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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4$\begingroup$ if multiplicity of eigenvalues is equal to dimension of corresponding eigenvector space, then it is diagonalizable $\endgroup$– dato datuashviliCommented Jan 3, 2017 at 10:15
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$\begingroup$ 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. $\endgroup$– mathreadlerCommented Jan 3, 2017 at 10:19
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1$\begingroup$ @Surb Also every rank one matrix. $\endgroup$– egregCommented Jan 3, 2017 at 10:31
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1$\begingroup$ @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!! $\endgroup$– Zero PancakesCommented Jan 3, 2017 at 13:39
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1$\begingroup$ @Michael Yes it is so (sorry I made a typo). You're welcome $\endgroup$– SurbCommented Jan 3, 2017 at 13:43
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2 Answers
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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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$\begingroup$ @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? $\endgroup$ Commented Jan 4, 2017 at 7:11
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1$\begingroup$ @Michael: Orthogonal means dot product is zero, yes. $\endgroup$– MvGCommented Jan 4, 2017 at 7:13