Is the transformation matrix of $\Phi$ in relation to $B$ correct?

[mathmari]

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)

 

[Fernando Revilla]

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)

If we prove that $\det A_{\Phi}\ne 0$ then, $\text{rank }A_{\Phi}=5$. So,
$$\text{rank }A_{\Phi}=5=\dim \text{Im }f =\dim V\Rightarrow \text{Im }f=V\Rightarrow f\text{ is surjective.}$$
On the other hand,
$$\dim \text{Im }f=5\Rightarrow \dim \ker f=\dim V-\dim \text{Im }f=0\Rightarrow f\text{ is injective.}$$

That is, we only need to prove that $\det A_{\Phi}\ne 0.$

 

[mathmari]

If we prove that $\det A_{\Phi}\ne 0$ then, $\text{rank }A_{\Phi}=5$. So,
$$\text{rank }A_{\Phi}=5=\dim \text{Im }f =\dim V\Rightarrow \text{Im }f=V\Rightarrow f\text{ is surjective.}$$
On the other hand,
$$\dim \text{Im }f=5\Rightarrow \dim \ker f=\dim V-\dim \text{Im }f=0\Rightarrow f\text{ is injective.}$$

Why do we have that $\dim \text{Im }f =\dim V$ ? (Wondering)

That is, we only need to prove that $\det A_{\Phi}\ne 0.$

Instead of calculating the determinant of the $5\times 5$-matrix, we could find the rank, right? (Wondering)

 

[HallsofIvy]

Why do we have that $\dim \text{Im }f =\dim V$ ? (Wondering)

Because we were given that dim V= 5 and since the determinant is non-zero, the transformation has 0 kernel meaning that the image has dimension 5.

Instead of calculating the determinant of the $5\times 5$-matrix, we could find the rank, right? (Wondering)

Finding that the determinant is non-zero is the simplest way to find the rank!

 

[mathmari]

Because we were given that dim V= 5 and since the determinant is non-zero, the transformation has 0 kernel meaning that the image has dimension 5.

Do we have that dim V= 5 because the basis B contains 5 elements? (Wondering)

 

[Fernando Revilla]

Do we have that dim V= 5 because the basis B contains 5 elements? (Wondering)

Right. :)

 

[mathmari]

Thanks a lot! (Happy)

 

[mathmari]

To calculate the transformation matrix of $\Phi\circ\Phi$ in relation to $B$, we have to calculate the images $\Phi^2(b_i), b_i\in B$, right? (Wondering)