Trace of 3x3 Matrix | Linear Function | Basis Set Representation

[blackbear]

Homework Statement

Let X denote the set of all real symmetric 3x3 matrices. The trace of a matrix, tr(x) is defined as the sum of the diagonal components and is a linear function. Define L(x) = tr(x), where x refers to X. Find the representation of this operator with respect to a basis set for the linear space (X,R) for X denote the set of all real symmetric 3x3 matrices.

2. Homework Equations [

The Attempt at a Solution

Let A= 2 3 4
3 1 0
4 0 5

So, tr(x) = a11+a22+a33
can I define L= 2 0 0
0 1 0
0 0 5

Thanks

 

[Dick]

No, it's tr(A)=2+1+5=8. The trace isn't another matrix. It's a number.

 

[blackbear]

No, it's tr(A)=2+1+5=8. The trace isn't another matrix. It's a number.

Thanks Dick...I have added the rest of the question. Eventually I think I have to find a matrix so I can represent this matrix with respect to a defined basis.

According to you L(x)=8; I need to find the representation of L(x) in the specified basis. First of all, can a point in space be represented in a specified basis? Any suggestions?

Thanks

 

[vela]

Find the representation of this operator with respect to [100],[010],[001] which is a basis set for the linear space (X,R) for X denote the set of all real symmetric 3x3 matrices.

Can you elaborate on your notation here? It doesn't make sense to me. The vector space consisting of real symmetric 3x3 matrices is 6-dimensional. You should have 6 basis vectors, each of which is a 3x3 matrix.

 

[blackbear]

Can you elaborate on your notation here? It doesn't make sense to me. The vector space consisting of real symmetric 3x3 matrices is 6-dimensional. You should have 6 basis vectors, each of which is a 3x3 matrix.

Thanks Vela...you are right there are 6 basis set but unfortunately I am not seeing that. Could you elaborate how you get the 6 vectors. Once I know this then representing into this basis will be clear to me.

 

[vela]

You are supposed to be able to do that.

The most general form of a 3x3 real symmetric matrix is

\begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f\end{pmatrix}

Express this as a linear combination of 6 matrices. Those 6 matrices will be a basis for X.

 

[blackbear]

You are supposed to be able to do that.

The most general form of a 3x3 real symmetric matrix is

\begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f\end{pmatrix}

Express this as a linear combination of 6 matrices. Those 6 matrices will be a basis for X.

If a matrix A:

\begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f\end{pmatrix}

is a linear combination of matrices B, C, D, E, & F then there exist scalars i, j, k, l, m such that A = iB + jC+kD+lE+mF; but how am I solve this...any help? Thanks

 

[vela]

No. Do you know what a linear combination is?

 

[blackbear]

No. Do you know what a linear combination is?

If a matrix A: <br /> \begin{pmatrix} a &amp; b &amp; c \\ b &amp; d &amp; e \\ c &amp; e &amp; f\end{pmatrix}<br />

is a linear combination of matrices B, C, D, E, & F then there exist scalars i, j, k, l, m such that A = iB + jC+kD+lE+mF; but how am I solve this...any help? Thanks

 

[Mark44]

This problem ties into the problem in the other thread you started, and you won't get very far in either one until you can come up with a basis for your space of 3X3 symmetric matrices.

From the other thread

These three matrices form part of a basis for your space:
\begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}

\begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}

\begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}

Can you come up with three more matrices to make a basis for this space?

 

[blackbear]

This problem ties into the problem in the other thread you started, and you won't get very far in either one until you can come up with a basis for your space of 3X3 symmetric matrices.

From the other thread

These three matrices form part of a basis for your space:
\begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}

\begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}

\begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}

Can you come up with three more matrices to make a basis for this space?

Out of the 9 matrices Mark Doom mentioned, 3 of them as you pointed here and there are symmetric having one element. The other 3 matrices must have minimum 2 elements. So there are as follows:

\begin{pmatrix} 0 &amp; 0 &amp; 1 \\ 0 &amp; 0 &amp; 0 \\ 1 &amp; 0 &amp; 0\end{pmatrix}

\begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}

\begin{pmatrix} 0 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{pmatrix}

Is this right?

Comment: I see that these and the other 3 metrics that you pointed are all linearly dependent! I thought in order to be a basis of a liner space, the vectors forming the metrics must be linearly independent. Please explain...thanks

 

[vela]

What does

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

equal? Can you generate every symmetric matrix with the right choices for a, b, c, d, e, and f?

 

[HallsofIvy]

One obvious point:
\begin{pmatrix}1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\0 &amp; 0 &amp; 0 \end{pmatrix}+ \begin{pmatrix}0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}= \begin{pmatrix}1 &amp; 0 &amp; 0 \\0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}
so those six matrices are not independent.

 

[blackbear]

What does

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

equal? Can you generate every symmetric matrix with the right choices for a, b, c, d, e, and f?

\begin<br /> {pmatrix} a+e &amp; f &amp; d \\ f &amp; b &amp; 0 \\d &amp; 0 &amp; c+e \end{pmatrix}<br />Yes...seems like you can generate any symmetric matrix with the proper choices of the scalars. So, coming back to the initial question:

L(x) = (a+e) + b + (c+e) = a+b+c+2e

How can I represent L(x) to the 6 symmetric basis set? I am to use the trivial basis set {1} for the range-space (R,R) to solve the problem.
Is the representation for the 1st basis set is as follows?

