# Invertibility Proof

• Nov 25th 2009, 07:04 PM
joe909
Invertibility Proof
The question is:

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

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 :)
• Nov 26th 2009, 01:11 AM
Opalg
Quote:

Originally Posted by joe909
The question is:

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

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

Now suppose that $(I_n+A)x=0$. Then $0 = \langle (I_n+A)x,x\rangle = \langle x,x\rangle + \langle Ax,x\rangle = \langle x,x\rangle$, and hence x=0. Thus $I_n+A$ is injective and therefore invertible.
• Nov 26th 2009, 07:56 AM
joe909
Quote:

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

Now suppose that $(I_n+A)x=0$. Then $0 = \langle (I_n+A)x,x\rangle = \langle x,x\rangle + \langle Ax,x\rangle = \langle x,x\rangle$, and hence x=0. Thus $I_n+A$ 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.
• Nov 26th 2009, 09:48 AM
Opalg
Quote:

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 $Ax\cdot y = x\cdot A^{\textsc t}y$.