Find the unique symmetric matrix A such that Y'AY=Y'GY

[TeenieBopper]

I asked this question here, however the title of the thread (and the thread itself) was sloppy and unclear.I could not find a way to delete or edit.

This is for a regression analysis course, and I've only taken one introductory course on linear algebra, so when I Google'd "finding a symmetric matrix" a lot of stuff that I didn't understand came up.

Homework Statement

Find the unique symmetric matrix A such that Y'AY=Y'GY

Homework Equations

The Attempt at a Solution

Y'AY = Y'GY
Y'AYY'=Y'GYY'
Y'AYY'(YY')^-1 = Y'GYY'(YY')^-1
Y'A=Y'G
YY'A=YY'G
(YY')^-1YY'A=YY')^-1YY'G
A=G

However, G is not symmetric:
G=
[1.0 1 1.0
-1.0 0 1.0
0.5 2 -0.5]

Other than the algebraic manipulation above, I don't know how else to approach this problem.

 

[Ray Vickson]

I asked this question here, however the title of the thread (and the thread itself) was sloppy and unclear.I could not find a way to delete or edit.

This is for a regression analysis course, and I've only taken one introductory course on linear algebra, so when I Google'd "finding a symmetric matrix" a lot of stuff that I didn't understand came up.

Homework Statement

Find the unique symmetric matrix A such that Y'AY=Y'GY

Homework Equations

The Attempt at a Solution

Y'AY = Y'GY
Y'AYY'=Y'GYY'
Y'AYY'(YY')^-1 = Y'GYY'(YY')^-1
Y'A=Y'G
YY'A=YY'G
(YY')^-1YY'A=YY')^-1YY'G
A=G

However, G is not symmetric:
G=
[1.0 1 1.0
-1.0 0 1.0
0.5 2 -0.5]

Other than the algebraic manipulation above, I don't know how else to approach this problem.

First: as regards your previous post on this matter, your friend is quite correct: the unique matrix ##G## that gives
[tex] GY = (Y_1 +Y_2 + Y_3, Y_3-Y_1, (1/2)Y_1 - (1/2)Y_3 +2Y_2)^T [/tex]
for ALL ##Y_1,Y_2,Y_3## is
[tex] G = \pmatrix{1&1&1 \\-1&0&1\\1/2&2&-1/2} [/tex]
You just read this off directly; no work is necessary. The given numerical values of the ##Y_i## are not relevant: they just get in the way. As far as I can see, they were given to you just to confuse you and lead you astray.

Now you want a symmetric matrix that gives the same quadratic form ##Q(Y) \equiv Y^T G Y##. To clarify, but in a much smaller example with two variables instead of three: you want to know what are the ##a_{ij}## that give
[tex] g_{11} Y_1^2 + g_{12} Y_1 Y_2 + g_{21} Y_2 Y_1 + g_{22} Y_2^2
= a_{11} Y_1^2 + 2 a_{12} Y_1 Y_2 + a_{22} Y_2^2[/tex]
for all ##Y_1,Y_2##.

 

[TeenieBopper]

First: as regards your previous post on this matter, your friend is quite correct: the unique matrix ##G## that gives
[tex] GY = (Y_1 +Y_2 + Y_3, Y_3-Y_1, (1/2)Y_1 - (1/2)Y_3 +2Y_2)^T [/tex]
for ALL ##Y_1,Y_2,Y_3## is
[tex] G = \pmatrix{1&1&1 \\-1&0&1\\1/2&2&-1/2} [/tex]
You just read this off directly; no work is necessary. The given numerical values of the ##Y_i## are not relevant: they just get in the way. As far as I can see, they were given to you just to confuse you and lead you astray.

Now you want a symmetric matrix that gives the same quadratic form ##Q(Y) \equiv Y^T G Y##. To clarify, but in a much smaller example with two variables instead of three: you want to know what are the ##a_{ij}## that give
[tex] g_{11} Y_1^2 + g_{12} Y_1 Y_2 + g_{21} Y_2 Y_1 + g_{22} Y_2^2
= a_{11} Y_1^2 + 2 a_{12} Y_1 Y_2 + a_{22} Y_2^2[/tex]
for all ##Y_1,Y_2##.

Yes, I realized today that YY' is a singular matrix, making it impossible to find the inverse.

as soon as you said 'quadratic form' I remembered something from the notes (I didn't recognize it as quadratic form because I don't remember learning about them in my linear algebra course. Anyways, in the notes it says "if the matrix A is not symmetric, one may replace [tex]a_{ij}[/tex] with [tex]\frac{a_{ij} + a_{ji}}{2}[/tex]. This gave me the symmetric matrix

[tex] A = \pmatrix{1&0&\frac{3}{4} \\0&0&\frac{3}{2}\\\frac{3}{4}&\frac{3}{2}&\frac{-1}{2}} [/tex]

That being said, I'm not really sure what you're doing with the equation

[tex] g_{11} Y_1^2 + g_{12} Y_1 Y_2 + g_{21} Y_2 Y_1 + g_{22} Y_2^2
= a_{11} Y_1^2 + 2 a_{12} Y_1 Y_2 + a_{22} Y_2^2[/tex]

Would it be essentially (for my problem with a 3x3) [tex]g_{11}=a_{11}, g_{12}+g_{21}=2a_{12}, g_{31}+g_{13}=2a_{13}, g_{22}=a_{22}, g_{23}+g_{32}=2a_{23}, and g_{33}=a_{33}[/tex]

 

[Ray Vickson]

Yes, I realized today that YY' is a singular matrix, making it impossible to find the inverse.

as soon as you said 'quadratic form' I remembered something from the notes (I didn't recognize it as quadratic form because I don't remember learning about them in my linear algebra course. Anyways, in the notes it says "if the matrix A is not symmetric, one may replace [tex]a_{ij}[/tex] with [tex]\frac{a_{ij} + a_{ji}}{2}[/tex]. This gave me the symmetric matrix

[tex] A = \pmatrix{1&0&\frac{3}{4} \\0&0&\frac{3}{2}\\\frac{3}{4}&\frac{3}{2}&\frac{-1}{2}} [/tex]

That being said, I'm not really sure what you're doing with the equation

[tex] g_{11} Y_1^2 + g_{12} Y_1 Y_2 + g_{21} Y_2 Y_1 + g_{22} Y_2^2
= a_{11} Y_1^2 + 2 a_{12} Y_1 Y_2 + a_{22} Y_2^2[/tex]

Would it be essentially (for my problem with a 3x3) [tex]g_{11}=a_{11}, g_{12}+g_{21}=2a_{12}, g_{31}+g_{13}=2a_{13}, g_{22}=a_{22}, g_{23}+g_{32}=2a_{23}, and g_{33}=a_{33}[/tex]

Yes, exactly: ##A = (G + G^T)/2##, as you have written.