<br /> a\begin{pmatrix}1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\0 &amp; 0 &amp; 0 \end{pmatrix}<br /> * (a+b+c+2e)

Thanks

 

[vela]

\begin<br /> {pmatrix} a+e &amp; f &amp; d \\ f &amp; b &amp; 0 \\d &amp; 0 &amp; c+e \end{pmatrix}<br />Yes...seems like you can generate any symmetric matrix with the proper choices of the scalars.

So take, for example, the symmetric matrix

\begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 2 \\ 0 &amp; 2 &amp; 0 \end{pmatrix}

For what values for a, b, c, d, e, and f does the linear combination equal that matrix?

 

[blackbear]

So take, for example, the symmetric matrix

\begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 2 \\ 0 &amp; 2 &amp; 0 \end{pmatrix}

For what values for a, b, c, d, e, and f does the linear combination equal that matrix?

One sets of value:

a=0 or any real number
b=0
c=0 or any real number
d=0
e=0 or -a=-c
f=0

 

[vela]

Really? If a=b=c=d=e=f=0, don't you get the 0 matrix? How do you get the 2s?

 

[blackbear]

Really? If a=b=c=d=e=f=0, don't you get the 0 matrix? How do you get the 2s?

Really? If a=b=c=d=e=f=0, don't you get the 0 matrix? How do you get the 2s?

I am not sure what you meant!

If a=b=c=d=e=f=0, and you put these values to the following matrix:

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

Then you do get a 0 matrix.

But, you gave the following matrix which had a 2 so for a=b=c=d=e=f=0, I don't understand how this matrix can become 0!

<br /> \begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 2 \\ 0 &amp; 2 &amp; 0 \end{pmatrix}<br />

2ndly, In the matrix below:

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

element a23 and a32 equals 0. So if there is a 2 for both the elements then obviously the matrix changes to the following: You are adding the following matrix

<br /> <br /> \begin<br /> {pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 2 \\0 &amp; 2 &amp; 0 \end{pmatrix}<br /> <br />

to get this matrix:<br /> <br /> <br /> \begin<br /> {pmatrix} a+e &amp; f &amp; d \\ f &amp; b &amp; 2 \\d &amp; 2 &amp; c+e \end{pmatrix}<br /> <br />

But the rest of the elements remains unchanged and there is no linear relation from this two elements with the rest of the elements. So, I am still confused to your question "How do you get the 2s". Please explain...thanks

 

[vela]

Your claim is that the six matrices form a basis for X. If that's true, then for any symmetric matrix A, you can find a linear combination of those six matrix equal to A. So I gave you a symmetric matrix, the one with the 2s, and asked you to tell me what a, b, c, d, e, and f had to equal so that the linear combination is equal to it. You asserted that a=b=c=d=e=f=0 works. I said it clearly doesn't.

So again, tell me what values of a, b, c, d, e, and f, if any, will result in the given matrix? If the six matrices form a basis, there must be a solution.

 

[blackbear]

Your claim is that the six matrices form a basis for X. If that's true, then for any symmetric matrix A, you can find a linear combination of those six matrix equal to A. So I gave you a symmetric matrix, the one with the 2s, and asked you to tell me what a, b, c, d, e, and f had to equal so that the linear combination is equal to it. You asserted that a=b=c=d=e=f=0 works. I said it clearly doesn't.

So again, tell me what values of a, b, c, d, e, and f, if any, will result in the given matrix? If the six matrices form a basis, there must be a solution.

There is none since elements a23 & a32 is not related to any values of a,b,c,d,e & f. So, following is not the basis matrix:

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

The following set of basis are:

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

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

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

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

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

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

Please verify...

 

[blackbear]

There is none since elements a23 & a32 is not related to any values of a,b,c,d,e & f. So, following is not the basis matrix:

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

The following set of basis are:

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

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

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

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

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

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

Please verify...

a=b=c=d=e=0
f=2

 

[Dick]

There is none since elements a23 & a32 is not related to any values of a,b,c,d,e & f. So, following is not the basis matrix:

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

The following set of basis are:

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

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

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

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

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

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

Please verify...

Sure, that's a fine basis for X.

 

[blackbear]

How can I represent L(x) to the 6 symmetric basis set? I am to use the trivial basis set {1} for the range-space (R,R) to solve the problem.

The symmetric matrix is:

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

Based o the above matrix, is the L(x) representation as follows:

L(x) = [a+b+c] = 1+0+0=1 for the 1st element

L(x) = [a+b+c] = 0+1+0=1 for the 2nd element

L(x) = [a+b+c] = 0+0+1=1 for the 3rd element

L(x) = [a+b+c] = 0+0+0=0 for the 4th 5th & 6th element

Please verify...

 

[Mark44]

Pretty much, except that you have one basis, the one whose elements you listed in post #21. If x is one of the basis elements, then L(x) = 1 or L(x) = 0.

 

[blackbear]

Pretty much, except that you have one basis, the one whose elements you listed in post #21. If x is one of the basis elements, then L(x) = 1 or L(x) = 0.

is the L(x) operator be as follows:L(x)= <br /> <br /> \begin{pmatrix} 1 &amp; 0 &amp; 0 \\0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}<br /> <br />

Please confirm...thanks

 

[Mark44]

Pretty much, except that you have one basis, the one whose elements you listed in post #21. If x is one of the basis elements, then L(x) = 1 or L(x) = 0.

is the L(x) operator be as follows:


L(x)= <br /> <br /> \begin{pmatrix} 1 &amp; 0 &amp; 0 \\0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}<br /> <br />

No. L(x) is a scalar (a number), not a matrix. If x is any matrix in your basis set, L(x) is either 0 or 1.[/