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i understand that i need to find a basis for the given set of matrices.

i think i have found one using the set of matrices shown below.

but i am having trouble figuring out how to show linear independence and generation/spanning.

$$ S_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}, S_2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}, S_3 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix},$$ $$S_4 = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}, S_5 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix}, S_6 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix} $$

linear dependence

for linear independence, i start with a linear combination of the six matrices above.

$a \cdot\!S_1 + b \cdot\!S_2 + c \cdot\!S_3 + d \cdot\!S_4 + e \cdot\!S_5 + f \cdot\!S_6 = 0$

putting each coefficient in the right place(s) creates a new matrix:

$$ A = \begin{pmatrix} a & d & e \\ d & b & f \\ e & f & c \\ \end{pmatrix} $$

i know this matrix is symmetric because $A = A^t$

$$ \begin{pmatrix} a & d & e \\ d & b & f \\ e & f & c \\ \end{pmatrix}^t = \begin{pmatrix} a & d & e \\ d & b & f \\ e & f & c \\ \end{pmatrix} $$

i am stuck here. i'm used to using row operations to show linear independence, but i am unsure of how to implement that here because i am not working with sets of vectors or polynomials. i do not know what to do with matrices because they are already matrices.

should i be using row operations at all though? if i set $a = b = c = d = e = f = 0$, that will give a 3x3 zero matrix and it will show that the only linear combination of the candidate basis matrices is the trivial representation, and that would indicate linear independence. but i am not sure that it's enough to just say this. i think i'm supposed to show it somehow.

generation/spanning

i am even more lost here. i would use row operations for this as well and i have with other problems involving sets of vectors and sets of polynomials, but i cannot figure out how to start this part of the problem either.

i realize that if the candidate basis i've chosen is indeed a basis then the dimension i'm looking for is 6, but i'm having trouble showing that this set meets the conditions for being a basis.

qwerty
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    As you noted, you get zero if and only if each coefficient is zero, so you are done with that part; linear independence is proved. This is not unexpected - the space of 3x3 matrices has 9 degrees of freedom (one for each entry). 3 of your matrices use up one degree of freedom, while the other 3 use up two degrees of freedom. Thus all 9 degrees are used up. – Brevan Ellefsen Jan 07 '20 at 12:10
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    Also, if they are independent then the smallest space they can form is exactly the space using each as a basis vector, which here is dimension 6. Clearly any 3x3 symmetric matrix is realizable this way, so we are done. The general case for $n \times n$ matrices follows almost the same (see here for more details) – Brevan Ellefsen Jan 07 '20 at 12:14

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when you say \begin{align} a \cdot\!S_1 + b \cdot\!S_2 + c \cdot\!S_3 + d \cdot\!S_4 + e \cdot\!S_5 + f \cdot\!S_6 = 0 \end{align} the $0$ on the right hand side is the zero vector in the vector space of $3\times3$ matrices. i.e. this equation says $$ \begin{pmatrix} a & d & e \\ d & b & f \\ e & f & c \\ \end{pmatrix} = \begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\\ \end{pmatrix}$$ Since two matrices are equal iff all their entries are, this immediately implies that $a=b=c=d=e=f=0$,

Spanning is also pretty simple, write down an arbitrary symmetric matrix. Say $$ A = \begin{pmatrix} a & d & e \\ d & b & f \\ e & f & c \\ \end{pmatrix}$$ it turns out that $A = a \cdot\!S_1 + b \cdot\!S_2 + c \cdot\!S_3 + d \cdot\!S_4 + e \cdot\!S_5 + f \cdot\!S_6 $. That's the end of the proof.

Calvin Khor
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