a basis of the real vector space $V$ and let $\Phi$ be an endomorphis of $V$ with
\begin{align*}\Phi (b_1)& =4b_1+2b_2 -2b_4-3b_5 \\ \Phi (b_2)& = -2b_3 +b_5 \\ \Phi (b_3)& = -4b_2+2b_3 -b_5 \\ \Phi (b_4)& =-2b_1 +3b_3+b_4-b_5 \\ \Phi (b_5)& = 3b_2 +2b_5\end{align*}
I have found the following transformation matrix of $\Phi$ in relation to $B$:
\begin{equation*}A_{\Phi}=\begin{pmatrix} \ 4& \ 0& \ 0& -2& \ 0\\ \ 2& \ 0& -4& \ 0& \ 3\\ \ 0& -2& \ 2& \ 3& \ 0 \\ -2& \ 0& \ 0& \ 1& \ 0 \\ -3& \ 1& -1& -1& \ 2\end{pmatrix}\end{equation*} Then I want to check if $\Phi$ is bijective. To check this do we use the transormation matrix? (Wondering) I want to show also that the set $C=\{c_1, c_2, c_3\}$, that consists of the vectors $c_1=b_2+b_3+b_5, \ c_2=-b_3+b_5, \ c_3=b_2+b_5$, is a basis of a vector subspace $U$ of $V$ with $\Phi (U)\subset U$.
To show that the set $C$ is a basis, we have to show that the vectors $c_1, c_2, c_3$ are linearly independent, right? (Wondering)
We have that \begin{equation*}x c_1+y c_2+z c_3=0 \Rightarrow x (b_2+b_3+b_5)+y (-b_3+b_5)+z (b_2+b_5)=0 \Rightarrow (x+y )b_2+(x -y )b_3+(x +y +z )b_5=0\end{equation*}
Since $b_2, b_3, b_5$ are elements of a basis we get $x+y =x -y =x +y +z =0$ and so $x=y=z=0$.
Therefore, $c_1, c_2, c_3$ are linearly independent and the set $C$ is a basis. To show that $\Phi (U)\subset U$ I have done the following:
Let $c\in U$. Simce $C$ is a basis, $c$ can be written as a linear combination of the $c_1, c_2, c_3$, $c=a_1c_1+a_2c_2+a_3c_3$.
To show that $\Phi (U)\subset U$, we have to show that $\Phi (c)\in U$.
We have that $\Phi (c)=a_1\Phi (c_1)+a_2\Phi (c_2)+a_3\Phi (c_3)$.
Calculating the $\Phi (c_1), \Phi (c_2), \Phi (c_3)$ we get:
\begin{align*}\Phi (c_1) = & \Phi (b_2+b_3+b_5)=\ldots =-b_2+2b_5 \\ \Phi (c_2) = & \Phi (-b_3+b_5)=\ldots =7b_2-2b_3+3b_5 \\ \Phi (c_3) = & \Phi (b_2+b_5)=\ldots =3b_2-2b_3+3b_5 \end{align*}From the definitions of $c_1,c_2,c_3$ we get that $c_1+c_2-c_3=b_5, 2c_3-(c_1+c_2)=b_2, b_3=c_1-c_3$.
The $c_1, c_2, c_3$ are elements of the vector subspace $U$. So, each linear combination will also be in $U$.
Since we can write the $b_2, b_3, b_5$ as a linear combination of the $c_1, c_2, c_3$, we get that $b_2, b_3, b_5\in U$.
So, every linear combination of the $b_2, b_3, b_5$ is also in $U$.
Therefore $\Phi (c_1), \Phi (c_2), \Phi (c_3)\in U$.
And from that we get that every linear combination of them is also in $U$, and so $\Phi (c)=a_1\Phi (c_1)+a_2\Phi (c_2)+\Phi (c_3)\in U$. Is everything correct? Could I improve something? (Wondering)