1
$\begingroup$

Let $B=\{b_1, \ldots , b_5\}$ be a basis of real vector space $V$ and $\Phi$ an endomorphism 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 transformation matrix of $\Phi$ and $\Phi\circ\Phi$ in relation to $B$ and I have shown that $\Phi$ is not bijective.

I want to show also the following:

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 $\Phi(U)\subset U$.

$$$$

First of all we have to show that the $c_1, c_2, c_3$ are linearly independent, or not?

What else do we have to show?

  • 1
    If you show that $C$ is linearly independent, it will autumatically be a basis for a subspace of $V$ with dimension $3$. Then only thing you need to check is that $U$ is invariant under $\Phi$, i.e. $\Phi(U)\subset U$ – 2017-01-25
  • 0
    We have that $Ac_1+Bc_2+Cc_3=0 \Rightarrow (A+B)b_2+(A-B)b_3+(A+B+C)b_5=0$ and we conclude that these coefficients are zero, $A+B=A-B=A+B+C=0$, since the the $b_i$'s are elements of the basis $B$, right? @user160738 – 2017-01-25
  • 1
    Not quite, $Ac_1+Bc_2+Cc_3=(A+C)b_2+(A-B)b_3+(A+B+C)b_5$, so$A+C=A-B=A+B+C=0$. But the result is the same; $A=B=C=0$ so $C$ is linearly independent. – 2017-01-25
  • 0
    Ah ok!! Let $c\in U$ then $c=a_1c_1+c_2c_2+a_3c_3$, so $$\Phi (c)=a_1\Phi (c_1)+a_2\Phi (c_2)+a_3\Phi (c_3)\\ =a_1\Phi (b_2+b_3+b_5)+a_2\Phi (-b_3+b_5)+a_3\Phi (b_2+b_5) \\ =a_1(-2b_3 +b_5-4b_2+2b_3 -b_5+3b_2 +2b_5)+a_2(4b_2-2b_3 +b_5+3b_2 +2b_5)+a_3(-2b_3 +b_5+3b_2 +2b_5)\\ = a_1( -b_2 +2b_5)+a_2(7b_2-2b_3 +7b_5)+a_3(-2b_3 +3b_2 +3b_5)$$ Do we have to show that we can write this as a linear combination of the $c_1,c_2,c_3$ to show that $\Phi (U)\subset U$ ? @user160738 – 2017-01-25
  • 1
    Yes, you'd have to, but in general there is much quicker way. For example if you can show that each $\Phi(c_i)$ is in $U$, then since $U$ is a subspace $a_i \Phi(c_i)$ is also in $U$, so their linear combination $\Phi(c)$ would also be in $U$. So showing that $\Phi(c_i)$ is in $U$ individually for each $i$ is much simpler. For example, $\Phi(c_1)=\Phi(b_2+b_3+b_5)=(-2b_3+b_5)+(-4b_2+2b_3-b_5)+(3b_2+2b_5)=-b_2+2b_5$. Also, here's a little trick that can greatly simplify your work. – 2017-01-25
  • 1
    Notice that $c_1+c_2-c_3=b_5$, $2c_3-(c_1+c_2)=b_2, b_3=c_1-c_3$ so in fact $b_2,b_3,b_5$ are all in $U$ hence any linear combination of them must be in $U$, and the result follows. – 2017-01-25
  • 0
    I see!! Thank you!! :-) @user160738 – 2017-01-26

1 Answers 1

1

You want to prove the claim that $\{c_1,c_2,c_3\}$ is a basis. To say this set is a basis means the vectors are linearly independent, so yes it is necessary to show this.

Next, to show that $\Phi(U)\subset U$, you need to take any vector $c=\alpha_1c_1+\alpha_2c_2+\alpha_3c_3$ in $U$ (here $\alpha_1,\alpha_2,\alpha_3\in\mathbb R$) and show that $T(c)\in U$. As a hint, you can show that $T(c_1),T(c_2),T(c_3)$ are all in $U$, as this would show that $T(U)\subset U$ (ask yourself why that is).

  • 0
    We have that $Ac_1+Bc_2+Cc_3=0 \Rightarrow (A+B)b_2+(A-B)b_3+(A+B+C)b_5=0$ and we conclude that these coefficients are zero, $A+B=A-B=A+B+C=0$, since the the $b_i$'s are elements of the basis $B$, right? – 2017-01-25
  • 1
    @MaryStar Yes. Now from here you have to show that $A,B,C$ are all zero. As a side note, you should use different representations for the coefficients as the characters $B$ and $C$ are already used. – 2017-01-25
  • 0
    Ah ok. For the second part: Let $c\in U$ then $c=a_1c_1+c_2c_2+a_3c_3$, so $$\Phi (c)=a_1\Phi (c_1)+a_2\Phi (c_2)+a_3\Phi (c_3)\\ =a_1\Phi (b_2+b_3+b_5)+a_2\Phi (-b_3+b_5)+a_3\Phi (b_2+b_5) \\ =a_1(-2b_3 +b_5-4b_2+2b_3 -b_5+3b_2 +2b_5)+a_2(4b_2-2b_3 +b_5+3b_2 +2b_5)+a_3(-2b_3 +b_5+3b_2 +2b_5)\\ = a_1( -b_2 +2b_5)+a_2(7b_2-2b_3 +7b_5)+a_3(-2b_3 +3b_2 +3b_5)$$ Do we have to show that we can write this as a linear combination of the $c_1, c_2, c_3$ ? – 2017-01-25
  • 1
    @MaryStar exactly. If we write $c=\alpha_1c_1+\alpha_2c_2+\alpha_3c_3$, then show that $\Phi(c)=\beta_1c_1+\beta_2c_2+\beta_3c_3$ for some scalars $\beta_1,\beta_2,\beta_3$. – 2017-01-25