# Change of basis of a linear operator.

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!

## 1 Answer

Here it is very easy to see how the standard basis vectors can be written as linear combinations of the vectors in B.

Example:

$(1,0,0)=(1,1,0)-(0,1,0)$

Therefore $T((1,0,0))=T((1,1,0))-T((0,1,0))$

Moreover we know that $T((1,1,0))=2(1,1,0)+n(0,1,0)=(2,2+n,0)$ and that $T((0,1,0))=(m,m,0)$, whence $T((1,0,0))=(2-m,2+n-m,0)$.

This is the first column in your sought after matrix, and you can do similarly with the other two standard basis vectors.

The more general approach is to first find the change of basis matrices $P_{\mathcal{B}\rightarrow{\mathcal{E}}}$ and $P_{\mathcal{E}\rightarrow{\mathcal{B}}}$, where $\mathcal{E}$ denotes the standard basis. One then has the following formula:

$[T]_{\mathcal{E}}=P_{\mathcal{E}\rightarrow{\mathcal{B}}}[T]_{\mathcal{B}}P_{\mathcal{B}\rightarrow{\mathcal{E}}}$