# Find the eigen values and the associated eigen vectors of the matrix,$~~~~$ $A=\begin{pmatrix} 3~~1~~1\\ 2~~4~~2\\ 1~~1~~3 \end{pmatrix}$

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.

• 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. – amd Oct 9 '18 at 18:38

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