Linear & Abstract algebra

Matrix $M$ which has the vectors $\xi_s$, $\xi_v$ as the columns is the matrix of the transformation which maps $(1,0)\mapsto\xi_s$ and $(0,1)\mapsto \xi_v$).

So your matrix is the inverse $M^{-1}$.

If needed, for $2\times2$ matrix there is a simple expression of the inverse using the entries of the matrix.

BAYMAX

6:34 AM

yes!

@MartinSleziak this is cool

NoLand'sMan

BAYMAX

I was doing usng angle slope and rotation matrix

would not that work?

Martin Sleziak

Rotation matrix does not change angles.

By "angle slope" you mean Shear mapping?

BAYMAX

Actually

we have two vectors $\xi_{s},\xi_{v}$

and they are $2 \times 1$ colu,mn vectors

so what I did is I calculated the slope of each of the vectors by dividing second entry by first entry

so I have two slopes

$m_{1}$ slope of $\xi_{s}$

$m_{2}$ slope of $\xi_{v}$

Tehn I know How much angle I have to rotate the vector in order to match with the $x$ axis for $\xi_{s}$ which is $\theta_{1}$ where $\tan(\theta_{1}) = m_{1}$

similalry I have to rotate the vector $v_{u}$ for an angle $\theta_{2}$ where $\tan(\theta_{2}) = m_{2}$

Is this approach good/ of any help t this problem?

Martin Sleziak

Of course, if you compose the two rotations that will "break" the other vector.

BAYMAX

6:42 AM

other vector?

Martin Sleziak

After the first rotation you moved $\xi_s$ into the right place. If you then apply the other rotation, it moves $\xi_s$ to a wrong place.

I am not sure whether you meant to compose the two rotations...?

BAYMAX

no I just applied rotation to two different vectors

Martin Sleziak

In any case, it should be obvious that you cannot do this only with rotations unless $\xi_s$ and $\xi_v$ are perpendicular and of unit length. Rotations do not change angles and lengths.

@BAYMAX From the picture you linked it seems you want to apply the same transformation on both vectors.

BAYMAX

oh ok

yes, therefore the idea I had will not owrk but yours was really cool one!

Martin Sleziak

BAYMAX

6:45 AM

thanks!

yup, so I cannot apply two different rotations

I have to ocnsider the system

@MartinSleziak how it did that?

Martin Sleziak

14 mins ago, by Martin Sleziak

Matrix $M$ which has the vectors $\xi_s$, $\xi_v$ as the columns is the matrix of the transformation which maps $(1,0)\mapsto\xi_s$ and $(0,1)\mapsto \xi_v$).

As far as I can tell from your diagram, you are using column vectors.

BAYMAX

$\begin{bmatrix} 1\\ 0\\ \end{bmatrix} \begin{bmatrix} \xi_{s} \xi_{v} \end{bmatrix} = \begin{bmatrix} (u-u^*)\\ (v-v^*)\\ \end{bmatrix}$

yes they are column vectors

Martin Sleziak

I.e., you are representing linear maps via matrices in this way $$\begin{pmatrix}x\\y\end{pmatrix} \mapsto M \begin{pmatrix}x\\y\end{pmatrix}.$$

BAYMAX

yes

Martin Sleziak

@BAYMAX You should multiply in such way that matrix is on the left, vector is on the right (for column vectors).

$\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{pmatrix}
\begin{pmatrix}1\\0\end{pmatrix}=
\begin{pmatrix}a_{11}\\a_{12}\end{pmatrix}$ and $\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{pmatrix}
\begin{pmatrix}0\\1\end{pmatrix}=
\begin{pmatrix}a_{21}\\a_{22}\end{pmatrix}$

BAYMAX

6:52 AM

yes

but now we did not get the first column as output right?

Martin Sleziak

Multiplying by $\begin{pmatrix}1\\0\end{pmatrix}$ returns the first column, multiplying $by \begin{pmatrix}0\\1\end{pmatrix}$ the second one.

BAYMAX

we got $\begin{bmatrix} a_{11}\\ a_{12}\\ \end{bmatrix}$

which is not the first column

of the matrix $M$

Martin Sleziak

Sorry, I typed it wrong.

BAYMAX

or we did not get $\xi_{s}$

Martin Sleziak

$\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{pmatrix}
\begin{pmatrix}1\\0\end{pmatrix}=
\begin{pmatrix}a_{11}\\a_{21}\end{pmatrix}$ and $\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{pmatrix}
\begin{pmatrix}0\\1\end{pmatrix}=
\begin{pmatrix}a_{12}\\a_{22}\end{pmatrix}$

BAYMAX

6:54 AM

oh ok

seems i did that too

gotchanow

Martin Sleziak

Of course, since you want $\xi_s\mapsto(1,0)$ and $\xi_v\mapsto(0,1)$, you need the inverse $M^{-1}$.

BAYMAX

cool got to my mind

yup exactly

the inverse

Martin Sleziak