Questions on symmetric matrices [closed]
HINT: If you know all of the elements on and above the diagonal of a symmetric matrix, you know the whole matrix. How many elements are there on or above the diagonal of an $n\times n$ matrix?
Added: I can see that you're having trouble getting a handle on the vector space in question; perhaps this will help. Let $S_n$ be the space of $n\times n$ symmetric matrices. In the simplest case that isn't completely trivial, $n=2$, the elements of $S_2$ are matrices of the form $$\pmatrix{a&b\\b&c}\;.$$ Vector addition in $S_2$ is just ordinary matrix addition: $$\pmatrix{a_1&b_1\\b_1&c_1}+\pmatrix{a_2&b_2\\b_2&c_2}=\pmatrix{a_1+a_2&b_1+b_2\\b_1+b_2&c_1+c_2}\;.$$ Note that the result of this addition is still symmetric, so it really is in $S_2$. If it weren't, $S_2$ wouldn't be closed under addition and therefore wouldn't be a vector space after all.
Scalar multiplication in $S_2$ is ordinary multiplication of a matrix by a scalar: $$\alpha\pmatrix{a&b\\b&c}=\pmatrix{\alpha a&\alpha b\\\alpha b&\alpha c}\;,$$ and again all's well, since the result is still in $S_2$.
Here's a simple exercise to help you get more accustomed to working with this vector space.
Let $V=\{\langle a,b,c,d\rangle\in\Bbb R^4:b=c\}$.
- Prove that $V$ is a subspace of $\Bbb R^4$.
- Prove that $V$ is isomorphic to $S_2$. That is, find a linear transformation $T:V\to S_2$ that is one-to-one and maps $V$ onto $S_2$.
If $A$ is a symmetric $n\times n$ matrix, then $A$ has the form $$ \begin{bmatrix} * \ & a_1 & a_2 & \cdots & a_k \\ a_1 & * \ & a_3 \\ a_2 & a_3 & * \ \\ \vdots & & & \ddots & \\ \\ a_k & & & & * \ \end{bmatrix} $$ where the $*$ entries are whatever you like them to be. You can see that we have $a_{ij}=a_{ji}$.
From this form you can see that we need $n$ elements in the basis to span the diagonal entries. For the remaining $n(n-1)$ entries, we need exactly $\frac{1}{2}n(n-1)$ elements in the basis to in order to span those entries (due to the fact that $a_{ij}=a_{ji}$). This gives a basis with $\frac{1}{2}n(n+1)$ elements.
Define $T_{ij}$ to be the matrix with $(T_{ij})_{ij}=1$ and all other entries equal to $0$. Then define $$ M_{ij} = T_{ij}+T_{ij}^\text t $$ where $i$ and $j$ range over $1,2,\dots, n$. Then for a given $n\times n$ symmetric matrix $A$, we can write it as $$ A = \sum_{i=1}^n\sum_{j=1}^n \frac{1+\delta_{ij}}{2}(A)_{ij}M_{ij} $$ where $(A)_{ij}$ denotes the $(i,j)^\text{th}$ entry of the given matrix $A$. The $\frac{1+\delta_{ij}}{2}$ in the sum is to correct for the fact that $M_{ij}=M_{ji}$.
The collection of the distinct $M_{ij}$ will form a basis for the space of $n\times n$ symmetric matrices. Of course, this is not proof, but provides a way that you might express the basis.