Question #254077

Let n∈\inN consider the set of nxn symmetric matrices over R with the usual addition and multiplication by a scalar

3.1. Show that this set with the given operations is a vector subspace of Mnn

3.2. What is the dimension of this vector subspace?

3.3. Find a basis for the vector space of 2x2 symmetric matrices.


1
Expert's answer
2021-10-21T10:12:20-0400

1)

We depend on that Mn,nM_{n,n} is a vector space with respect to usual matrix operation addition and multiplication by a scalar. To prove that space Sn,nS_{n,n} of all symmetric matrices is a such vector space also is sufficient to prove that Sn,nS_{n,n} is closed for mentioned operations. Let A,B∈Sn,nA,B\in S_{n,n} , so âˆ€(i,j,1≤i,j≤n)(Ai,j=Aj,i,Bi,j=Bj,i)\forall(i,j,1\le i,j \le n)(A_{i,j}=A_{j,i},B_{i,j}=B_{j,i})

Let C=A+B,or âˆ€(i,j,1≤i,j≤n)(Ci,j=Ai,j+Bi,j)\forall(i,j,1\le i,j \le n)(C_{i,j}=A_{i,j}+B_{i,j})

We must show that C∈Sn,nC\in S_{n,n} . Let i,j be any indexes such that 1≤i,j≤n1\le i,j\le n

Then Cj,i=[by definition]=Aj,i+Bj,iC_{j,i}=[by\space definition ]=A_{j,i}+B_{j,i} But A,B∈Sn,nA,B\in S_{n,n} , therefore

Aj,i=Ai,j,Bj,i=Bi,j=>Aj,i+Bj,i=Ai,j+Bi,j=Ci,jA_{j,i}=A_{i,j},B_{j,i}=B_{i,j}=>A_{j,i}+B_{j,i}=A_{i,j}+B_{i,j}=C_{i,j}

Thus C∈Sn,nC\in S_{n,n}

Further, let A be any matrix from Sn,nS_{n,n} and c∈Rc\in R any number, our task is to prove that (cA)∈Sn,n(cA)\in S_{n,n} or is symmetric. Let i,j- be any coefficients such, that 1≤i,j≤n1\le i,j\le n , then (cA)i,j=câ‹…Aj,i=[A∈Sn,n]=câ‹…Ai,j=(câ‹…A)i,j(cA)_{i,j}=c\cdot A_{j,i}=[A\in S_{n,n}]=c\cdot A_{i,j}=(c\cdot A)_{i,j} that is equivalent to (cA)∈Sn,n(cA)\in S_{n,n}

2)

For linear space Sn,nS_{n,n} basis may be formed from elementary matrices Ek,m∈Sn,nE^{k,m}\in S_{n,n} defined as (Ei,jk,m=1)≡((i=k,j=m)∨(i=m,j=k))(E_{i,j}^{k,m}=1)\equiv((i=k,j=m)\lor (i=m,j=k))

Basis in Sn,nS_{n,n} is {Ek,m,1≤k≤m≤n}\lbrace E^{k,m}, 1\le k\le m\le n \rbrace with cardinalyty N=1+2+...+n=nâ‹…(n+1)2\frac{n\cdot (n+1)}{2} and therefore dim(Sn,n)=nâ‹…(n+1)2dim(S_{n,n})=\frac{n\cdot (n+1)}{2} .

if X=(xi,j)1≤i,j≤nX=(x_{i,j})_{1\le i,j\le n} any matrix from Sn,nS_{n,n} then

X=∑1≤i≤j≤nxi,j⋅Ei,jX=\sum_{1\le i \le j \le n}x_{i,j}\cdot E^{i,j} - decomposition X by baisis.

3) If n=2 we have basis in S2,2={(1000).(0110),(0001)}S_{2,2}=\lbrace \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}. \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \rbrace


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