If $B \in \mathbb{M}_n (\mathbb{R})$ is a symmetric matrix (i.e. $B = B^T$), how many entries of B can be chosen independently.
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$\begingroup$ Totally on the wrong track. This is extremely basic. $\endgroup$– Matt SamuelCommented Mar 6, 2016 at 20:23
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$\begingroup$ @MattSamuel Any hints, then? $\endgroup$– RatonCommented Mar 6, 2016 at 20:23
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$\begingroup$ HINT: Being symmetric means $a_{ij}=a_{ji}$. $\endgroup$– user112358Commented Mar 6, 2016 at 20:24
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1$\begingroup$ If you pick an off-diagonal entry, what happens on the other side of the diagonal? Does this affect any other position? $\endgroup$– Matt SamuelCommented Mar 6, 2016 at 20:25
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1$\begingroup$ No. The point is that the matrix is symmetric along the diagonal. Pick one entry it determines another one. One of the mysteriously downvoted answers hints that $a_{ij}=a_{ji}$. That's really all you need to know. $\endgroup$– Matt SamuelCommented Mar 6, 2016 at 20:29
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2 Answers
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The upper (or lower) triangular part of a symmetric matrix completely determines the other half. It is generally enlightening to look at low-dimensional examples to see a pattern:
$$\begin{pmatrix}a_1 & a_2 \\ * & a_3 \end{pmatrix}, \begin{pmatrix}a_1 & a_2 & a_3 \\ * & a_4 & a_5 \\ * & * & a_6\end{pmatrix}, \cdots$$
There is complete freedom in choosing the $a_i$ and no freedom in choosing $*$. Therefore the space of symmetric $n \times n$ matrices is $1 + 2 + \cdots + n = \binom{n+1}{2}$-dimensional.
answered Mar 6, 2016 at 20:27
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In a symmetric matrix $A=(a_{i,j})_{1\le i,j\le n}$ which has $n^2$ entries we have $$a_{i,j}=a_{j,i}$$ so we can choose independently the diagonal entries $a_{i,i}$ for $i=1,\ldots,n$ and the entries $a_{i,j}$ for $1\le i<j\le n$ (all off diagonal entries half). Hence we have $n+\frac{n^2-n}{2}=\frac{n(n+1)}{2}$ independent entries.
