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I am aware that a similar question has been asked here, among other questions, but I feel that my question is different because I am actually trying to write up a very rigorous proof that such a set spans the $n \times n$ symmetric matrices and that it is linearly independent.

I am having trouble with the proof because I know exactly what the basis is, but I'm having a difficult time proving it to the reader.

$\textbf{Problem}$: Find a basis for the space of $n \times n$ symmetric matrices.

$\textbf{Solution:}$ Let $\mathbf{B}$ be the set containing $n$ $n \times n$ matrices, each with a single 1 in a different entry of the diagonal and a 0 elsewhere. Now for an $n \times n$ matrix, there are $n^2 - n$ entries not on the diagonal. Half of these, or $\frac{n(n - 1)}{2}$, are above the diagonal. Also let $\mathbf{B}$ contain the $\frac{n(n - 1)}{2}$ matrices with a 1 in an entry above the diagonal and its reflection across the diagonal and 0's elsewhere.

We claim that $\mathbf{B}$ is a basis. First we show that $\mathbf{B}$ spans the set of $n \times n$ symmetric matrices. Let $A$ be an $n \times n$ symmetric matrix. Let $a_{i, j}$ be the $i$th entry of the $j$th row. Select $V_{i, j}$ in $\mathbf{B}$ such that the $(V_{i, j})_{i, j} = 1$.

Then $A = \sum_{i = 1}^{n}\sum_{j = 1}^n a_{i, j}V_{i, j}$.

[Now one thing I'm not sure how to do here is to account for the fact that I've added in too much due to the symmetry of the matrices. Could I divide by 2?]

Now we show that $\mathbf{B}$ is linearly independent. There are $k = n + \frac{n(n - 1)}{2}$ elements of $\mathbb{B}$. Suppose that $a_1v_1 + a_2v_2 + \ldots + a_kv_k = 0$.

[I'm having trouble going from here. Once I assume that this is equal to 0, I'm not sure how to conclude that the $a_i$ must be 0.]

...

As you can see, I'm having quite a difficult time articulating this idea to the reader.

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  • $\begingroup$ What are you asking aside from [Now one thing I'm not sure how to do here is to account for the fact that I've added in too much due to the symmetry of the matrices. Could I divide by 2?]? $\endgroup$ – shooting-squirrel Aug 16 '14 at 3:31
  • $\begingroup$ I'm not sure how to show that $\mathbb{B}$ is linearly independent after I set the sum I have written equal to 0. I am trying to show that $a_i = 0$ for all $i$. $\endgroup$ – Joseph DiNatale Aug 16 '14 at 3:34
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Choose a symmetric $A$. Note that you only need to choose $a_{ij}$ where $j \ge i$. In particular, you have $A = \sum_{i\le j} [A]_{ij} V_{ij}$ (where $[A]_{ij}$ denotes the $ij$ entry of $A$). This shows that the $V_{ij}$ span the set of symmetric matrices.

In the following, $e_n$ denotes the vector of zeros with a one in the $n$th position. Note that $e_m^T A e_n = [A]_{mn}$ for any matrix, symmetric or not.

Now suppose $\sum_{i\le j} \alpha_{ij} V_{ij} = 0$. If we have $m \le n$, then $e_m^T V_{ij} e_n = \delta_{mi} \delta_{jn}$, so $e_m^T(\sum_{i\le j} \alpha_{ij} V_{ij}) e_n = 0 = \sum_{i\le j} \alpha_{ij} e_m^T V_{ij} e_n = \alpha_{mn} $, and so $\alpha_{mn} = 0$. Hence the $V_{ij}$ (with $i \le j$) are linearly independent.

Hence $\{ V_{ij} \}_{i \le j}$ forms a basis for the set of symmetric matrices.

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