Show this is a projection on a vector space

JaysFan31

Homework Statement


Let V=Mn(F) be the space of all nxn matrices over F; define TA=(1/2)(A+transpose(A)) for A in V.
Verify that T is not only a linear operator on V, but is also a projection.


Homework Equations


A is a projection when A squared=A.


The Attempt at a Solution


I don't see how this works since clearly (1/2)(A+transpose(A)) squared does not equal (1/2)(A+transpose(A)) for all matrices.

What am I doing wrong?
 
Physics news on Phys.org
A is a projection when A2 = A, not when (Ax)2 = Ax. So you don't need to look at whether

\left [\frac{1}{2}(A + A^t)\right ]^2 = \frac{1}{2}(A + A^t)

You need to look at whether T2 = T, i.e. whether T(TA) = TA for all A, i.e. whether:

\frac{1}{2}\left [\left (\frac{1}{2}(A + A^t)\right ) + \left (\frac{1}{2}(A + A^t)\right )^t\right ] = \frac{1}{2}(A + A^t)

Remember, you're used to writing A for your linear operators, and vectors in your vector space V are normally written as x or v or something. But now you have matrices AS THE VECTORS IN YOUR VECTOR SPACE, so you'll probably use A to stand for a vector, and now T is the operator.

And you still need to check linearity.
 
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...
Back
Top