Proving partitioned matrix involving conditional inverse

[cielo]

Homework Statement

For matrix X partitioned as $$\underline{X}$$ = [ $$\underline{X}$$1 $$\underline{X}$$2 ] with $$\underline{X}$$1 matrix full column rank rx,

Homework Equations

prove that $$\underline{X}$$($$\underline{X}$$'$$\underline{X}$$)$$^{C}$$$$\underline{X}$$ = $$\underline{X'}$$$$_{1}$$'($$\underline{X'}$$$$_{1}$$$$\underline{X}$$$$_{1}$$)$$^{-1}$$$$\underline{X'}$$$$_{1}$$

The Attempt at a Solution

I would like to, but I don't have an idea how to solve this problem. I hope somebody can help me here.

 

[mighty2000]

Hello! I can help you with this problem. Let's start by breaking down the equations and understanding what they mean.

First, let's define the different parts of the equation:

- \underline{X} is a matrix that has been partitioned into two parts: \underline{X}1 and \underline{X}2.
- \underline{X}' is the transpose of \underline{X}.
- \underline{X}'\underline{X} is the product of \underline{X}' and \underline{X}.
- (\underline{X}'\underline{X})^{C} is the inverse of (\underline{X}'\underline{X}).
- \underline{X} is the product of \underline{X} and (\underline{X}'\underline{X})^{C}.
- \underline{X}1 is the first part of the partitioned matrix \underline{X}.
- \underline{X'}_{1} is the transpose of \underline{X}1.
- \underline{X'}_{1}\underline{X}_{1} is the product of \underline{X'}_{1} and \underline{X}_{1}.
- (\underline{X'}_{1}\underline{X}_{1})^{-1} is the inverse of (\underline{X'}_{1}\underline{X}_{1}).
- \underline{X'}_{1}'(\underline{X'}_{1}\underline{X}_{1})^{-1}\underline{X'}_{1} is the product of \underline{X'}_{1}', (\underline{X'}_{1}\underline{X}_{1})^{-1}, and \underline{X'}_{1}.

Now, let's look at the equations and see how they are related:

- \underline{X}(\underline{X}'\underline{X})^{C}\underline{X} = \underline{X'}_{1}'(\underline{X'}_{1}\underline{X}_{1})^{-1}\underline{X'}_{1}: This equation is showing that the product of \underline{X} and (\underline{X}'\underline{X})^{C} is equal to the product of \underline{X'}_{1}', (\underline{X'}_{1}\underline{X}_{1})^{-1}, and \underline{X'}_{1}. In other words, the