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?


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.


**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 $n(n+1)/2$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.