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Let $M = \{ A \in \mathbb{R}^{n \times n} | A^{T} = A \}$ be the space of symmetrical matrices. How do I calculate the dimension of this space?

As far as I understand the dimension of a space is equal to the number of vectors of the basis of the space. So for this space the Basis must have $n$ vectors, therefore my intuition is that the $dim(M)$ should be $=n$.

According to the solution of this exercise, the dimension is $\frac{n(n+1)}{2}$. I don't understand how to get there.

Can someone give a intuitive answer why my simple approach is wrong and how the solution makes sense?

Chris
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    But why do you think the basis must have $n$ vectors? – Stuck Jun 11 '21 at 09:09
  • I was under the impression that in order to span the Space $\mathbb{R}^{n \times n}$ $n$ vectors are needed. If I had $n+1$ vectors then it wouldn't be a basis anymore because at least one of those would not be linear independent anymore, right? – Chris Jun 11 '21 at 09:12
  • You are thinking of $\mathbb{R}^n$. Here, you have $\mathbb{R}^{n\times n}=\mathbb{R}^{n^2}$. But you can't just say it's $n^2$, because you aren't using the fact that they are symmetric matrices. What you should ask yourself is: how many parameters do I need to specify a symmetric matrix? – Stuck Jun 11 '21 at 09:15
  • Ah okay thank you a lot for the intuition to think about "how many parameters do I need to specify the matrix". That helps. So I just look at the number of entries that I need. Therefore I need $n^2$ parameters but since there's duplicates below the diagonal I need to subtract those. I didn't understand how to come up with that formula yet but I think I will get there somehow. – Chris Jun 11 '21 at 09:27

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Use the identification $\mathbb{R}^{n\times n} = \mathbb{R}^{2n}$. Hence, you want to find the number of independent entries of your symmetric matrices. The lower triangle part is determined by the upper triangular part. Therefore, only the upper triangle and the diagonal entirely determine the matrix. This reduces to counting the number of elements of the upper triangle and the diagonal. On the first line there are $n$ elements, on the second line there are $n-1$, and so. You thus have a dimension of $$ \sum_{k = n}^{1} k = \frac{n(n+1)}{2} $$

Rundasice
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Let $n=3$. The simplest basis for this space is given by $$\left\{\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \begin{pmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\right\} $$

As you can, see you need more than $n=3$ basis vectors. Can you generalize from here?

Cade Reinberger
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