Finding a Basis Set for a Real Symmetrical 3x3 Matrix Space

[blackbear]

Let X denote the set of real symmetrical 3X3 matrix. Then (X,R) forms a linear space. What will be a basis set for this linear space?

I would appreciate if someone can help me with the question. My understanding is in R3 space there could be many 3X3 matrix that could be the basis set for the space as long they are linearly independent. Is there any special rule for symmetrical matrix for reals?

I also know for every symmetric real matrix A there exists a real orthogonal matrix Q such that D = Q(T)AQ is a diagonal matrix. Every symmetric matrix is thus, up to choice of an orthonormal basis, a diagonal matrix.

D = Q(T)*A*Q, where Q(T)= Q transpose

A= Symmetric Matrix
Q = Orthogonal Matrix
D = Diagonal Matrix

Let's say A= 2 3 4
3 1 0
4 0 5

Is it possible to find Q from here?

Thanks much.

 

[Mark44]

Let X denote the set of real symmetrical 3X3 matrix. Then (X,R) forms a linear space. What will be a basis set for this linear space?

I would appreciate if someone can help me with the question. My understanding is in R3 space there could be many 3X3 matrix that could be the basis set for the space as long they are linearly independent. Is there any special rule for symmetrical matrix for reals?

R³ doesn't have anything to do with this problem. Each matrix in your basis is going to be a 3 X 3 matrix. Yes, there are an infinite number of basis sets, but you only have to find one set of matrices.

I also know for every symmetric real matrix A there exists a real orthogonal matrix Q such that D = Q(T)AQ is a diagonal matrix. Every symmetric matrix is thus, up to choice of an orthonormal basis, a diagonal matrix.

D = Q(T)*A*Q, where Q(T)= Q transpose

A= Symmetric Matrix
Q = Orthogonal Matrix
D = Diagonal Matrix

I don't see that any of this is relevant, except for the fact that you're dealing with symmetric matrices.

Let's say A= 2 3 4
3 1 0
4 0 5

Is it possible to find Q from here?

Maybe, but the real question is, is it necessary?

Notice that for your symmetric matrix example, that a_{2, 1} = a_{1, 2}. In general for a symmetric matrix, a_{m, n} = a_{m, n}. How many elements of a symmetric matrix like this must you know before you know all the other elements of that matrix? Does that give you some idea of the dimension of the subspace of 3 X 3 matrices?

 

[blackbear]

Thanks Mark44 for your reply;

Few things:


The entries of a symmetric matrix are symmetric with respect to the main diagonal (top left to bottom right). So if the entries are written as A = (aij), then

a_{ij} = a_{ji}

for all indices i and j. The following 3×3 matrix is symmetric: So, I am not sure what you meant by

" In general for a symmetric matrix, am, n = am, n. How many elements of a symmetric matrix like this must you know before you know all the other elements of that matrix? Does that give you some idea of the dimension of the subspace of 3 X 3 matrices? "

The dimension is 3;

2ndly, if a matrix is 3X3, isn't it obvious that it is from a R3 space? So I am also not sure when you wrote:

" R3 doesn't have anything to do with this problem."

 

[Moo Of Doom]

While the space 3x3 matrices operate on is 3 dimensional, the space of all 3x3 matrices is 9-dimensional, one example of a basis being the set of nine matrices with one 1 entry, and the rest zeros:

[tex]\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 1 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}\right)
[/tex]

The set of symmetric matrices are a subspace of this 9-dimensional space. What is its dimension?

 

[HallsofIvy]

Moo of Doom's post deserves a lot of thought. But notice that NONE of the basis members he gives for the space of all 3 by 3 matrices is in the set of symmetric 3 by 3 matrices.

One thing to think about to answer his question- if you wanted to write down an arbitrary symmetric matrix, how many numbers could you choose independently? (for example, if you already have the number in the third column of the first row, you cannot choose a different number for the first column of the third row.)

