Is Every Skew Symmetric Matrix Invertible When Added to the Identity Matrix?

[weasel3000]

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

 

[micromass]

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.

 

[weasel3000]

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.

 

[micromass]

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

 

[weasel3000]

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

 

[micromass]

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...

 

[weasel3000]

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

 

[micromass]

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.