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Define $T ∈ L(F 3 )$ by $T(z_1, z_2, z_3) = (2z_2, 0, 5z_3)$. Find all eigenvalues and eigenvectors for T. Explain how you know you have found them all

I was able to find two eigenvalues for the following question and their respective eigenvectors. However, I do not know if I have found them all. I will use x and y to denote my eigenvalues (even though they are commonly denoted with lambda)

$x = 5$ with the associated eigenvector $(0,0,z_3)$

$y= 0$ with the associated eigenvector $(z_1, 0, 0)$

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If $T$ is a linear application in a 3-dimension space, it has to have $3$ or less different (linearly independent) vectors, (and they can have different eigenvalues or not).

From the equality $T(z_1, z_2, z_3)=(2z_2, 0, 5z_3)=\lambda (z_1, z_2, z_3)$ you get $$\left. \begin{matrix} 2z_2=\lambda z_1 \\ 0=\lambda z_2 \\ 5z_3= \lambda z_3 \end{matrix} \right\}$$

Now, from the second equation, it has to be $\lambda=0$ or $z_2=0$. In the first case, $2z_2=0$ and $5z_3=0$, so $x=0$ is an eigenvalue whose eigen vector is $(z_1, 0,0)$. In the second case, if $\lambda\neq 0$, then $z_2=0$, which implies $\lambda z_1=0$, so $z_1=0$, and $5z_3=\lambda z_3$. If we let $z_3=0$, then we obtain the zero vector $(0,0,0)$, which is a trivial solution. So $z_3\neq0$ and $\lambda=5$ is the other eigenvalue.

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    Am I missing the eigenvalue of 2? But it didn't show up in my calculations lol so I'm confused... – 2017-02-22
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    Yes, that would be the case. Please show your calculations, I think it would be easier to detect the problem. – 2017-02-22
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    @amd thanks for noticing, it actualltpy changes it all. – 2017-02-23
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    Not quite complete, I think. Since you’ve found two real eigenvalues, there must be a third real one. It doesn’t look like there are any other linearly independent eigenvectors, which means that one of the two eigenvalues you’ve found must have algebraic multiplicity 2 but geometric multiplicity 1, but which one? – 2017-02-23
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    @amd Well, that is true. To know the algebraic multiplicity, you'll need to pick up a basis and write the matrix of $T$. If you pick the usual basis $(1,0,0), (0,1,0), (0,0,1)$, that will be: \begin{pmatrix} 0 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 &5 \end{pmatrix} Now, you just have to calculate the characteristic polynomial, which is $\lambda^2(\lambda-5)$. So, $\lambda=0$ is the eigenvalue with algebraic multiplicity $2$. – 2017-02-23
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    You could also examine the trace of this matrix or note that it’s upper-triangular, so has its eigenvalues along the main diagonal. Using the trace is nice because it can be defined for a linear operator without reference to a matrix. – 2017-02-23