A "general" 3 by 3 matrix would look like
[tex]\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}[/tex]
where a, b, c, d, e, f, g, h, and i can be any numbers.

What would a "general" symmetric matrix look like?

 

[blackbear]

Moo of Doom's post deserves a lot of thought. But notice that NONE of the basis members he gives for the space of all 3 by 3 matrices is in the set of symmetric 3 by 3 matrices.

One thing to think about to answer his question- if you wanted to write down an arbitrary symmetric matrix, how many numbers could you choose independently? (for example, if you already have the number in the third column of the first row, you cannot choose a different number for the first column of the third row.)

A "general" 3 by 3 matrix would look like
[tex]\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}[/tex]
where a, b, c, d, e, f, g, h, and i can be any numbers.

What would a "general" symmetric matrix look like?

Symmetric set may look:

[tex]\begin{bmatrix}a & b & c \\ b & d & e \\ c & e & f\end{bmatrix}[/tex]

or

[tex]\begin{bmatrix}a & b & d \\ b & e & f \\ d & f & g\end{bmatrix}[/tex]

although these don't have all the elements! I thought for a 3X3 matrix defined in R3 space and real has a dimension of 3. Since 3 (minimum) elements are required to represent each column or vectors. Does this make sense...please explain.

Thanks

 

[Mark44]

While the space 3x3 matrices operate on is 3 dimensional, the space of all 3x3 matrices is 9-dimensional, one example of a basis being the set of nine matrices with one 1 entry, and the rest zeros:

[tex]\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 1 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}\right)
[/tex]

The set of symmetric matrices are a subspace of this 9-dimensional space. What is its dimension?

Moo of Doom's post deserves a lot of thought. But notice that NONE of the basis members he gives for the space of all 3 by 3 matrices is in the set of symmetric 3 by 3 matrices.

Three of them are - the 1st, 5th, and 9th matrices listed are symmetric.

 

[Mark44]

Symmetric set may look:

[tex]\begin{bmatrix}a & b & c \\ d & d & e \\ c & e & f\end{bmatrix}[/tex]

The one above isn't symmetric unless b = d, in which case you should write it as
[tex]\begin{bmatrix}a & b & c \\ b & d & e \\ c & e & f\end{bmatrix}[/tex]

or

[tex]\begin{bmatrix}a & b & d \\ b & e & f \\ d & f & g\end{bmatrix}[/tex]

although these don't have all the elements! I thought for a 3X3 matrix defined in R3 space and real has a dimension of 3. Since 3 (minimum) elements are required to represent each column or vectors. Does this make sense...please explain.

"3X3 matrix defined in R3 space and real has a dimension of 3" doesn't make any sense to me. Your 3X3 matrix has 9 real entries.

 

[blackbear]

While the space 3x3 matrices operate on is 3 dimensional, the space of all 3x3 matrices is 9-dimensional, one example of a basis being the set of nine matrices with one 1 entry, and the rest zeros:

[tex]\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 1 & 0 \\
\end{array}\right)
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}\right)
[/tex]

The set of symmetric matrices are a subspace of this 9-dimensional space. What is its dimension?

As Mark pointed out 3 symmetric matrices are the subspace of this 9-dimensional space. So the 3x3 symmetrical matrices operates on 3D but the space for all 3x3 symmetrical matrices is 9D? Please verify.

Thanks

 

[Mark44]

As Mark pointed out 3 symmetric matrices are the subspace of this 9-dimensional space.

No, that's not what I said at all. Moo of Doom listed nine matrices that form a basis for the vector space of 3X3 matrices. Of those nine, three were symmetric.

So the 3x3 symmetrical matrices operates on 3D but the space for all 3x3 symmetrical matrices is 9D? Please verify.

Let's get the terminology straight. Any 3X3 matrix, symmetric or not, can multiply a vector in R³; i.e., a 3D vector. Since every basis for the vector space of 3X3 matrices must contain nine matrices, the dimension of this space is 9.