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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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closed as off-topic by Did, Siminore, Ivo Terek, Claude Leibovici, Carl Mummert Sep 22 '14 at 10:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Siminore, Ivo Terek, Claude Leibovici, Carl Mummert
If this question can be reworded to fit the rules in the help center, please edit the question.

Is this homework? If it is, it is customary to label it so, and to mention any attempts you have made. – Martin Argerami Apr 30 '12 at 16:08
@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. – Michael Hardy Apr 30 '12 at 16:16
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. – Jim_CS Apr 30 '12 at 16:20
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? – Did Apr 30 '12 at 18:13
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) – Jim_CS Apr 30 '12 at 22:19
up vote 10 down vote accepted

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? – Jim_CS Apr 30 '12 at 17:01
@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. – Brian M. Scott Apr 30 '12 at 17:03
@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) – Andrea Mori Apr 30 '12 at 17:13
@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. – Brian M. Scott Apr 30 '12 at 17:15
@Jim_CS: I'll expand my answer a bit to try to give you a better idea of what the space itself is like. – Brian M. Scott Apr 30 '12 at 22:24

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.

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