Question #253084

let n€N consider the set of nxn symmetric matrices over R with the usual addition and multiplication by a scalar. show that this set with the given operations is a vector subspace of Mnn

1
Expert's answer
2021-11-03T18:33:09-0400



We have

V=Mn(R)V=M_n(\R)


is the vector space over R\R .


Now,

denote AijA_{ij} is the matrix whose (i,j)th(i,j)^{th} entry is 11 if i=ji=j

and ;if


ij,i,j{1,,n}i\ne j,\,\forall i,j\in \{1,\dots,n\}

Also denote the collection of all such matrix by XX ,

thus;


W=span(X)W=\text{span}(X) is the smallest subspace


such that In×nWI_{n\times n}\in W .


Define


Mns(R):={AMn(R):A=AT}M_n^s(\R):=\{A\in M_n(\R):A=A^T\}



Clearly, Mns(R)M_n^s(\R) is the set of all symmetric matrix. we show that it is subspace of Mn(R)M_n(\R) .


Let, a,bRa,b\in \R and A,BMns(R)A,B\in M_n^s(\R) , thus AT=A,BT=BA^T=A,B^T=B


Now, consider the liner combination aA+bBaA+bB .


Note that


(aA+bB)T=(bB)T+(aA)T=(aA)T+(bB)T=aAT+bBT=aA+bB(aA+bB)^T=(bB)^T+(aA)^T\\ =(aA)^T+(bB)^T\\ =aA^T+bB^T\\ =aA+bB

Thus,we can see that


aA+bBMns(R)aA+bB\in M_n^s(\R)



Hence,it can be seen that



Mns(R)M_n^s(\R) is subspace of Mn(R)M_n(\R)



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