Calculating Representation of Linear Operator for Symmetric Matrix

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Homework Statement

Let L(x) a linear operator defined by setting the diagonal elements of x to zero. What will be the representation of this operator to the following basis set? x E X. X denote the set of all real symmetric 3x3 matrices.

Homework Equations


L*y=x
L=x*inv(y)

<br /> \begin{pmatrix} a &amp; e &amp; d \\ e &amp; b &amp; f \\d &amp; f &amp; c \end{pmatrix} <br /> X1:<br /> \begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}<br />

X2:<br /> \begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}<br />

X3<br /> \begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}<br />

X4<br /> \begin{pmatrix} 0 &amp; 0 &amp; 1 \\ 0 &amp; 0 &amp; 0 \\ 1 &amp; 0 &amp; 0\end{pmatrix}<br />

X5<br /> \begin{pmatrix} 0 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}<br />

X6<br /> \begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 0\end{pmatrix}<br />

The Attempt at a Solution


To get the question set right, I am assuming I am to find a linear operator whose functionality is as follows:

Brain storming: L(symmetric matrix) * Y( Any symmetric matrix) = X(matrix, whose diagonal is zero)...is this right? So Y has to be symmetric and L, Y and X has to be from the same basis. all I know is the basis set. Any hints please?
 
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blackbear said:

Homework Statement



Re: Trace of a Matrix
Let L(x) a linear operator defined by setting the diagonal elements of X to zero. What will be the representation of this operator to the following basis set? x E X. X denote the set of all real symmetric 3x3 matrices.


Homework Equations


L*y=x
L=inv(y)*x
The 2nd equation doesn't follow from the first.
If LY = X, and Y is an invertible matrix,
then L = X Y-1
Note that X Y-1 and Y-1X don't have to be equal.


blackbear said:

The Attempt at a Solution


To get the question set right, I am assuming I am to find a linear operator whose functionality is as follows:

Brain storming: L(symmetric matrix) * Y( Any symmetric matrix) = X(matrix, whose diagonal is zero)...is this right? So Y has to be symmetric and L, Y and X has to be from the same basis. all I know is the basis set. Any hits please?


You have a basis consisting of the six matrices in your other thread. What does this transformation do to each of these matrices? It might be helpful to give each of the basis matrices a name, say X1 through X6. What is L(X1)? L(X2)? And so on.
 
Mark44 said:
The 2nd equation doesn't follow from the first.
If LY = X, and Y is an invertible matrix,
then L = X Y-1
Note that X Y-1 and Y-1X don't have to be equal.You have a basis consisting of the six matrices in your other thread. What does this transformation do to each of these matrices? It might be helpful to give each of the basis matrices a name, say X1 through X6. What is L(X1)? L(X2)? And so on.
let Y =

<br /> <br /> \begin{pmatrix} 2 &amp; -1 &amp; 0 \\ -1 &amp; 1 &amp; 5 \\ 0 &amp; 5 &amp; 3\end{pmatrix}<br /> <br />

then LX1=

<br /> <br /> \begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}<br /> <br />

So, the transformation to brings back X1, X2 and so on...but the diagonal is not zero! Also, my understanding is L should work for any matrix ..right?
 
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From post #1, the domain for L seems the set of 3x3 symmetric matrices.
blackbear said:
Let L(x) a linear operator defined by setting the diagonal elements of x to zero. What will be the representation of this operator to the following basis set? x E X. X denote the set of all real symmetric 3x3 matrices.

To be able to say what L does to an arbitrary symmetric 3x3 matrix in E, it suffices to know what L does to each matrix in a basis for E. Then, since any matrix in E is a linear combination of the six basis elements, you can find L(e) for any matrix e in E.

BTW, I don't think this is useful - Brain storming: L(symmetric matrix) * Y( Any symmetric matrix) = X(matrix, whose diagonal is zero)... At least, it doesn't seem so to me.
 
Mark44 said:
it suffices to know what L does to each matrix in a basis for E.
Question: I know the 6 basis set so x in the equation is given. But I am not sure how get L for the basis set since "Y" is not given. I only found LX1 using the Y matrix but that is not a arbitrary matrix.
L = X Y-1

The basis matrix is:

B= <br /> \begin{pmatrix} 1 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 1\end{pmatrix}<br />
 
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blackbear said:
Question: I know the 6 basis set so x in the equation is given. But I am not sure how get L for the basis set since "Y" is not given. I only found LX1 using the M matrix, which was "Y" for that particular matrix.

