Is there any $a\in\Bbb R$ s. t. $A$ is a Hermitian operator on $\Bbb R^3$ with the standard scalar product? If so, find all such $a$.

Question

Matrix representation of a linear operator $A\in\mathcal L\left(\Bbb R^3\right)$ is given in the basis: $$f=\{(1,0,0),(1,1,0),(1,1,1)\}$$ $$[A]_f=\begin{bmatrix}1-a&3-2a&5-2a\\a-3&2a-4&2a-3\\3&4&4\end{bmatrix}$$ Is there any $a\in\Bbb R$ s. t. $A$ is a Hermitian operator on $\Bbb R^3$ with the standard scalar product? If so, find all such $a$.

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My attempt:

Since $f$ isn't orthonormal, I used the following proposition:

Linear operator $T\in\mathcal L(U)$ is Hermitian iff its matrix representation in an orthonormal basis is a hermitian matrix.

I chose the standard canonical basis for the sake of simplicity.

Let $F=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}$ and $T=F^{-1}I=F^{-1}$ be a transition matrix representing the change of basis $f$ into the standard canonical basis $e$.

Then $[A]_e=T^{-1}[A]_fT$.

After I computed the inverse $F^{-1}=\begin{bmatrix}1&-1&0\\0&1&-1\\0&0&1\end{bmatrix}$, I got:

$\begin{aligned}\ [A]_e=F[A]_fF^{-1}&=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}1-a&3-2a&5-2a\\a-3&2a-4&2a-3\\3&4&4\end{bmatrix}\cdot\begin{bmatrix}1&-1&0\\0&1&-1\\0&0&1\end{bmatrix}\\&=\begin{bmatrix}1&3&6\\a&2a&2a+1\\3&4&4\end{bmatrix}\cdot\begin{bmatrix}1&-1&0\\0&1&-1\\0&0&1\end{bmatrix}\\&=\begin{bmatrix}1&2&3\\a&a&1\\3&1&0\end{bmatrix}\end{aligned}$

$A\in M_3(\Bbb R)\implies A=A^T$, so $A$ only has to be symmetric $\implies a=2$

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Is this correct? If so, is there any faster method?

Thank you in advance!

Answer 1

Your solution is right, and is probably the shortest.

Here is another way. You need to check that $\langle Ax,y\rangle=\langle x,Ay\rangle$ for all $x,y$. Then it is enough to check the equality on basis elements. As $\langle Af_j,f_j\rangle=\langle f_j,Af_j\rangle$, we just need to check $\langle Af_j,f_k\rangle=\langle f_j,Af_k\rangle$ for $j\ne k$. This reduces us to check that \begin{align} \langle Af_1,f_2\rangle&=\langle f_1,Af_2\rangle\\ \langle Af_1,f_3\rangle&=\langle f_1,Af_3\rangle\\ \langle Af_2,f_3\rangle&=\langle f_2,Af_3\rangle. \end{align} From the matrix form of $A$ and the form of $f_1,f_2,f_3$ we get, in the canonical basis, $$ Af_1=\begin{bmatrix} 1\\ a\\3\end{bmatrix},\qquad Af_2=\begin{bmatrix} 3\\ 2a\\4\end{bmatrix},\qquad Af_3=\begin{bmatrix} 6\\ 2a+1\\4\end{bmatrix}. $$ Then the three equalities above become \begin{align} 1+a&=3\\ 4+a&=6\\ 7+2a&=2a+7 \end{align} The first two equalities require $a=2$, so this is the only solution.