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  1. MHB Master
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    #1
    Hey!!

    I want to show that for $A,B\in \mathbb{R}^{2\times 2}$ the $U=\{X\in \mathbb{R}^{2\times 2}\mid AX=XB\}$ is a vector subspace of $\mathbb{R}^{2\times 2}$.

    We have that it is non-empty, since the zero matrix belongs to $U$ : $AO=O=OB$.

    Let $X_1, X_2\in U$ then $AX_1=X_1B$ and $AX_2=X_2B$.
    We have that $A(X_1+X_2)=AX_1+AX_2=X_1B+X_2B=(X_1+X_2)B$, and so $X_1+X_2\in U$.

    Let $\lambda \in \mathbb{R}$.
    We have that $A(\lambda X_1)=\lambda (AX_1)=\lambda (X_1B)=(\lambda X_1)B$,and so $\lambda X_1\in U$.

    Is everything correct?


    Then when $A=\begin{pmatrix}a_1 & 0 \\ a_3 & a_4\end{pmatrix}$ and $B=\begin{pmatrix}b_1 & b_2 \\ 0 & b_4\end{pmatrix}$ I want to show that $$U=\{0\} \iff \{a_1, a_4\}\cap \{b_1, b_4\}=\emptyset$$

    We have that $$AX=\begin{pmatrix}a_1 & 0 \\ a_3 & a_4\end{pmatrix}\begin{pmatrix}x_1 & x_2 \\ x_3 & x_4\end{pmatrix}=\begin{pmatrix}a_1x_1 & a_1x_2 \\ a_3x_1+a_4x_3 & a_3x_2+a_4x_4\end{pmatrix}$$
    and
    $$XB=\begin{pmatrix}x_1 & x_2 \\ x_3 & x_4\end{pmatrix}\begin{pmatrix}b_1 & b_2 \\ 0 & b_4\end{pmatrix}=\begin{pmatrix}x_1b_1 & x_1b_2+x_2b_4 \\ x_3b_1 & x_3b_2+x_4b_4\end{pmatrix}$$

    So, $$AX=XB \Rightarrow \begin{pmatrix}a_1x_1 & a_1x_2 \\ a_3x_1+a_4x_3 & a_3x_2+a_4x_4\end{pmatrix}=\begin{pmatrix}x_1b_1 & x_1b_2+x_2b_4 \\ x_3b_1 & x_3b_2+x_4b_4\end{pmatrix} \\ \Rightarrow \left\{\begin{matrix}
    a_1x_1=x_1b_1 \\
    a_1x_2=x_1b_2+x_2b_4 \\
    a_3x_1+a_4x_3=x_3b_1 \\
    a_3x_2+a_4x_4=x_3b_2+x_4b_4
    \end{matrix}\right.\Rightarrow \left\{\begin{matrix}
    (a_1-b_1)x_1=0 \\
    (a_1-b_4)x_2=x_1b_2 \\
    a_3x_1=x_3(b_1-a_4) \\
    a_3x_2=x_3b_2+x_4(b_4-a_4)
    \end{matrix}\right.$$

    For the direction $\Leftarrow$ we have that $\{a_1, a_4\}\cap \{b_1, b_4\}=\emptyset$, so we have the following:
    Since $a_1\neq b_1$ it must be $(a_1-b_1)x_1=0 \Rightarrow x_1=0$.
    From the second equation we have $(a_1-b_4)x_2=x_1b_2\Rightarrow (a_1-b_4)x_2=0$, since $a_1\neq b_4$ it follows that $x_2=0$.
    From the third equation we have that $a_3x_1=x_3(b_1-a_4)\Rightarrow x_3(b_1-a_4)=0$, since $b_1\neq a_4$ it follows that $x_3=0$.
    From the last equation we have $a_3x_2=x_3b_2+x_4(b_4-a_4)\Rightarrow 0=x_4(b_4-a_4)$, since $b_4\neq a_4$ it follows that $x_4=0$.

    Therefore, the matrix $X$ is the zero matrix, and so $U=\{0\}$, right?


    How can we show the other direction?
    When we have the zero matrix, doesn't it hold for every $A$ and $B$ ?
    Last edited by mathmari; December 13th, 2016 at 08:57.

  2. Indicium Physicus
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    #2
    Quote Originally Posted by mathmari View Post
    Hey!!

    I want to show that for $A,B\in \mathbb{R}^{2\times 2}$ the $U=\{X\in \mathbb{R}^{2\times 2}\mid AX=XB\}$ is a vector subspace of $\mathbb{R}^{2\times 2}$.