L = X Y-1
You're thinking of L as being a matrix, with the idea of solving LX = Y for L. I don't think that it will be possible for you to do that. I believe that it is much more useful to think of L as a transformation, and investigate what it does to each of the six matrices in your basis.

If you know L(X1), L(X2), ..., L(X6), you know what L will do to any matrix in X, the set of 3x3 symmetric matrices.

For example, L(X1) = 0, where 0 represents the 3x3 matrix whose entries are all zero, and X1 is the first matrix you listed in the first post of this thread.

Your basis X1, X2, ..., X6 spans set X and is linearly independent, so every matrix in X can be written as a linear combination of X1, X2, ..., X6. If you know L(X1), L(X2), ..., and L(X6), then you can find L(x) for any matrix x in X.

The idea here is that with a linear transformation, L(au + bv) = aL(u) + bL(v). This notion can be extended to sums with more than two terms.
 
Mark44 said:
If you know L(X1), L(X2), ..., L(X6), you know what L will do to any matrix in X, the set of 3x3 symmetric matrices.

I am still not sure how to get L(X1) but logically speaking is L(X1)= X1 the 1st matrix from the basis set?
 
blackbear said:
I am still not sure how to get L(X1) but logically speaking is L(X1)= X1 the 1st matrix from the basis set?
X1 is the first matrix you listed among the six basis members, but L(X1) \neq X1. Think about how L is defined.

These calculations really aren't rocket science. You should be able to figure out what L(X1) is without having to write anything. Besides, I already told you what L(X1) is.
 
LX1, LX2, LX3=

<br /> <br /> \begin{pmatrix} 0 &amp; 1 &amp; 1 \\ 1 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 0\end{pmatrix}<br /> <br />

LX4, LX4, LX6=

<br /> <br /> \begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}<br /> <br />

Is this correct?
 
  • #10
You have gone back and edited your earlier posts, so it's difficult keeping up with you.
blackbear said:
let Y =

<br /> <br /> \begin{pmatrix} 2 &amp; -1 &amp; 0 \\ -1 &amp; 1 &amp; 5 \\ 0 &amp; 5 &amp; 3\end{pmatrix}<br /> <br />

then LX1=

<br /> <br /> \begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}<br /> <br />

So, the transformation to brings back X1, X2 and so on...but the diagonal is not zero! Also, my understanding is L should work for any matrix ..right?
Forget Y!

What you wrote as LX1 above is X1. The question is what is L(X1). I have already said that L(X1) is the 3x3 zero matrix.
Remember what L does to a matrix: It changes then entries on the main diagonal to 0. So
L\left(\begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}\right) = \begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}


blackbear said:
Question: I know the 6 basis set so x in the equation is given. But I am not sure how get L for the basis set since "Y" is not given.
Again, forget about Y.
blackbear said:
I only found LX1 using the Y matrix but that is not a arbitrary matrix.
L = X Y-1

The basis matrix is:

B= <br /> \begin{pmatrix} 1 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 1\end{pmatrix}<br />
?
There is no "basis matrix." A basis for X, the space of 3x3 symmetric matrices, is the six matrices you show in post #1.
blackbear said:
LX1, LX2, LX3=

<br /> <br /> \begin{pmatrix} 0 &amp; 1 &amp; 1 \\ 1 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 0\end{pmatrix}<br /> <br />

LX4, LX4, LX6=

<br /> <br /> \begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}<br /> <br />

Is this correct?
Not even close.
L takes a matrix as input, and produces another matrix as its output. Please stop writing things like LX1, LX2, etc., since this is reinforcing the idea that you are multiplying L and some matrix.

You are completely forgetting how L is defined, which you have in your first post.
blackbear said:
Let L(x) a linear operator defined by setting the diagonal elements of x to zero.
Use that, and that alone to figure out L(X2), L(X3), L(X4), L(X5), and L(X6). I already did L(X1) for you.
 
  • #11
L(X2), L(X3)

<br /> \begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}<br />

L(X4)

<br /> \begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}<br />

Having difficulty to compute L(X5) & L(X6)...
 
  • #12
blackbear said:
L(X2), L(X3)

<br /> \begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}<br />

L(X4)

<br /> \begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}<br />

Having difficulty to compute L(X5) & L(X6)...

