Answer to Question #253088 in Linear Algebra for Sabelo Xulu

Question

let n€N consider the set of nxn symmetric matrices over R with the usual addition and multiplication by a scalar.

a) Show that this set with the given operations is a vector subspace of M_nn

_{b) What is the dimension of this vector subspace?}

_{c) Find a basis for the vector space of 2x2 symmetric matrices.}

Answer 1

a)

We have

$V=M_n(\R)$

is the vector space over $\R$ .

Now,

denote $A_{ij}$ is the matrix whose $(i,j)^{th}$ entry is $1$ if $i=j$

and ;if

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

Also denote the collection of all such matrix by $X$ ,

thus;

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

such that $I_{n\times n}\in W$ .

Define

$$M_n^s(\R):=\{A\in M_n(\R):A=A^T\}$$

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

Let, $a,b\in \R$ and $A,B\in M_n^s(\R)$ , thus $A^T=A,B^T=B$

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

Note that

$$(aA+bB)^T=(bB)^T+(aA)^T\\ =(aA)^T+(bB)^T\\ =aA^T+bB^T\\ =aA+bB$$

Thus,we can see that

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

Hence,it can be seen that

$M_n^s(\R)$ is subspace of $M_n(\R)$

b)

Let

$A=\begin{bmatrix} a_{11} & a_{12}\\ a_{21}& a_{22} \end{bmatrix}$

be an arbitrary element in the subspace W.

Then since

AT=A, we have

$\begin{bmatrix} a_{11} & a_{21}\\ a_{12}& a_{22} \end{bmatrix}=\begin{bmatrix} a_{11} & a_{12}\\ a_{21}& a_{22} \end{bmatrix}.$

This implies that

$a_{12}=a_{21}$ , and hence

$A= \begin{bmatrix} a_{11} & a_{12}\\ a_{12}& a_{22} \end{bmatrix}\\$

$=a_{11}\begin{bmatrix} 1 & 0\\ 0& 0 \end{bmatrix}+a_{12}\begin{bmatrix} 0 & 1\\ 1& 0 \end{bmatrix}+a_{22}\begin{bmatrix} 0 & 0\\ 0& 1 \end{bmatrix}.$

Let

B={v1,v2,v3}, where 

v1,v2,v3 are 2×2

 matrices appearing in the above linear combination of A.

Note that these matrices are symmetric.

Hence we showed that any element in W

W is a linear combination of matrices in B.

Thus Bi s a spanning set for the subspace W.

We show that B is linearly independent.

Suppose that we have

$c_1v_1+c_2v_2+c_3v_3=\begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix}.$

Then it follows that

[$\begin{bmatrix} c_1 & c_2\\ c_2& c_3 \end{bmatrix}=\begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix}.$

Thus

c1=c2=c3=and the setB

is linearly independent.As 

B is a linearly independent spanning set, we conclude that 

B is a basis for the subspace 

W.

c)

Recall that the dimension of a subspace is the number of vectors in a basis of the subspace.

In part (b), we found that 

B={v1,v2,v3} is a basis for the subspace W

W.

As 

B consists of three vectors, the dimension of W

W is 3.