A is a skew symmetric nxn matrix with entries in R. AT=-A.
1). Prove uTAu=0 for every u in R. (I made it)
2). Prove In + A is a invertible matrix.
I have no idea how to prove the 2nd.
Thanks

Hmm, I guess one of your classmates also asked this question recently, cause I can remember it.

Anyway, I'll give a hint. For which vectors u, does (I+A)u=0. Use (1) for this.

Hmm, I guess one of your classmates also asked this question recently, cause I can remember it.

Anyway, I'll give a hint. For which vectors u, does (I+A)u=0. Use (1) for this.

??? not get it. 0*u=0; (I+A)*0=0; non-zero vector *non-zero vector=0 also possible.

You're right, if u=0, then (I+A)u=0. Are there other possibilities for u?

thank you for your hint. However I am so dumb ,and unable to get it. R u gonna get det(I+A) not 0 or find (I+A)^-1

No, we are trying to establish that the kernel of I+A is trivial. This implies that I+A is an injective operator. And by some theorem of linear algebra, this implies that I+A is invertible...

No, we are trying to establish that the kernel of I+A is trivial. This implies that I+A is an injective operator. And by some theorem of linear algebra, this implies that I+A is invertible...

oh my god. i am gonna find a place to cry. i just know the null space. never learn what kernel and injective operators are. Thank you again. I am gonna learn that damn THINGS tonight.

Don't worry The kernel is exactly thesame as the null space. And injective (or one-to-one or monomorphism) just means that the null space is trivial.