Prove that P and (1/2)(I + P) have the same fixed vectors.
First we will show that \(\frac{1}{2}(I+P)x=x\) implies \(Px=x\).
Expanding the formula: \(\frac{1}{2}(I+P)x \rightarrow \frac{1}{2}Ix + \frac{1}{2}Px =x\)
As per the definition of the identity matrix, \(Ix=x\), so we can simplify to: \[ \frac{1}{2}x +\frac{1}{2} Px = x\\ \frac{1}{2} x = \frac{1}{2}Px\\ \text{Multiplying both sides by } 2:\\ x=Px \]
Therefore, if \(x\) is a fixed vector of \(\frac{1}{2}(I+P)\), it is also a fixed vector of P.
Next, we will show that that \(Px=x\) implies \(\frac{1}{2}(I+P)x=x\).
We will substitute \(Px=x\) into the equation for \(\frac{1}{2}(I+P)x\): \[ \frac{1}{2}(I+P)x = \frac{1}{2}(Ix+Px)\\ \text{Since } Ix=x:\\ \frac{1}{2}(x+x) = \frac{1}{2}(2x) = x \]
So if \(x\) is a fixed vector of P, then it is also a fixed vector of \(\frac{1}{2}(I+P)\).
In conclusion, we’ve proven that if \(x\) is a fixed vector of P, then it is also a fixed vector of \(\frac{1}{2}(I+P)\). We’ve also proven that if \(x\) is a fixed vector \(\frac{1}{2}(I+P)\), then it is also a fixed vector of P. So we can conclude that P and \(\frac{1}{2}(I+P)\) have the same fixed vectors.