1. ## Invertibility Proof

The question is:

Let
A be a skew–symmetric n×n–matrix with entries in $\displaystyle R$, i.e. $\displaystyle A^T=-A$

Prove that In+A is an invertible matrix.

I have tried solving this both algebraically and non-algebraically. Algebraically I have started with the fact In+A=In-A, played around starting with there, including multiplying both sides by a nxn invertible matrix B, however never really get anywhere there.

Other then that, things that I have noticed that probably are of importance is the fact that the diagonals of the matrix In+A will be all ones, however im not sure if i need to use that and if so how how.

Any help would be greatly appreciated

2. Originally Posted by joe909
The question is:

Let
A be a skew–symmetric n×n–matrix with entries in $\displaystyle R$, i.e. $\displaystyle A^T=-A$

Prove that In+A is an invertible matrix.

I have tried solving this both algebraically and non-algebraically. Algebraically I have started with the fact In+A=In-A, played around starting with there, including multiplying both sides by a nxn invertible matrix B, however never really get anywhere there.

Other then that, things that I have noticed that probably are of importance is the fact that the diagonals of the matrix In+A will be all ones, however im not sure if i need to use that and if so how how.

Any help would be greatly appreciated
If $\displaystyle A^{\textsc t} = -A$ then, for $\displaystyle x,y\in\mathbb{R}^n$, $\displaystyle \langle Ax,y\rangle = \langle x,A^{\textsc t}y\rangle = -\langle x, Ay\rangle = -\langle Ay,x\rangle$. In particular, if x=y it follows that $\displaystyle \langle Ax,x\rangle = 0$.

Now suppose that $\displaystyle (I_n+A)x=0$. Then $\displaystyle 0 = \langle (I_n+A)x,x\rangle = \langle x,x\rangle + \langle Ax,x\rangle = \langle x,x\rangle$, and hence x=0. Thus $\displaystyle I_n+A$ is injective and therefore invertible.

3. Originally Posted by Opalg
If $\displaystyle A^{\textsc t} = -A$ then, for $\displaystyle x,y\in\mathbb{R}^n$, $\displaystyle \langle Ax,y\rangle = \langle x,A^{\textsc t}y\rangle = -\langle x, Ay\rangle = -\langle Ay,x\rangle$. In particular, if x=y it follows that $\displaystyle \langle Ax,x\rangle = 0$.

Now suppose that $\displaystyle (I_n+A)x=0$. Then $\displaystyle 0 = \langle (I_n+A)x,x\rangle = \langle x,x\rangle + \langle Ax,x\rangle = \langle x,x\rangle$, and hence x=0. Thus $\displaystyle I_n+A$ is injective and therefore invertible.
I would like to start by thanking you for time and help.

Im slightly confused by the explanation though, mainly because I dont think I have been taught the material that used for the proof. What exactly does the notation <x,y> represent, I have never seen that before? Thanks again.

4. Originally Posted by joe909
What exactly does the notation <x,y> represent, I have never seen that before?
It is the inner product, or dot product. Maybe you are used to the alternative notation x.y ? In the dot notation, the key thing about the adjoint is the fact that $\displaystyle Ax\cdot y = x\cdot A^{\textsc t}y$.