I have this silly question, but I don't know what I'm doing wrong at all!
Considerate the vector space $\mathbb {R^3}$ with usual dot product and let $T:\mathbb{R^3}\rightarrow \mathbb {R^3}$ the linear operator with matrix $[T]_B$ related to the basis $B=\{(1,1,0),(0,1,0),(0,0,2)\}$ is: $[T]_B = \begin{pmatrix} 2 & m & 2n \\ n & 0 & 0 \\ 0 & 0 & 0\\ \end{pmatrix}$
The question follows:
Determine:
(a)The matrix $T$ related to the canonical basis of $\mathbb {R^3}$.
(b)$m,n$ that make $T$ symmetric operator.
(c)Using the values of $m,n$ you found above, give an orthonormal basis of $\mathbb {R^3}$ that diagonalizes $T$ and the matrix $T$ in this basis.
I did the following:
to change the basis:
$T(1,0,0)=(2,n,0)=a(1,0,0)+b(0,1,0)+c(0,0,1)$
$T(0,1,0)=(0,m,0)=d(1,0,0)+e(0,1,0)+f(0,0,1)$
$T(0,0,1)=(2n,0,0)=g(1,0,0)+h(0,1,0)+i(0,0,1)$
Then i found the matrix :
$[T]_{can} = \begin{pmatrix} 2 & 0 & 2n \\ n & m & 0 \\ 0 & 0 & 0\\ \end{pmatrix}$
But that's clearly wrong...I can't proceed from that,it's not right...where exactly is wrong?
I appreciate any efforts, Thanks for attention!