I thought L was supposed to set diagonal elements to zero. X4 already has diagonal elements zero, doesn't it? Shouldn't L(X4)=X4? What do you think L does and how did you get what you got?
 
  • #13
Dick said:
I thought L was supposed to set diagonal elements to zero. X4 already has diagonal elements zero, doesn't it? Shouldn't L(X4)=X4? What do you think L does and how did you get what you got?

L multiplies with a matrix (X) and get a matrix (Y) which has diagonal elements zero.

L*X=Y

so L(X4)*X4 = X4 so I had to compute a matrix L(X4) which multiplies with X4 matrix so the result becomes same matrix X4.

Guess this is not the way to go then...
 
  • #14
blackbear said:
L multiplies with a matrix (X) and get a matrix (Y) which has diagonal elements zero.

L*X=Y

so L(X4)*X4 = X4 so I had to compute a matrix L(X4) which multiplies with X4 matrix so the result becomes same matrix X4.

Guess this is not the way to go then...

You are putting the cart before the horse. L will be a 6x6 matrix defined on your basis for X. Just figure out what L does to each basis element before you start multiplying by anything. I'll agree L(X2)=L(X3)=0. Now figure out the rest of them. Do not multiply anything by a matrix. Just set diagonal elements to zero.
 
  • #15
Dick said:
You are putting the cart before the horse. L will be a 6x6 matrix defined on your basis for X. Just figure out what L does to each basis element before you start multiplying by anything. I'll agree L(X2)=L(X3)=0. Now figure out the rest of them. Do not multiply anything by a matrix. Just set diagonal elements to zero.

In that case L(X4) = X4; L(X5)= X5; L(X6)=X6 since all the elements have zero diagonals.
 
  • #16
blackbear said:
In that case L(X4) = X4; L(X5)= X5; L(X6)=X6 since all the elements have zero diagonals.
YES!

Now you are able to say what L does to an arbitrary matrix x in X.

For example, if
x = \begin{pmatrix} a &amp; e &amp; d \\ e &amp; b &amp; f \\d &amp; f &amp; c \end{pmatrix}

then x = aX1 + bX2 + cX3 + dX4 + eX5 + fX6,
so L(x) = L(aX1 + bX2 + cX3 + dX4 + eX5 + fX6)
= aL(X1) + bL(X2) + cL(X3) + dL(X4) + eL(X5) + fL(X6)

Since you now know what L(Xk) is for each matrix Xk in your basis, it's a simple matter to plug things into the equation above to find L(x).
 
  • #17
Mark44 said:
YES!

Now you are able to say what L does to an arbitrary matrix x in X.

For example, if
x = \begin{pmatrix} a &amp; e &amp; d \\ e &amp; b &amp; f \\d &amp; f &amp; c \end{pmatrix}

then x = aX1 + bX2 + cX3 + dX4 + eX5 + fX6,
so L(x) = L(aX1 + bX2 + cX3 + dX4 + eX5 + fX6)
= aL(X1) + bL(X2) + cL(X3) + dL(X4) + eL(X5) + fL(X6)

Since you now know what L(Xk) is for each matrix Xk in your basis, it's a simple matter to plug things into the equation above to find L(x).

so L(x)= aL(X1) + bL(X2) + cL(X3) + dL(X4) + eL(X5) + fL(X6)

but the first 3 terms will always be zero...if that's true I just tried to multiply with a Y matrix posted below and did not get the zero diagonal matrix.

Y= <br /> \begin{pmatrix} 2 &amp; -1 &amp; 0 \\ -1 &amp; 1 &amp; 5 \\ 0 &amp; 5 &amp; 3\end{pmatrix}<br />

L = <br /> \begin{pmatrix} 0 &amp; -1 &amp; 0 \\ -1 &amp; 0 &amp; 5 \\ 0 &amp; 5 &amp; 0\end{pmatrix}<br />

so I guess L is the required matrix with a zero diagonal...so I don't have to perform any additional multiplication i.e. L*Y...right?
 
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  • #18
blackbear said:
so L(x)= aL(X1) + bL(X2) + cL(X3) + dL(X4) + eL(X5) + fL(X6)

but the first 3 terms will always be zero...
Yes, but so what?
blackbear said:
if that's true I just tried to multiply with a Y matrix posted below and did not get the zero diagonal matrix.

