# Find a basis $T$ for $R^3$ such that $_T[f]_T$ is diagonal

The linear transformation $$g: R^3 \rightarrow R^3$$ given by $$g(x,y,z) = (x+y,~2y+z,~3z)$$ Find a basis $$T$$ for $$R^3$$ such that $$_T[f]_T$$ is diagonal.

First I found a matrix of a linear transformation

$$_S[g]_S= \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}$$

and $$g$$ has eigenvalues $$1,2,3$$.

But know I am not sure how to find a basis $$T$$.

Should I try to find the eigenspaces $$V(\lambda)$$-s corresponding to eigenvalues?

I'm not really sure how, to begin with, this one. Anything would help thanks.

Try to find three non-zero vectors $$v_1,v_2,v_3$$ in $$\mathbb R^3$$ such that $$g(v_1)=v_1$$, $$g(v_2) = 2v_2$$ and $$g(v_3) = 3v_3$$. Then put $$T = \{v_1,v_2,v_3\}$$.
Now, you find eigenvectors corresponding to those eigenvalues. It is clear that $$(1,0,0)$$ is an eigenvector corresponding to the eigenvalue $$1$$. With a few calculations, you get that $$(1,1,0)$$ is an eigenvector corresponding to the eigenvalue $$2$$ and that $$(1,2,2)$$ is an eigenvector corresponding to the eigenvalue $$3$$. So, take$$T=\bigl\{(1,0,0),(1,1,0),(1,2,2)\bigr\}.$$Then, the matrix of $$g$$ with respect to $$T$$ is$$\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}.$$