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My book asks for the dimensions of the vector spaces for the following two cases:

1)vector space of all upper triangular $n × n$ matrices, and

2)vector space of all symmetric $n × n$ matrices

The answer for both is $n(n+1)/2$ and this is easy enough to verify with arbitrary instances but what is the formal way to conclude this in the general case, as per the question?

Thanks!

EDIT: I understand how bases and dimensions and linear independe relate. I'm asking how to find the number of dimensions for those two cases above WITHOUT just counting examples where n is equal to, say, 4, 5, 6...

Krpcannon
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1 Answers1

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One way to formally prove that a vector space has a given dimension is to exhibit a basis and count the number of elements. In fact, the number of basis elements is one definition of dimension.

So, for instance, in the case of symmetric matrices, let $E_{ij}$ be the matrix which is all zeros except for the $i,j$ and $j,i$ entries, which are one. The set $S=\{ E_{ij}, i=1,...,n, i\leq j\}$ is a basis for the space, which can be shown by observing that an arbitrary symmetric matrix $A=(a_{ij})$ can be written as $\sum_{i\leq j} a_{ij} E_{ij}$ and that this is the unique decomposition of $A$ as a linear combination of the $E_{ij}$.

Zach Boyd
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