T invariant subspace (intro lin alg class undergrad)

david118
Messages
2
Reaction score
0

Homework Statement


V=Matrix (2x2), T(A) = (0 1 ) A , and W = {A\epsilon V: A^{}t = A
(1 0)


Homework Equations


So T(A) transformation, multiplies a 2x2 matix with entries 0 1 1 0 by A with A on the right side


The Attempt at a Solution



I said let A be any arbitrary symmetric matrix, for example a 2x2 matrix with entries
a b b a

(a b)
(b a), this that matrix multiplied on the right of 0 1 1 0, = (b a)
(a b) , also a symmetric matrix, and therefore this matrix is also an element of W, this W is a T-invariant subspace.

but the back of the book does not say it is T-invariant

please point out if i am making a mistake
test tomorrow

thanks!
 
Physics news on Phys.org
Remember, a symmetric matrix A satisfies A=A^T. That places no restriction on the main diagonal entries. An arbitrary symmetric 2x2 should be of the form

\begin{bmatrix} x & y \\ y & z \end{bmatrix}
 
Last edited:
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