# Checking if matrix is diagonalizable

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?

• if multiplicity of eigenvalues is equal to dimension of corresponding eigenvector space, then it is diagonalizable – dato datuashvili Jan 3 '17 at 10:15
• 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. – mathreadler Jan 3 '17 at 10:19
• @Surb Also every rank one matrix. – egreg Jan 3 '17 at 10:31
• @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!! – Zero Pancakes Jan 3 '17 at 13:39
• @Michael Yes it is so (sorry I made a typo). You're welcome – Surb Jan 3 '17 at 13:43

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 $\Bbb R^{3\times3}$
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 $C$.