0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.