    We have that it is non-empty, since the zero matrix belongs to $U$ : $AO=O=OB$.

    Let $X_1, X_2\in U$ then $AX_1=X_1B$ and $AX_2=X_2B$.
    We have that $A(X_1+X_2)=AX_1+AX_2=X_1B+X_2B=(X_1+X_2)B$, and so $X_1+X_2\in U$.

    Let $\lambda \in \mathbb{R}$.
    We have that $A(\lambda X_1)=\lambda (AX_1)=\lambda (X_1B)=(\lambda X_1)B$,and so $\lambda X_1\in U$.

    Is everything correct?
    Looks good to me!


    Quote Quote:
    Then when $A=\begin{pmatrix}a_1 & 0 \\ a_3 & a_4\end{pmatrix}$ and $B=\begin{pmatrix}b_1 & b_2 \\ 0 & b_4\end{pmatrix}$ I want to show that $$U=\{0\} \iff \{a_1, a_4\}\cap \{b_1, b_4\}=\emptyset$$

    We have that $$AX=\begin{pmatrix}a_1 & 0 \\ a_3 & a_4\end{pmatrix}\begin{pmatrix}x_1 & x_2 \\ x_3 & x_4\end{pmatrix}=\begin{pmatrix}a_1x_1 & a_1x_2 \\ a_3x_1+a_4x_3 & a_3x_2+a_4x_4\end{pmatrix}$$
    and
    $$XB=\begin{pmatrix}x_1 & x_2 \\ x_3 & x_4\end{pmatrix}\begin{pmatrix}b_1 & b_2 \\ 0 & b_4\end{pmatrix}=\begin{pmatrix}x_1b_1 & x_1b_2+x_2b_4 \\ x_3b_1 & x_3b_2+x_4b_4\end{pmatrix}$$

    So, $$AX=XB \Rightarrow \begin{pmatrix}a_1x_1 & a_1x_2 \\ a_3x_1+a_4x_3 & a_3x_2+a_4x_4\end{pmatrix}=\begin{pmatrix}x_1b_1 & x_1b_2+x_2b_4 \\ x_3b_1 & x_3b_2+x_4b_4\end{pmatrix} \\ \Rightarrow \left\{\begin{matrix}
    a_1x_1=x_1b_1 \\
    a_1x_2=x_1b_2+x_2b_4 \\
    a_3x_1+a_4x_3=x_3b_1 \\
    a_3x_2+a_4x_4=x_3b_2+x_4b_4
    \end{matrix}\right.\Rightarrow \left\{\begin{matrix}
    (a_1-b_1)x_1=0 \\
    (a_1-b_4)x_2=x_1b_2 \\
    a_3x_1=x_3(b_1-a_4) \\
    a_3x_2=x_3b_2+x_4(b_4-a_4)
    \end{matrix}\right.$$

    For the direction $\Leftarrow$ we have that $\{a_1, a_4\}\cap \{b_1, b_4\}=\emptyset$, so we have the following:
    Since $a_1\neq b_1$ it must be $(a_1-b_1)x_1=0 \Rightarrow x_1=0$.
    From the second equation we have $(a_1-b_4)x_2=x_1b_2\Rightarrow (a_1-b_4)x_2=0$, since $a_1\neq b_4$ it follows that $x_2=0$.
    From the third equation we have that $a_3x_1=x_3(b_1-a_4)\Rightarrow x_3(b_1-a_4)=0$, since $b_1\neq a_4$ it follows that $x_3=0$.
    From the last equation we have $a_3x_2=x_3b_2+x_4(b_4-a_4)\Rightarrow 0=x_4(b_4-a_4)$, since $b_4\neq a_4$ it follows that $x_4=0$.

    Therefore, the matrix $X$ is the zero matrix, and so $U=\{0\}$, right?
    Looks great!

    Quote Quote:
    How can we show the other direction?
    When we have the zero matrix, doesn't it hold for every $A$ and $B$ ?
    That's not quite the right way of looking at it, so far as I can make out. The thing is, your $U$ is dependent on your choice of $A$ and $B$. That is, you really have this:
    $$U(A,B)=\left\{X\in \mathbb{R}^{2\times 2}\mid AX=XB\right\}.$$
    Saying that $U=\{0\}$ is really saying that ONLY the zero matrix allows $AX=XB$ for that choice of $A$ and $B$. In other words, for your choice of $A$ and $B$, $AX=XB$ implies $X=0$. Does that help?