Y= <br /> <br /> <br /> \begin{pmatrix} 2 &amp; -1 &amp; 0 \\ -1 &amp; 1 &amp; 5 \\ 0 &amp; 5 &amp; 3\end{pmatrix}<br /> <br /> <br />
What you said makes no sense. You tried to multiply what with Y? Where did this matrix Y come from?

Are you trying to find L(Y)? That makes more sense, but you aren't doing matrix multiplication to get L(Y). Write Y as a linear combination of the basis matrices, and then find what L of that linear combination is, just as I described in my previous post.
 
  • #19
Mark44 said:
Yes, but so what?

What you said makes no sense. You tried to multiply what with Y? Where did this matrix Y come from?

Are you trying to find L(Y)? That makes more sense, but you aren't doing matrix multiplication to get L(Y). Write Y as a linear combination of the basis matrices, and then find what L of that linear combination is, just as I described in my previous post.

Got it...thank you;
 
  • #20
blackbear said:
so L(x)= aL(X1) + bL(X2) + cL(X3) + dL(X4) + eL(X5) + fL(X6)

but the first 3 terms will always be zero...if that's true I just tried to multiply with a Y matrix posted below and did not get the zero diagonal matrix.

Y= <br /> \begin{pmatrix} 2 &amp; -1 &amp; 0 \\ -1 &amp; 1 &amp; 5 \\ 0 &amp; 5 &amp; 3\end{pmatrix}<br />

L = <br /> \begin{pmatrix} 0 &amp; -1 &amp; 0 \\ -1 &amp; 0 &amp; 5 \\ 0 &amp; 5 &amp; 0\end{pmatrix}<br />

so I guess L is the required matrix with a zero diagonal...so I don't have to perform any additional multiplication i.e. L*Y...right?
Let's get the terminology right, and to lessen confusion, let's rename your matrices.

X = \begin{pmatrix} 2 &amp; -1 &amp; 0 \\ -1 &amp; 1 &amp; 5 \\ 0 &amp; 5 &amp; 3\end{pmatrix}


Then
L(X) = \begin{pmatrix} 0 &amp; -1 &amp; 0 \\ -1 &amp; 0 &amp; 5 \\ 0 &amp; 5 &amp; 0\end{pmatrix}

You did NOT multiply L and Y (or L and X) to get the 2nd matrix above.

Got it?
 
  • #21
Mark44 said:
Let's get the terminology right, and to lessen confusion, let's rename your matrices.

X = \begin{pmatrix} 2 &amp; -1 &amp; 0 \\ -1 &amp; 1 &amp; 5 \\ 0 &amp; 5 &amp; 3\end{pmatrix}


Then
L(X) = \begin{pmatrix} 0 &amp; -1 &amp; 0 \\ -1 &amp; 0 &amp; 5 \\ 0 &amp; 5 &amp; 0\end{pmatrix}

You did NOT multiply L and Y (or L and X) to get the 2nd matrix above.

Got it?

Yes...
 
  • #22
In general, if L maps a vector space U with a given basis, into a vector space V with a given basis, one can express U as a matrix by: Apply L to each basis vector for U then write the result as a linear combination of the vectors in the basis for V. The coefficients in that linear transformation give the coefficients for one column of the matrix representation for L.

If L maps a vector space to itself, you can use the same basis for both "U" and "V".

Here, using the basis you give, {X1, X2, X3, X4, X5, X6}, you have determined that L(X1)= 0, L(X2)= 0, L(X3)= 0, L(X4)= X4, L(X5)= X5, and L(X6)= X6.

It should be easy to just write down the matrix representation of L from that.
 
  • #23
Linear Least Squares

I have a question about Linear least squares:

In Linear least squares, For any critical point "x" it must follow the linear system:
A(Transpose) * Ax = b * A(Transpose) where x is the critical point.

But here x is an n vector, so does that mean there are as many critical points (x) as there are columns?

So in case of an quadratic polynomial : 1 + X2*t + X3*t2 with three parameters, would we have three critical points?

Thanks in advance
 
  • #24
Please start a new thread rather than tack an unrelated question to an existing thread.
Thanks...
 
  • #25
Mark44 said:
Please start a new thread rather than tack an unrelated question to an existing thread.
Thanks...

How do I do that? I am new to physics forums.
 
  • #26
Navigate to the section this thread is in (Calculus & Beyond), and then click New Topic.
 

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