$A,B$ are orthogonal projections and $\|Ax\|^2+\|Bx\|^2=\|x\|^2$ show $A+B=I$

Question

Here is the problem: $A,B:\mathbb{C}^n\to\mathbb{C}^n$ are two orthogonal projections satisfying for any $x\in\mathbb{C}^n$, $$\|Ax\|^2+\|Bx\|^2=\|x\|^2$$ Show that $A+B=I$.

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I know that $\|Ax\|^2+\|Bx\|^2=\|x\|^2$ tells that $(Ax,Ax)+(Bx,Bx)=(x,x)$.

Since $$\|(A+B)x\|^2=((A+B)x,(A+B)x)$$$$=(Ax,Ax)+(Bx,Bx)+(Ax,Bx)+(Bx,Ax)$$$$=(x,x)+(Ax,Bx)+(Bx,Ax)$$

It remian to show that $(Ax,Bx)+(Bx,Ax)=0$, but I am not sure how to show it.

Please help, thanks a lot!

Answer 1

Let me state two hints and one proposal:

• One has $\,\|Ax\|^2=(Ax,x)\,$ by hypothesis.

• What does $\,(Tx,x)=0\;\forall x\in\mathbb C^n\,$ imply for $T$?
  (The conclusion would not hold when working in $\mathbb R^n$!)

• Please use the command "\|" to produce nice(-r) norm delimiters, cf $\,\|\,$ versus $\,||$.