# Show that the set of all symmetric, real matrices is a subspace, determine the dimension

Question: Let $V \subset M(n,n,\mathbb{R})$ be the set of all symmetric, real $(n \times n)$ matrices, that is $a_{ij} = a_{ji}$ for all $i,j$. Show that $V$ is a subspace of $M(n,n,\mathbb{R})$ and calculate dim$(V)$.

My attempt so far: First part: To show that $V$ is a subspace I need to show: (a) $0 \in V$ and (b) $\forall A,B \in V: (i) A + B \in V (ii) \lambda A \in V$

For (a) I would say: Let $a_{ij} \in 0$(this should represent a zero matrix, is that how to write it?)

$a_{ij} = 0 = a_{ji} \Rightarrow 0 \in V$

For (b) I am actually confused since I would first think: both a $(2 \times 2)$ matrix and a $(3 \times 3)$ matrix belong to $V$ but addition of matrices of different size is undefined $\Rightarrow$ $V$ is not closed under addition $\Rightarrow$ $V$ is not a subspace of $M(n,n,\mathbb{R})$... what am I missing here? (To start I don't really understand the notation $M(n,n,\mathbb{R})$... what exactly does the $\mathbb{R}$ represent there?).

Disregarding my confusion I would still try to show (b), but my mathematical notation is still lacking competence... Is the following somewhat clear? Would anyone ever use "$\in$" to denote "is an element of matrix"?

(i)Let $a_{ij},a_{ji} \in A$ and $b_{ij}, b_{ji} \in B$. Let $A,B \in V$

$\Rightarrow a_{ij} = a_{ji}, b_{ij} = b_{ji}$

$A + B = C \Rightarrow c_{ij} = (a_{ij}+b_{ij}) = (a_{ji} + b_{ij}) = (a_{ij} + b_{ji}) = c_{ji} = (a_{ji} + b_{ji})$

$\Rightarrow C \in V$

(ii) Let $A\in V, \lambda \in \mathbb{R}$. Let $a_{ij},a_{ji} \in A$.

$\Rightarrow a_{ij} = a_{ji}$

$\lambda \cdot A = A'$ with $\lambda a_{ij} = \lambda a_{ji} \Rightarrow A' \in V$

Second part: I feel that I understand the answer... For an $(n \times n)$ matrix, the diagonal length $= n$ and these are the elements which have no counterpart and are not critical to the symmetry. When these elements are subtracted from the total$(n^{2})$, half of the remainder can be independently selected and the other half will follow as a result. Therefore I think it makes sense to write that dim$(V) = n + \frac{n^{2}-n}{2}$.

Is this correct? If so, given the context of the exercise, how could I make my answer more acceptable?

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+1 for showing your work. –  Arturo Magidin Nov 30 '10 at 20:50

To write that the matrix is the zero matrix, you should write "let $a_{ij}=0$ for all $i$ and $j$", not "$a_{ij}\in 0$". (Nothing is an element of $0$).

For (b): No, notice that the $n$ is fixed. You are only considering matrices that are symmetric of a fixed size. If $n=2$, then you only consider $2\times 2$ matrices; if $n=3$, then you only consider $3\times 3$ matrices. You never consider both $2\times 2$ and $3\times 3$ matrices at the same time.

$M(n,n,\mathbb{R})$ means:

• Matrices (that's the "$M$");
• with $n$ rows (that's the first $n$);
• with $n$ columns (that's the second $n$);
• and each entry is a real number (that is the $\mathbb{R}$).

So $M(2,3,\mathbb{Z})$ would mean "matrices with 2 rows and 3 columns each, and each entry is an integer." $M(7,2,\mathbb{C})$ means "matrices with 7 rows and 2 columns, and every entry is a complex number. Etc.

The way you want to say (b) is: Let $A=(a_{ij})$ and $B=(b_{ij})$ (that is, let's call the entries of $A$ "$a_{ij}$", and let's call the entries of $B$ "$b_{ij}$"). Because $A$ is symmetric, we know that for all $i$ and $j$, $a_{ij}=a_{ji}$; and since $B$ is symmetric we know that for all $i$ and $j$, $b_{ij}=b_{ji}$. Now, let $C=A+B$. If we call the $(i,j)$th entry of $C$ "$c_{ij}$", then you want to show that $c_{ij}=c_{ji}$ for all $i$ and $j$. How do you do that? You use the fact that you can express $c_{ij}$ in terms of the entries of $A$ and of $B$, and that $A$ and $B$ are symmetric, pretty much how you did; it's just a matter of writing it clearly. Same with the scalar multiplication.

Your argument for the second part is essentially correct. To make it water tight, you should produce a basis of the appropriate size. Each element would be used to determine a particular entry as you describe them.

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**Let $A$ and $B$ be symmetric matrices of the same size. Consider $A+B$. We need to prove that $A+B$ is symmetric. This means $(A+B)^\mathrm{T}=A+B$. Recall a property of transposes: the transpose of a sum is the sum of transposes. Thus $(A+B)^\mathrm{T}=A^\mathrm{T}+B^\mathrm{T}$. But $A$ and $B$ are symmetric. Thus $A^\mathrm{T}=A$ and $B^\mathrm{T}=B$. So $(A+B)^\mathrm{T}=A+B$ and the proof is complete. For second part all $n\times n$ symmetric matrices form a vector space of dimension $n(n+1)/2$. Proof- Let $A= (a_{ij})$ be an $n\times n$ symmetric matrix. then $a_{ij}+=a_{ji}$, for $i$ not equal to $j$. Thus the independent entries are $a_{ij}$ (i less than j) and $a_{ii}$ . where $i$ varies from 1 to $n$. And these are $n(n-1)/2+ n = n(n+1)/2$. Hence the space has dimension $n(n+1)/2$

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