A space of matrices
Task number: 2546
Let \(V\) ve the set of all real symmetric matrices of order three with zeros on the diagonal.
Show that \(V\) is a subspace of \(\mathbb R^{3\times 3}\). Determine the dimension of \(V\) and find some basis of this subspace.
Resolution
When \(A,B\in V\) then both matrices \(\alpha A\) and \(A+B\) have zeros on the diagonal: \((\alpha A)_{i,i}=\alpha a_{i,i}=\alpha 0 =0\), \((A+B)_{i,i}=a_{i,i}+b_{i,i}=0+0=0\).
The sum and a scalar multiple remain symmetric: \((\alpha A)_{i,j}=\alpha a_{i,j}=\alpha a_{j,i}=(\alpha A)_{j,i}\), \((A+B)_{i,j}=a_{i,j}+b_{i,j}=a_{j,i}+b_{j,i}=(A+B)_{j,i}\).
As the sum and the scalar multiple are preserved in the set \(V\), it is a subspace.
When \(A \in V\), then its elements satisfy: \(a_{1{,}1}=a_{2{,}2}=a_{3{,}3}=0\), \(a_{1{,}2}=a_{2{,}1}\), \(a_{1{,}3}=a_{3{,}1}\) a \(a_{2{,}3}=a_{3{,}2}\).
The solution of this system could be expressed by using three parameters as follows:
\(A= a_{1{,}2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} + a_{1{,}3} \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix} + a_{2{,}3} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix} \).Hence \(\dim(V)=3\) and the three matrices above could be chosen as a basis of \(V\).
