I have a question from my proffesor that I can not figure it out.
V will be inner product system above R2.
Let E some basis with the gram matrix (E={v1,v2})
This is the gram matrix:
\begin{pmatrix}
2 & -1\\
-1 & 1\\
\end{pmatrix} =
\begin{pmatrix}
{(v_{1},v_{1})} & {(v_{1},v_{2})}\\
{(v_{2},v_{1})} & {(v_{2},v_{2})}\\
\end{pmatrix}
Let T:V->V be a linear map with a matrix represent T according to basis E
\begin{pmatrix}
1 & 2\\
2 & 1\\
\end{pmatrix}
is T is self adjoint ?
well first of all according to gram matrix it is very easy to see that E is not a orthonormal set and therefore T* is necessarily equal to the matrix of T with a transpose and conjunction.
So where can I go from here because I can not assume that the inner product is the standard one and therefore I do not know How to look for it, if exists ?
and if it is not exist how can I proof that if I do not know inner product and by that I can not find the adjoint
Best Answer
Let us call $P$ the matrix of the basis $E$ expressed in an orthonormal basis, so that $$ P^*P=\pmatrix{2&-1\\-1&1}=Q $$ and, if $A$ denotes the matrix of $T$ in the same orthonormal basis, $$ P^{-1}AP=\pmatrix{1&2\\2&1}=B. $$ So the question amounts to: do we have $A^*=A$? Now, note that $B^*=B$ and $$ A^*=A\;\iff\;(PBP^{-1})^*=PBP^{-1}\;\iff\; (P^*)^{-1}BP^*=PBP^{-1}\;\iff\;BQ=QB. $$ I let you check whether the latter is true.