Linear Transformation, eigenvalues ​and eigenvectors

Question

Could someone help me with this linear algebra question:

Let $T:\mathbb R^3 → \mathbb R^3$ be a linear transformation, $B = \{v_1, v_2, v_3\}$ a basis of $\mathbb R^3$ and $U = [v_1, v_3].$ Knowing that $T (v) = v$, for all $v ∈ U$ and that $T (v_2) = v_1 + 2v_2 + 3v_3$, determine the eigenvalues ​​and eigenvectors of $T$.

Answer 1

Here is a matrix for $T$:
$$ \Biggl(\begin{matrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 3 & 1 \\ \end{matrix}\Biggr) $$

(Maybe I have it transposed, but that doesn't really matter.)

Do you see where the entries came from?

To find the eigenvalues, solve $\det|T-\lambda I|=0.$

That's $(1-\lambda)(2-\lambda)(1-\lambda)=0$.

Thus, the eigenvalues are $1, 1,$ and $2$.

Eigenvectors for values $1$ and $1$ are $v_1$ and $v_3,$ since we're given $T(v)=v$ when $v \in \{v_1,v_3\}$.

Eigenvector for value $2$ is found from $T(v)=2v$. Say $v=av_1+bv_2+cv_3 = (a,b,c)^T.$

Then $T(v)=(a+b,2b,3b+c)^T=(2a,2b,2c)^T$, so $a=b$ and $c=3b$;

i.e., $v$ is a scalar multiple of $(1,1,3)^T=v_1+v_2+3v_3.$

Answer 2

Clearly $v_1$ and $v_3$ are eigenvectors corresponding to $\lambda_1 = 1$. From there, we look for an eigenvector in the form $av_1 + v_2 + bv_3$. Then

\begin{align*} T(av_1 + v_2 + bv_3) &= aT(v_1)+T(v_2)+bT(v_3) \\ &= (a+1)v_1 + 2v_2+(b+3)v_3 \end{align*} Now since $v_2 \to 2v_2$, we notice that a possibility for $\lambda_2$ is when $\lambda_2 =2$ , meaning that $$2(av_1 + v_2 + bv_3) = (a+1)v_1 + 2v_2+(b+3)v_3$$ If we want the above to be true, then $2a = a+1$ and $2b = b+3$, so $a = 1$ and $b=3$. Thus, we find that $$T(v_1 + v_2 + 3v_3) = 2v_1+2v_2+6v_3$$ So our last eigenvector is $v_1 + v_2 + 3v_3$ corresponding to $\lambda_2 = 2$.