- #1
mathmari
Gold Member
MHB
- 5,049
- 7
Hey!
We have that a matrix $A$ is idempotent if it holds that $A^2=A$.
We suppose that $X$ is a $m\times n$-matrix and that $(X^TX)^{-1}$ exists.
I want to show that $A=I_m-X(X^TX)^{-1}X^T$ is idempotent. I have done the following:
$$A^2 =A\cdot A=(I_m-X(X^TX)^{-1}X^T)\cdot (I_m-X(X^TX)^{-1}X^T) \\ =I_m(I_m-X(X^TX)^{-1}X^T)-X(X^TX)^{-1}X^T(I_m-X(X^TX)^{-1}X^T) \\ =I_mI_m-I_mX(X^TX)^{-1}X^T-X(X^TX)^{-1}X^TI_m+(X(X^TX)^{-1}X^T)(X(X^TX)^{-1}X^T) \\ =I_m-X(X^TX)^{-1}X^T-X(X^TX)^{-1}X^T+X(X^TX)^{-1}(X^TX)(X^TX)^{-1}X^T \\ =I_m-2X(X^TX)^{-1}X^T+XI_n(X^TX)^{-1}X^T \\ =I_m-2X(X^TX)^{-1}X^T+X(X^TX)^{-1}X^T \\ =I_m-X(X^TX)^{-1}X^T =A$$
Is this correct? (Wondering)
Is the step from the $3$th to the $4$th line correct? Or doesn't it hold that $(X(X^TX)^{-1}X^T)(X(X^TX)^{-1}X^T)=X(X^TX)^{-1}(X^TX)(X^TX)^{-1}X^T$ ? (Wondering)
We have that a matrix $A$ is idempotent if it holds that $A^2=A$.
We suppose that $X$ is a $m\times n$-matrix and that $(X^TX)^{-1}$ exists.
I want to show that $A=I_m-X(X^TX)^{-1}X^T$ is idempotent. I have done the following:
$$A^2 =A\cdot A=(I_m-X(X^TX)^{-1}X^T)\cdot (I_m-X(X^TX)^{-1}X^T) \\ =I_m(I_m-X(X^TX)^{-1}X^T)-X(X^TX)^{-1}X^T(I_m-X(X^TX)^{-1}X^T) \\ =I_mI_m-I_mX(X^TX)^{-1}X^T-X(X^TX)^{-1}X^TI_m+(X(X^TX)^{-1}X^T)(X(X^TX)^{-1}X^T) \\ =I_m-X(X^TX)^{-1}X^T-X(X^TX)^{-1}X^T+X(X^TX)^{-1}(X^TX)(X^TX)^{-1}X^T \\ =I_m-2X(X^TX)^{-1}X^T+XI_n(X^TX)^{-1}X^T \\ =I_m-2X(X^TX)^{-1}X^T+X(X^TX)^{-1}X^T \\ =I_m-X(X^TX)^{-1}X^T =A$$
Is this correct? (Wondering)
Is the step from the $3$th to the $4$th line correct? Or doesn't it hold that $(X(X^TX)^{-1}X^T)(X(X^TX)^{-1}X^T)=X(X^TX)^{-1}(X^TX)(X^TX)^{-1}X^T$ ? (Wondering)