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Let $T : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by : $$T(x,y,z) = (x-y+4z,3x+2y-z,2x+y-z)$$

How can i find the vector $(x,y,z) \in \mathbb{R}^3$ and the constants $\lambda \in \mathbb{R}$ such that $T(x,y,z) = (\lambda x, \lambda y, \lambda z )$, i stuck i don't know how to start this exercise , please help.

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  • $\begingroup$ Hint: $T u = \lambda u$ is called an eigenvalue problem, with eigenvalue (german for "self value") $\lambda$ and eigenvector $u$. $\endgroup$ – mvw Jun 12 '14 at 20:49
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I'll assume you don't know about eigenvalues, yet.

What you have to find are the numbers $\lambda\in\mathbb{R}$ for which a nonzero vector $(x,y,z)$ exists such that $T(x,y,z)=\lambda(x,y,z)$. Why nonzero? Because $(0,0,0)$ is good for any $\lambda$.

This is equivalent to saying $$ \begin{cases} x-y+4z=\lambda x\\ 3x+2y-z=\lambda y\\ 2x+y-z=\lambda z \end{cases} $$ or, which is the same, $$ \begin{cases} (1-\lambda)x-y+4z=0\\ 3x+(2-\lambda)y-z=0\\ 2x+y+(-1-\lambda)z=0 \end{cases} $$ and this linear system has a nonzero solution if and only if the rank of the matrix $$ \begin{bmatrix} 1-\lambda & -1 & 4\\ 3 & 2-\lambda & -1\\ 2 & 1 & -1-\lambda \end{bmatrix} $$ is less than $3$.

Can you go on from here?

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  • $\begingroup$ then i need to find the characteristic polynomial then find its roots ? $\endgroup$ – Rachel Jun 12 '14 at 20:55
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    $\begingroup$ @Knight So you do know about eigenvalues, don't you? $\endgroup$ – egreg Jun 12 '14 at 20:56
  • $\begingroup$ yes but for me its complicated understand when i need to apply this process , for example in books this process is for diagonalize a matrix and when i have a linear transformation like that i´m confused $\endgroup$ – Rachel Jun 12 '14 at 20:57
  • $\begingroup$ @Knight The problem is to find $v\ne0$ such that $T(v)=\lambda v$, for some $\lambda$. This is exactly the definition of eigenvector (and eigenvalue). $\endgroup$ – egreg Jun 12 '14 at 20:59
  • $\begingroup$ ok, thanxs for the explication really ty, im studying so hard this topics. $\endgroup$ – Rachel Jun 12 '14 at 21:00
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Note that $$ T\begin{bmatrix}x\\y\\z\end{bmatrix} = A\begin{bmatrix}x\\y\\z\end{bmatrix} $$ where $$ A= \begin{bmatrix} 1 & -1 & 4\\ 3 & 2 & -1\\ 2 & 1 & -1 \end{bmatrix} $$ Thus we wish to compute the eigenvectors of $A$. Do you know where to go from here?

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  • $\begingroup$ yes from the canonical basis of R^3 $\endgroup$ – Rachel Jun 12 '14 at 20:53
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Hint: $$ T u = \lambda u \iff (T - \lambda E) u = 0 $$ where $T$ is the matrix form of your map $T$ and $E$ is the unit / identity matrix with $E_{ij} = \delta_{ij}$.

$$ u \in \mbox{ker} (T - \lambda E) $$

and you want a non-zero vector $u$, so you know the matrix $T - \lambda E$ can not be invertible.

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