Calculating Mean Square Error with Pseudo Inverse Approach

nikki92
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Find the mean square error using the pseudo inverse approach.

I am given a 11X9 matrix A, a 11X1 vector F and R = 11X11 diagonal matrix

so Rhat = A[(A'A)^-1 ]A' R . Then I get a 11X11 matrix. Shouldn't I get getting a 8X11 matrix How do I get the most optimum vector F?
 
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nikki92 said:
Find the mean square error using the pseudo inverse approach.

I am given a 11X9 matrix A, a 11X1 vector F and R = 11X11 diagonal matrix

so Rhat = A[(A'A)^-1 ]A' R . Then I get a 11X11 matrix.

Correct.

Shouldn't I get getting a 8X11 matrix?

No.

How do I get the most optimum vector F?

I don't know.
 
I have a linear question A*F =R where R is the diagonal 11 x 11 matrix A is 9X11 and F is 9X1. This system is over determined. I am confused on how to get the values of F .

I get that F =[(A'A)^-1 ]A' R which gives me a 9 x 11 matrix which does not make sense .
 
nikki92 said:
I have a linear question A*F =R where R is the diagonal 11 x 11 matrix A is 9X11 and F is 9X1. This system is over determined. I am confused on how to get the values of F .

I get that F =[(A'A)^-1 ]A' R which gives me a 9 x 11 matrix which does not make sense .
Your equation AF=R is not overdetermined. It doesn't even make sense. There is no such thing as the product of a 9x11 matrix with a 9x1 matrix. If your A was 11x9, then the product AF would be defined, but it's dimensionality would be 11x1, not 11x11.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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