UnkleRhaukus
  • UnkleRhaukus
By explicit construction of the matrices in question, show that any matrix T can be written as the sum of matrices: a) symmetric matrix S and antisymmetric matrix A b) real matrix R and imaginary matrix M c) hermitian matrix H and skew-hermitian matrix K
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
JamesJ
  • JamesJ
are you stuck on this, really? Which bit? part b is trivial yes?
UnkleRhaukus
  • UnkleRhaukus
i have the answer i just dont find it particularly illuminating
JamesJ
  • JamesJ
Hmm ... for part a it's A = (1/2) (A + A^t) + (1/2) (A - A^t) I think that's quite intuitive.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

JamesJ
  • JamesJ
In fact, all three parts have the 'same' answer if you replace transpose for conjugate for complex transpose
JamesJ
  • JamesJ
It's analogous also to writing a real function as the sum of an even and an odd function.
UnkleRhaukus
  • UnkleRhaukus
the answer for a) i have is \[S=\frac{1}{2}(T+\tilde{T}); \quad A=\frac{1}{2}(T-\tilde{T})\]
JamesJ
  • JamesJ
right, where ~ means transpose
UnkleRhaukus
  • UnkleRhaukus
yeah
UnkleRhaukus
  • UnkleRhaukus
ok well is there a step before this that i can not see for some reason
JamesJ
  • JamesJ
Um, I arrived at it by remembering that A + A^t is always symmetric. Then I just had to scale the thing appropriately and find it's complement. Then I remembered the pattern works everywhere
JamesJ
  • JamesJ
*its
UnkleRhaukus
  • UnkleRhaukus
So i guess my question should be why is \[A +\tilde {A} \]aka\[A+A^T\] necessarily symmetric
JamesJ
  • JamesJ
Because (A^t)^t = A, hence .... see it now?
UnkleRhaukus
  • UnkleRhaukus
\[A+A^T=(A^T+A)^T\]\[A^T+A = (A^T+A)^T\]
UnkleRhaukus
  • UnkleRhaukus
the bottom right corner i mean to have\[(A+A^T)^T\]
UnkleRhaukus
  • UnkleRhaukus
and the point is transposes do noting to the sum.
UnkleRhaukus
  • UnkleRhaukus
have i got it?
JamesJ
  • JamesJ
Yes. A little more abstractly, the operation of taking the transpose is order 2. This is why the same method will also work with the other operations.
JamesJ
  • JamesJ
Remember when we were kids in primary school we'd make painting of butterflies
JamesJ
  • JamesJ
We'd put a vertical fold in the paper and put some paint on both sides. Then we'd fold the two sides together and open it up again. The analogy isn't perfect, but that's sort of what's going on here.
JamesJ
  • JamesJ
On that colourful note, I'm out of here.
UnkleRhaukus
  • UnkleRhaukus
thanks i think it makes more sense to me now
UnkleRhaukus
  • UnkleRhaukus
this is a bit like odd and even functions isnt it

Looking for something else?

Not the answer you are looking for? Search for more explanations.