Understanding Matrix Dimensions: Calculating Entries and Recognizing Symmetry

[pgandalf]

How do we calculate the dimension of a matrix? Is it the number of entries? Or is it the number of different entries? For instance if I have a matrix 2x2 the dimension would be 4 but if the matrix is simetrical it would be 3. Is this correct? Thanks for your help.

 

[CompuChip]

I'm not really sure that you are referring to a general definition. As I learned it, the dimensions of a matrix are the number of rows and columns, e.g. 2x2, 4x1 or 16x38.
Would it be possible you are referring to some other dimension (e.g. the dimension of the column space, row space, null space, kernel, etc.?)

 

[pgandalf]

I'm not really sure that you are referring to a general definition. As I learned it, the dimensions of a matrix are the number of rows and columns, e.g. 2x2, 4x1 or 16x38.
Would it be possible you are referring to some other dimension (e.g. the dimension of the column space, row space, null space, kernel, etc.?)

I mean the dimension like R3 is a vectorial space of dimension 3 and it's isomorph to the three dimensional space, R2 is has dimension 2 and is isomorph to the plane (2D), the space of polynomials of grade lower or equal to n has a dimension of n+1 and so on...but how do we know the dimension of the space of matrixes nxn? Because I have a problem in which there is a linear transformation from R3 into the matrixes of 2x2 and I think that it contadicts the theorem of dimensions (for a linear transformation T :V-->W , dim (V)= dim kernel (T) + dim image (T) and dim image (T) less or equal to dim W ) so 3 = dim kernel + dim matrixes 2x2 and this is absurd because the kernel cannot be less than 0 . Here is the problem (now I realize I should have used latex):

Consider a linear transformation T:R^3\rightarrow{}S where S is the vectorial space of all the real matrixes 2x2 symmetricals. So T cannot be biyective.

And this is false but I do not understand why unless the dimension of a symmetrical matrix is the number of different entries so this one of 2x2 would have dimension 3 and so dim image (T) less or equal to 3 and there I conclude that as dim image (T) = dim W the transformation is supraeyective and since the kernel is 0 it is inyective so in the end it can be biyective. But is it correct to assume that the dimension of a symmetrical matrix is the number of different entries?

 

[JosephR]

If You have an Matrix nxn ~> It's dimension= nxn
for example the dimension of a 3x5 matrix is 15

if you have a polynimial it's dimension is n+1

And if You have for Example R^n it's dimension is n.

 

[CompuChip]

Ah, you mean the dimension of the vector space of 2x2 symmetric matrices. Well, that's easily verified by writing down a basis for it
How many matrices do you need to express and symmetric 2x2 (real) matrix as a linear combination of them? Of course,
\{ <br /> B_1 = \begin{pmatrix} 1 &amp; 0 \\ 0 &amp; 0 \end{pmatrix},<br /> B_2 = \begin{pmatrix} 0 &amp; 1 \\ 0 &amp; 0 \end{pmatrix},<br /> B_3 = \begin{pmatrix} 0 &amp; 0 \\ 1 &amp; 0 \end{pmatrix},<br /> B_4 = \begin{pmatrix} 0 &amp; 0 \\ 0 &amp; 1 \end{pmatrix}<br /> \}<br />
is a basis for all 2x2 matrices, of which the symmetric ones are a subspace... so it will be at most 4.
But based on your previous post you expect 3 (because you claim to have an isomorphism from \mathbb R^3 to these matrices, which means that every element (x, y, z) of \mathbb R^3 is uniquely mapped to a linear combination x A_1 + y A_2 + z A_3 of 2x2 symmetric matrices, where \{A_1, A_2, A_3\} is the basis for the vector space of symmetric matrices).

 

[pgandalf]

Yes, that's it. Now I understand. Thank you both!

 

[CompuChip]

So what basis did you find?

 

[pgandalf]

\left\{{\begin{bmatrix}{1}&amp;{0}\\{0}&amp;{0}\end{bmatrix},\begin{bmatrix}{0}&amp;{1}\\{1}&amp;{0}\end{bmatrix},\begin{bmatrix}{0}&amp;{0}\\{0}&amp;{1}\end{bmatrix}}\right\}

However, what I needed to know was if the dimension was 3 because 3= Dim N(T) + Dim Im(T) and Dim Im(T)< or equal to Dim W that is equal to 3
so in the case that it is equal to 3 the dimension of the kernel is 0 and is
inyective and suprayective-->biyective.

 

[CompuChip]

Yes very nice.
Even dim(ker(T)) = 0 you can check, the only element mapped to the 0 matrix is the origin of R^3.