Linear Algebra Symmetric Matrix Set Question

[Gamble93]

First of all, I apologize if this is in the wrong place. I didn't really know where it should be placed and if it is in the wrong place I am sorry.

This question was on my recent Linear Algebra I final exam and I had no idea how to do it when I was writing the exam and I'm still stumped by it. I'm simply curious as how to go about obtaining the solution mostly because it has been bugging me for a few weeks now. Also, this is the first time using LaTeX typesetting so forgive me if it's sloppy.

Let [itex]A[/itex] be any [itex] n \mbox{ by } n [/itex] matrix.

Let [itex]S[/itex] be defined as:

[tex] \displaystyle{S=\{A|A=A^T\}} [/tex]
​

What is the size of [itex]S[/itex] ?

Thanks in advance for any help.

 

[DivisionByZro]

I suppose you're asking for the dimension of the basis set when you say "size"? I'll assume so. The way to approach this problem is to determine if your set of all symmetric matrices forms a subspace of R^n and then find a suitable basis. The dimension will be the number of elements in your basis set. This is what I think you're asking, but I find your question vague. Do you know how to do this?

 

[micromass]

In addition to division's answer: the dimension of a space can often be seen as the number of free variables. For example, a 2x2-matrix

[tex]\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)[/tex]

has 4 free variables so has dimension 4 (this is NOT a rigorous argument, but merely an intuition). However, if you want the matrix to be symmetric, then it must happen that b=c. So now there are only 3 free variables, so the dimension is 3.

The question now asks you to calculate this for nxn matrices.

 

[Gamble93]

Thank you both for your answers. The question I posted was word for word what was on my exam. I also assumed that it was the dimension of the set he was asking for. When I get an opportunity I'll see if I can figure out the answer. Thanks again both of you.

 

[vish_maths]

In addition to division's answer: the dimension of a space can often be seen as the number of free variables. For example, a 2x2-matrix

[tex]\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)[/tex]

has 4 free variables so has dimension 4 (this is NOT a rigorous argument, but merely an intuition). However, if you want the matrix to be symmetric, then it must happen that b=c. So now there are only 3 free variables, so the dimension is 3.

The question now asks you to calculate this for nxn matrices.

Hello, i find a paradox here. We know that dimension implies the number of vectors in a basis set needed to traverse the whole space. And the basis theorem states that a matrix of size nxn ( be it symmetric or not ) , needs only n vectors in it's basis set to traverse the space ?

Then, where could i be going wrong ?

 

[Deveno]

one basis for the set of all 2x2 matrices is:

[tex]\begin{bmatrix}1&0\\0&0 \end{bmatrix},\begin{bmatrix}0&1\\0&0 \end{bmatrix},\begin{bmatrix}0&0\\1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1 \end{bmatrix}[/tex]

however, since for a symmetric matrix the 2,1 entry must equal the 1,2 entry, we only need the following 3 matrices as a basis:

[tex]\begin{bmatrix}1&0\\0&0 \end{bmatrix}, \begin{bmatrix}0&1\\1&0 \end{bmatrix}, \begin{bmatrix}0&0\\0&1 \end{bmatrix}[/tex]

for the general nxn case, it helps to divide the matrix entries into 2 disjoint sets:

diagonal entries, off-diagonal entries

if a matrix is symmetric, it tells us something about the off-diagonal enties (which are of course, divided into two equal subsets: above diagonal, and below diagonal entires). can you think of what this might be?

Hello, i find a paradox here. We know that dimension implies the number of vectors in a basis set needed to traverse the whole space. And the basis theorem states that a matrix of size nxn ( be it symmetric or not ) , needs only n vectors in it's basis set to traverse the space ?

Then, where could i be going wrong ?

dim(Mat(n,F)) = n², not n.

 

[vish_maths]

one basis for the set of all 2x2 matrices is:

[tex]\begin{bmatrix}1&0\\0&0 \end{bmatrix},\begin{bmatrix}0&1\\0&0 \end{bmatrix},\begin{bmatrix}0&0\\1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1 \end{bmatrix}[/tex]

however, since for a symmetric matrix the 2,1 entry must equal the 1,2 entry, we only need the following 3 matrices as a basis:

[tex]\begin{bmatrix}1&0\\0&0 \end{bmatrix}, \begin{bmatrix}0&1\\1&0 \end{bmatrix}, \begin{bmatrix}0&0\\0&1 \end{bmatrix}[/tex]

for the general nxn case, it helps to divide the matrix entries into 2 disjoint sets:

diagonal entries, off-diagonal entries

if a matrix is symmetric, it tells us something about the off-diagonal enties (which are of course, divided into two equal subsets: above diagonal, and below diagonal entires). can you think of what this might be?



dim(Mat(n,F)) = n², not n.

Yep got it :) was confusing the column space :)