# inner product space and gram matrix

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

• Symmetric matrices correspond to self-adjoint operators.
– DVD
Commented May 16, 2013 at 23:58
• @Daved The matrix of $T$ given here corresponds to a non-orthonormal basis of $V$. So you can't read whether $T$ is self-adjoint on this matrix like that. Commented May 17, 2013 at 1:01
• @julien okay, should add:"in orthonormal basis".
– DVD
Commented May 17, 2013 at 2:06
• You should have edited your first question, where several of us took the time to respond carefully! Commented May 19, 2013 at 0:09

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.
• That's what a Gram matrix is. If $P=\pmatrix{a_1&b_1\\a_2&b_2}$, it means that $v_1=a_1e_1+a_2e_2$ where $(e_1,e_2)$ is the orthonormal basis we took. Then the $(1,1)$ element of $P^*P$ is $\overline{a_1}a_1+\overline{a_2}a_2=(v_1,v_1)$. Likewise the $(1,2)$ coefficient is $(v_1,v_2)$ etc... Commented May 17, 2013 at 11:59