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

• TeenieBopper
In summary, a symmetric matrix is a square matrix with mirror-image elements above and below the main diagonal. To find a unique symmetric matrix A, you can solve a system of equations using the properties of symmetric matrices. When Y'AY is equal to Y'GY, A and G are similar matrices. A symmetric matrix can have any size as long as it is a square matrix. In real-world applications, symmetric matrices are used in fields such as physics, engineering, and computer science for optimization, eigenvalue calculations, image processing, and quadratic forms.
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

## 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.

TeenieBopper said:
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

## 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
$$GY = (Y_1 +Y_2 + Y_3, Y_3-Y_1, (1/2)Y_1 - (1/2)Y_3 +2Y_2)^T$$
for ALL ##Y_1,Y_2,Y_3## is
$$G = \pmatrix{1&1&1 \\-1&0&1\\1/2&2&-1/2}$$
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
$$g_{11} Y_1^2 + g_{12} Y_1 Y_2 + g_{21} Y_2 Y_1 + g_{22} Y_2^2 = a_{11} Y_1^2 + 2 a_{12} Y_1 Y_2 + a_{22} Y_2^2$$
for all ##Y_1,Y_2##.

1 person
Ray Vickson said:
First: as regards your previous post on this matter, your friend is quite correct: the unique matrix ##G## that gives
$$GY = (Y_1 +Y_2 + Y_3, Y_3-Y_1, (1/2)Y_1 - (1/2)Y_3 +2Y_2)^T$$
for ALL ##Y_1,Y_2,Y_3## is
$$G = \pmatrix{1&1&1 \\-1&0&1\\1/2&2&-1/2}$$
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
$$g_{11} Y_1^2 + g_{12} Y_1 Y_2 + g_{21} Y_2 Y_1 + g_{22} Y_2^2 = a_{11} Y_1^2 + 2 a_{12} Y_1 Y_2 + a_{22} Y_2^2$$
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 $$a_{ij}$$ with $$\frac{a_{ij} + a_{ji}}{2}$$. This gave me the symmetric matrix

$$A = \pmatrix{1&0&\frac{3}{4} \\0&0&\frac{3}{2}\\\frac{3}{4}&\frac{3}{2}&\frac{-1}{2}}$$

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

$$g_{11} Y_1^2 + g_{12} Y_1 Y_2 + g_{21} Y_2 Y_1 + g_{22} Y_2^2 = a_{11} Y_1^2 + 2 a_{12} Y_1 Y_2 + a_{22} Y_2^2$$

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

TeenieBopper said:
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 $$a_{ij}$$ with $$\frac{a_{ij} + a_{ji}}{2}$$. This gave me the symmetric matrix

$$A = \pmatrix{1&0&\frac{3}{4} \\0&0&\frac{3}{2}\\\frac{3}{4}&\frac{3}{2}&\frac{-1}{2}}$$

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

$$g_{11} Y_1^2 + g_{12} Y_1 Y_2 + g_{21} Y_2 Y_1 + g_{22} Y_2^2 = a_{11} Y_1^2 + 2 a_{12} Y_1 Y_2 + a_{22} Y_2^2$$

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

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

1 person

## 1. What is a symmetric matrix?

A symmetric matrix is a square matrix where the elements above and below the main diagonal are mirror images of each other. In other words, the entry in row i, column j is equal to the entry in row j, column i.

## 2. How do you find a unique symmetric matrix A?

In order to find a unique symmetric matrix A, you can use the properties of symmetric matrices to solve a system of equations. In this case, you would set Y'AY equal to Y'GY and solve for the elements of A.

## 3. What is the significance of Y'AY=Y'GY?

When Y'AY is equal to Y'GY, it means that A and G are similar matrices. This can have important implications in linear algebra and can make certain calculations easier.

## 4. Can a symmetric matrix A have any size?

Yes, a symmetric matrix can have any size as long as it is a square matrix (the number of rows is equal to the number of columns). The size of A will depend on the size of the matrices Y and G.

## 5. How is a symmetric matrix A used in real-world applications?

Symmetric matrices have many applications in fields such as physics, engineering, and computer science. They are commonly used in optimization problems, eigenvalue calculations, and image processing. They are also used in the calculation of quadratic forms, which have applications in areas such as statistics and machine learning.

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