Standard Matrix of Linear Transformation T: R2--R2

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The question I am attempting to understand (my book absolutely sux) is this:

Give the standard matrix of the linear transformation T: R2--R2 which switches the two coordinates of every point?

Just not understanding what i am trying to accomplish here
 
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Sorry about the notation.

Find a,b,c,d such that


| a b |
| c d | < x, y> = <y, x>
 
Let us think about what matrix multiplication encodes. It is linear, so to understand what it does you just need to understand what it does to a basis (i.e. A(x+y)=Ax + Ay, so knowing what happens to x and y tells you what happens to x+y).

What happens if I multiply a matrix A into the column vector (1,0)^t? I get out the first column of A (check this!). Now find out what you get from multiplying into (0,1)^t

So if I know that A sends (1,0)^t to c and (0,1)^t to d, then I can write down A's matrix immediately.
 
Here is the simplest way to find the matrix of a linear tranformation, in a given basis.

Apply the transformation to the first basis vector: write the result in terms of the basis. The coefficients are the first column in the matrix.

Now do the same to the second basis vector. That gives the second column.

I notice, now, that this is exactly what matt grime said!

That is, write the matrix, as you do:
\begin{bmatrix}a &amp; b \\ c &amp; d\end{bmatrix}
and apply it to [1 0].
\begin{bmatrix}a &amp; b \\ c &amp; d\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ c\end{bmatrix}
But since this transformation "swaps the two coordinates", [1 0] is taken to [0 1] so you must have [a c]= [0 1].
 
ok so my answer should be a 2x2 matrix?

I am to someway use the standard matrix
|1 0 |
|0 1 |
to get to my answer?

It seems that i should use the standard matrix muliplied by |0 1| to get a c, is this correct?
And then i could multiply standard matrix by |1 0| to get b d?
 
Of course the matrix of the transformation T is a 2 x 2 matrix. It maps a vector in R2 to a different vector in R2.

Look at HallsOfIvy's post; he has done half the problem for you.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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