  3. MHB Master
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    #3 Thread Author
    Quote Originally Posted by Ackbach View Post
    That's not quite the right way of looking at it, so far as I can make out. The thing is, your $U$ is dependent on your choice of $A$ and $B$. That is, you really have this:
    $$U(A,B)=\left\{X\in \mathbb{R}^{2\times 2}\mid AX=XB\right\}.$$
    Saying that $U=\{0\}$ is really saying that ONLY the zero matrix allows $AX=XB$ for that choice of $A$ and $B$. In other words, for your choice of $A$ and $B$, $AX=XB$ implies $X=0$. Does that help?
    So, do we want to show that $X=0$ is the only solution that allows $AX=XB$ for the given matrices $A$ and $B$ and with the restriction $\{a_1, a_4\}\cap \{b_1, b_2\}=\emptyset$ ?

  4. Indicium Physicus
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    #4
    Quote Originally Posted by mathmari View Post
    So, do we want to show that $X=0$ is the only solution that allows $AX=XB$ for the given matrices $A$ and $B$ and with the restriction $\{a_1, a_4\}\cap \{b_1, b_2\}=\emptyset$ ?
    Actually, I think that's what you've already shown. I think what you want to show now is this:

    Given that $A$ and $B$ are of the form above, and that the set $U=\{X\mid AX=XB\}$ is the singleton set $U=\{0\}$, prove that $\{a_1, a_4\} \cap \{b_1, b_2\} = \emptyset$.

  5. MHB Master
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    #5 Thread Author
    Quote Originally Posted by Ackbach View Post
    Actually, I think that's what you've already shown. I think what you want to show now is this:

    Given that $A$ and $B$ are of the form above, and that the set $U=\{X\mid AX=XB\}$ is the singleton set $U=\{0\}$, prove that $\{a_1, a_4\} \cap \{b_1, b_2\} = \emptyset$.
    So, for the direction $\Rightarrow$ we do the following:

    We have the system \begin{equation*}\left\{\begin{matrix}
    (a_1-b_1)x_1=0 \\
    b_2x_1+(b_4-a_1)x_2=0 \\
    a_3x_1+(a_4-b_1)x_3=0 \\
    a_3x_2+(-b_2)x_3+(a_4-b_4)x_4=0
    \end{matrix}\right.\end{equation*}

    From the first equation we have that if $a_1=b_1$ then it would exist infinitely many $x_1$ such that $(a_1-b_1)x_1=0$. But $x_1$ must have only the value $0$, therefore it must be $a_1\neq b_1$.

    From the second equation we have that, since $x_1=0$, then if $a_1=b_4$ then it would exist infinitely many $x_2$ such that $(b_4-a_1)x_2=0$. But $x_2$ must have only the value $0$, therefore it must be $a_1\neq b_4$.

    From the third equation we have that, since $x_1=x_2=0$, then if $a_4=b_1$ then it would exist infinitely many $x_3$ such that $(a_4-b_1)x_3=0$. But $x_3$ must have only the value $0$, therefore it must be $a_4\neq b_1$.

    From the last equation we have that, since $x_1=x_2=x_3=0$, then if $a_4=b_4$ then it would exist infinitely many $x_4$ such that $(a_4-b_4)x_4=0$. But $x_4$ must have only the value $0$, therefore it must be $a_4\neq b_4$.


    Is everything correct? Could I improve something?

  6. Indicium Physicus
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    #6
    I think you've got it! Yeah, there might be some cutesy, ridiculously short way to do it, but I'm not always so much in favor of those sorts of methods as some. If you can do the straight-forward method, just do that!

  7. MHB Master
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    #7 Thread Author
    Quote Originally Posted by Ackbach View Post
    I think you've got it! Yeah, there might be some cutesy, ridiculously short way to do it, but I'm not always so much in favor of those sorts of methods as some. If you can do the straight-forward method, just do that!
    Ok!! Thank you very much!!

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    #8
    OK, one economy I can think of is that your four paragraphs are all using essentially the same argument. So, after your first paragraph, you could say something like, "Similarly, given that the factors $(b_4-a_1), \; (a_4-b_1), \;$ and $(a_4-b_4)$ all appear in the linear system, the quantities $a_1$ and $b_4$, $a_4$ and $b_1$, and $a_4$ and $b_4$ must be distinct." Whether this shortcut is acceptable to your teacher or not is something you have to gauge.

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