Dimension of a set of symmetric matrices & prove it's a vector space

[kesun]

Prove: the set of 3x3 symmetric matrices is a vector space and find its dimension.

Well in class my prof has done this question, but I still don't quite get it..

Ok, first off, I need to prove that it's a vector space. The easy way is probably to prove that it contains the zero space and is closed under the two operations. Certainly it contains the zero vector, which I understand; but when my prof went for the second part, he wrote something in this fashion:

[a d e]___[1 0 0]____[0 0 0]
[d b f] = a [0 0 0] + b [0 1 0] + c [...] + d [...] + e [...] + f [...] => dimension of 6.
[e f c]___[0 0 0]____[0 0 0]

(sorry I have to use the underscore to line them up properly..)

How is that closed under the two operations? I am not getting it completely..

Also, how they add up to the original matrix certainly makes sense to me, but how he devided them up in this way sort of puzzles me...Since every matrix is reducible, so this one should be as well. Since the dimension of the number of elements in a basis, then shouldn't this have a basis of max. 3? I'm pretty sure I am not thinking correctly, but someone please straighten this stuff out for me please!

 

[yyat]

Prove: the set of 3x3 symmetric matrices is a vector space and find its dimension.

Well in class my prof has done this question, but I still don't quite get it..

Ok, first off, I need to prove that it's a vector space. The easy way is probably to prove that it contains the zero space and is closed under the two operations. Certainly it contains the zero vector, which I understand; but when my prof went for the second part, he wrote something in this fashion:

[a d e]___[1 0 0]____[0 0 0]
[d b f] = a [0 0 0] + b [0 1 0] + c [...] + d [...] + e [...] + f [...] => dimension of 6.
[e f c]___[0 0 0]____[0 0 0]

(sorry I have to use the underscore to line them up properly..)

Here your prof is already proving that the dimension is 6. The point is that the six matrices on the right hand side form a basis of the vector space of 3x3 symmetric matrices. The formula proves that: 1) They generate the space, because any element can be written as a linear combination. 2) They are linearly independent because if the left hand side is 0, then certainly the scalar factors a,b,c,d,e,f are zero.

How is that closed under the two operations? I am not getting it completely..

Your professor may have skipped that. One way to prove this is just check "by hand" that the sum and scalar multiples of symmetric matrices are again symmetric, another way is to use the fact that a matrix is symmetric if and only if it equals its transpose (and use linearity of transposition).

Also, how they add up to the original matrix certainly makes sense to me, but how he devided them up in this way sort of puzzles me...Since every matrix is reducible, so this one should be as well. Since the dimension of the number of elements in a basis, then shouldn't this have a basis of max. 3? I'm pretty sure I am not thinking correctly, but someone please straighten this stuff out for me please!

 

[kesun]

Ah! So say for matrix:

[1 4]
[9 0]

(which is the same as [1 4, 9 0] to make representation easier)
It can have a basis like this:

1[1 0, 0 0]+4[0 1, 0 0]+9[0 0, 1 0]+0[0 0, 0 1]

right?

Since this matrix reduces to the identity matrix, should the original matrix itself be the basis by definition? Should its dimension be 2 or 4? When we talk about things like basis and dimension, shouldn't we be talking about the reduced version of the matrix? I'm still a bit confused..

 

[yyat]

A basis is something a vector space can have, for example (1,0,0), (0,1,0), (0,0,1) is a basis of R^3. The matrices

\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}

form a basis of the symmetric 2x2 matrices.

On the other hand, a vector can have a representation in terms of a basis, so for example

(2,-1,0) = 2(1,0,0) + (-1)(0,1,0) + 0(0,0,1)

or

\begin{bmatrix}3&7\\7&1\end{bmatrix}=3\begin{bmatrix}1&0\\0&0\end{bmatrix} + 7\begin{bmatrix}0&1\\1&0\end{bmatrix}+1\begin{bmatrix}0&0\\0&1\end{bmatrix}.

The properties of a basis guarantee that such a representation always exists and that it is unique, i.e. there is only one possible choice of coefficients in front of the basis vectors. These coefficients are also called the coordinates of the vector on the left hand side with respect to the basis.

One source of confusion maybe that a matrix can also be considered as a vector (in a vector space of matrices).

 

[kesun]

So for example, a 2x2 skew-symmetric matrix should have a basis of:

[1 0] [0 1],[0 0]
[0 0],[-1 0],[0 1]

is that correct?

 

[yyat]

So for example, a 2x2 skew-symmetric matrix should have a basis of:

[1 0] [0 1],[0 0]
[0 0],[-1 0],[0 1]

is that correct?

No, only the second matrix is skew symmetric.

 

[kesun]

Oh, I forgot the important fact that the diagonal consists only zero in a skew-symmetric matrix. Yes, the only basis for it would be [0 1, -1 0] then.

So to prove that a matrix is a basis, I just need to show that its vectors(columns) are linearly independent, right?

 

[yyat]

Oh, I forgot the important fact that the diagonal consists only zero in a skew-symmetric matrix. Yes, the only basis for it would be [0 1, -1 0] then.

So to prove that a matrix is a basis, I just need to show that its vectors(columns) are linearly independent, right?

No, this would prove that (0,-1), (1,0) is a basis of R^2. If you want to show that [0 1, -1 0] is a basis for the skew-symmetric 2x2-matrices you need to show two things:

1) Every skew-symmetric 2x2 matrix can be written in the form a*[0 1, -1 0] for some a (in other words this proves that the vector space of skew symmetric 2x2 matrices is generated by [0 1, -1 0]). This should be easy.

2) The set of vectors {[0 1, -1 0]} (in this case consisting only of one element) is linearly independent, this is also rather trivial if you take a look at the definition of "linearly independent" and follows from the fact that [0 1, -1 0] is not the matrix with all zeros.