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Sol. det($A-\lambda I$)$=0$ $\implies \lambda = 2, 2, 6.$ Determining eigen vector with respect to the eigen value $\lambda=2:$ $ \begin{pmatrix} 1~~1~~1\\ 2~~2~~2\\ 1~~1~~1 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$ implies that

$x+y+z=0\\ ~~~~~~x+y+z=0\\ ~~~~~~x+y+z=0$. Solving this equation by cross multiplication, we get $\frac{x}{1-1}=\frac{-y}{1-1}=\frac{z}{1-1}= k\text{(say)} \implies x=y=z=0$. Therefore, the eigen vector is $\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$. Now, my question is why the eigen vector turns out to be zero here? The definition of the eigen vector itself says that $Av=\lambda v$, where $\lambda$ is an scalar called eigen value and {\bf non-zero} vector $v$ is called eigen vector corresponding to $\lambda$. But i found some literature, where they seemingly consider zero vector also as an eigen value. Please clearify me where did i go wrong? Though similar questions might be asked in this forum already but my problem couln't be fixed by those problems yet.

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  • $\begingroup$ The expressions $x/(1-1)$ etc. that you’ve come up with are undefined, so you can’t draw any conclusions about the values of $x$, $y$ and $z$ from them. $\endgroup$ – amd Oct 9 '18 at 18:38
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no! actually we have one single equation, so arbitrarily take two values and then find one

For, take $y=t$ and $z=k$ to see the eigenspace corresponding to $2$ is $$\Bigg\{t \begin{pmatrix} -1\\1\\0 \end{pmatrix}+k \begin{pmatrix} -1\\0\\1 \end{pmatrix}: k,t \in \Bbb{R}\Bigg\}$$

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All you need is $x+y+z=0$ For example $(2,-1,-1)$ works as an eigenvector for eigenvalue $2$ You can find another one as well such as $(1,1,-2)$

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The eigenspace$E_2$ for the eigenvalue $2$ has equation $\; x+yz=0$, hence it is isomorphic to $K^2$ ($K$ being the base field) by the isomorphism \begin{align} K^2&\longrightarrow E_2\\ (x,y)&\longmapsto (x,y,-x-y) \end{align} The image by this isomorphism of the canonical basis of $K^2$: $\;(1,0,-1)$ and $(0,1,-1)$, is a basis of the eigenspace.

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