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

Question

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.

Answer 1

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 $T = \{v_1,v_2,v_3\}$.

Answer 2

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$$\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}.$$