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What 2 × 2 matrix M satisfies M ∗ [x, y]=[y, x] for all vectors [x, y]?

Could someone explain to me, step by step, what I should do here? I have an exam but I am stuck with this one. I am a newbie, but I like it to be honest, although the teachers explanation are way too complex for me at the moment.

I would appreciate a step by step approach, because that's what I am missing at the moment.

4 Answers 4

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You're looking for a $2 \times 2$ matrix so that

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} y \\ x \end{pmatrix}$$

You could start by seeing what happens when this matrix acts on the basis vectors of $\mathbb{R}^2$, $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$.

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ c \end{pmatrix}$$

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} b \\ d \end{pmatrix}$$

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    Oeh, I think I am close now... Thanks for the clear explanation, yet I don't understand the last one. Okay, we have a,c and b,d now, but then what? Why not use x,y instead?2017-02-19
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    The vector $\begin{pmatrix} x \\ y \end{pmatrix}$ can be expressed in terms of $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$: $\begin{pmatrix} x \\ y \end{pmatrix} = x \begin{pmatrix} 1 \\ 0 \end{pmatrix} + y \begin{pmatrix} 0 \\ 1 \end{pmatrix}$2017-02-19
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    Yes, I got that, but what are a,c and b,d then? How does that answer the question, because the question tells me to get = x,y...2017-02-19
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    $a,b,c,d$ are the entries in the matrix. You're trying to find their values so that the transformation works for ANY vector.2017-02-19
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    So basically, you need to prove the columns AC and BD and get their outcome based on x = 1, x = 0 and y = 1 and y = 0, correct?2017-02-19
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    Basically, yes.2017-02-19
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The assignment is probably $M \times \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} y\\x\end{pmatrix}$. (Note, that vectors must be column, otherwise the multiplivation does not make sense.)

A general $2\times 2$ matrix matrix is of the form $\begin{pmatrix} a&b\\c&d\end{pmatrix}$, so the LHS of the equation is $M \times \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} a&b\\c&d\end{pmatrix} \times \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} ax+by\\cx+dy\end{pmatrix} $ and comparing this to the RHS you get system of 2 equations and 4 parameters $$ax + by = y \, \&\, cx+ dy = x.$$

Now, the equation has to work for column vectors $\begin{pmatrix} 0\\1\end{pmatrix}$ and $\begin{pmatrix} 1\\0\end{pmatrix}$, which gives the result $a=d=0$ and $b=c=1$.

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    Love the example. Thanks man, appreciated and upvoted!2017-02-19
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$$M = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$ This is because for any $x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \in \mathbb{R}^{2}$ we have that $$Mx = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} x_2 \\ x_1 \end{pmatrix}$$

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    This does not seem like "step-by-step" to me..2017-02-19
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    It is not step by step indeed, but I think I get it. I still don't understand why we use 0,1 and 1,0 in M, though. The fact that we are putting numbers that we didn't get in the question, confuses me.2017-02-19
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For any ($2 \times 2$) matrix $M$ you can see from the definition of matrix multiplication that $M \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ equals the first column of $M$. Also, $M \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ equals the second column of $M$. But the requirements say what these should equal, namely the vector with interchanged coordinates. So we get

$$ M = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

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    Thanks, Henno. You said that "you can see from the definition that M (1,0), yet I can't see that haha. How do I know that? I am missing this piece of information.2017-02-19
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    Write out the definition equations for $M$ times a vector $\begin{pmatrix} x \\ y \end{pmatrix}$. What happens if $x=1$ and $y=0$?2017-02-19