Question #144988
Consider the real vector space Mn(R), of all
n x n matrices with entries from the set of
real numbers with respect to the usual
addition and scalar multiplication of
matrices. Find the smallest subspace of
Mn(R) which contains the identity matrix.
Also show that the set of all symmetric
matrices is a subspace of Mn(R).
1
Expert's answer
2020-11-19T17:10:07-0500

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 linear 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,

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

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS