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Describe a basis for the vector space of symmetric n x n matrices. What is the dimension of this space?

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    Is this homework? If it is, it is customary to label it so, and to mention any attempts you have made.2012-04-30
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    @Jim_CS : My guess is that whoever down-voted this question did so because it's written as if you copied a question written by someone other than yourself.2012-04-30
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    It was in an exam i had today and i never came across it during the course or pre exam study so i had no answer for it. well i said the dimension was n as that seemed obvious.2012-04-30
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    Reading the comments, I feel you could first try to answer the following questions: (i) Describe a basis for the vector space of **all** the $n\times n$ matrices? (ii) What is its dimension?2012-04-30
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    I dont know what all the downvotes are for...what else am I supposed to put in the post when I wasnt even able to make an attempt at this question in the exam? (apart from putting dim = n, which seems wrong in any case)2012-04-30
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    The title of this question should be "Example basis for space of $n \times n$ symmetric matrices".2014-09-22

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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 $V=\{\langle a,b,c,d\rangle\in\Bbb R^4:b=c\}$.

  1. Prove that $V$ is a subspace of $\Bbb R^4$.
  2. 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 $S_2$.
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    Well there are n elements on the diagonal so the dimension is n, yes?2012-04-30
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    What about the part "Describe a basis for the vector space of symmetric n x n matrices." This looks ambiguous to me, is it asking about the vector space that contains all symmetric nxn matrices (the matrix space in other words) or it is asking about the vector space that is represented by the columns in a symmetric matrix?2012-04-30
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    @Jim_CS: There are $n$ elements on the diagonal, but specifying them isn't enough to specify the whole matrix, so the dimension of the space is *more* than $n$. You also have to specify the elements *above* the diagonal. How many of those are there? As for your other question, it means exactly what it says: the space of $n\times n$ symmetric matrices, i.e., the space whose elements are these matrices.2012-04-30
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    Say I have the 3x3 identity matrix. That has three 1's on the diagonal, three linearly independent columns, so its dimension is 3. Why is the same not true of a symmetric matrices in general?2012-04-30
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    @Jim_CS : it seems to me that you're making a confusion between the dimension of the space spanned by the column of a *single* matrix (i.e. its rank) and the dimension of the space of *all* (symmetric) matrices, which is a vector space itself (the "vectors" are the matrices)2012-04-30
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    @Jim_CS: The $3\times 3$ identity matrix isn't a vector space, so it doesn't even *have* a dimension. Its *rank* is $3$. That aside, you're not thinking about what the question actually asks. You have a vector space $V$ whose elements $-$ the actual vectors in $V$ $-$ are $n\times n$ symmetric matrices. Each matrix is one vector in $V$. You could write it out as a single row of $n^2$ numbers instead of as a square array, except that it would be much harder to tell that it was symmetric.2012-04-30
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    Hmmm, cheers guys I will have to look into this, I have next to know experience with vector spaces with 'matrix vectors'.2012-04-30
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    @Jim_CS: I'll expand my answer a bit to try to give you a better idea of what the space itself is like.2012-04-30
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    Thanks alot for that extra info mate, I have a better concept of what's happening now.2012-05-01
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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.