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When trying to check if $$C=\begin{pmatrix} 1&1&1\\ 1&1&1\\ 1&1&1\\ \end{pmatrix}\in\Bbb R^{3\times3}$$ is diagonalizable, we find $c_C(x)=-x^2(x-3) \Rightarrow λ=0,3$ with $m(0)=2,m(3)=1$. So to find eigenvectors we solve $(C-λI)X=0$ which give $$\begin{pmatrix}x\\y\\z\\\end{pmatrix}\in\Bbb R^{3\times1}\setminus\{0\},\begin{pmatrix}x\\x\\x\\\end{pmatrix}\in\Bbb R^{3\times1}\setminus\{0\}$$ for $0$ and $3$ respectively. Is it this so far correct? If so then how do I proceed to finding if there exists a basis of $\Bbb R^{3\times1}$ from these eigenvectors?

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    if multiplicity of eigenvalues is equal to dimension of corresponding eigenvector space, then it is diagonalizable – 2017-01-03
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    You can use the definition of eigenvalue decomposition $C = SDS^{-1}$, with eigenvectors stuffed in $S$. If you can find a diagonal $D$ which satisfies that equation then it is by definition diagonalizable. – 2017-01-03
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    Continuing the comment by dato: the dimension of the corresponding eigenspace = the geometric multiplicity of the corres. eigenvalue = the dimension of the solution space of the homogenoeus sytem $\;(\lambda I-C)\mathbf x=\mathbf0\;,\;\;\lambda=$ the corres. eigenvalue – 2017-01-03
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    Every symmetric matrix is diagonalisable. – 2017-01-03
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    @Surb Also every rank one matrix. – 2017-01-03
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    Ok, so by using $dimV_C(λ)=dim(\Bbb R^{3x3})-r(C-λI)$ we get that $dimV_C(0)=2$ and $dimV_C(3)=1$ so it is diagonalizable. Now how can I find a base for $\Bbb R^{3x1}$ composed by eigenvectors of $C$ ? – 2017-01-03
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    @Michael $(1,1,1)$, $(1,-1,0)$, $(1,0,-1)$ – 2017-01-03
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    @Surb How did you get these? – 2017-01-03
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    @Michael Honestly, I simply guessed them (but I should admit that I'm used to work with spectral problems). If you want to go formally about it. You know that the eigenvalues are $3$,$0$ and $0$. So, solve the system $Cx = 3x$ for $x$. This will give you $x=(1,1,1)$. Then, find a basis of the kernel of $C$, i.e. solve $Cy=0y=0$ for $y$. This will give you the other ones. – 2017-01-03
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    @Surb so for $λ=0$ we have $$\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\\ \end{pmatrix} \begin{pmatrix}x\\y\\z\\ \end{pmatrix}=0$$ which gives $$\{\begin{pmatrix}x\\y\\z\\\end{pmatrix} \in \Bbb R^{3x1} \setminus \{0\}\}=V_C(0) \setminus \{0\}$$ right? So how do $$\begin{pmatrix}1\\-1\\0\\\end{pmatrix},\begin{pmatrix}1\\0\\-1\\\end{pmatrix}$$ occur from this? This is basically the most important part. – 2017-01-03
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    @Michael .... can you solve the equation $x+y+z = 0$? – 2017-01-03
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    @Surb so we just put randomly values on (xyz) that satisfy the equation and are elements of the basis? – 2017-01-03
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    No. This system is underdetermined and it has a whole vector space of solutions namely $U =\{(x,y,z)\mid x = -(y+z)\}$. Now you should find a basis of $U$. Any basis will do the job, I proposed you one possible choice. – 2017-01-03
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    @Surb (I'm assuming that is (x,y,z) and not (x,y,y) right?) If so then it all makes sense now thank you!! – 2017-01-03
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    @Michael Yes it is so (sorry I made a typo). You're welcome – 2017-01-03

2 Answers 2

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By using $\dim V_C(λ)=\dim(\Bbb R^{3\times3})−r(C−λI)$ we get that $\dim V_C(0)=2$ and $\dim V_C(3)=1$ so it is diagonalizable. Now from the solutions of the two systems we find a basis of $V_C(0)\setminus \{0\}$, say $$\begin{pmatrix}1\\-1\\0\end{pmatrix}, \begin{pmatrix}1\\0\\-1\end{pmatrix}\;,$$ and a basis of $V_C(3)\setminus\{0\}$, say $$\begin{pmatrix}1\\1\\1\end{pmatrix}\;.$$ These three vectors are a basis of $\Bbb R^{3\times3}$

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Observe the vector of all ones gives the eigenvalue $3$. So the other two eigenvectors are orthogonal to it. You can choose the vectors $$\begin{pmatrix}1\\-1\\0\end{pmatrix}, \begin{pmatrix}1\\0\\-1\end{pmatrix}$$ as the other two eigenvectors of $C$.

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    @G_o_pi_i_e I don't know what orthogonal vectors are yet but I'm assuming this answer is correct. Does is have something to do with dot product? – 2017-01-04
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    @Michael: Orthogonal means dot product is zero, yes. – 2017-01-04