My question is related to this question discussed in MSE.
$J$ be a $3\times 3$ matrix with all entries $1\,\,$. Then prove that $J$ is
diagonalizable.
Can someone explain it in terms of A.M. and G.M. (algebraic and geometric multiplicity) concept ? Thanks in advance for your time.
So we have $$ J = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} $$ Let's compute the characteristic polynomial $\chi_J(X) = \det(J - X)$: \begin{align*} \chi_J(X) &= \det \begin{pmatrix} 1-X & 1 & 1 \\ 1 & 1-X & 1 \\ 1 & 1 & 1-X \end{pmatrix}\\ &= (1-X)^3 + 2 - 3(1-X)\\ &= 1 - 3X + 3X^2 - X^3 + 2 - 3 + 3X\\ &= -X^3 + 3X^2\\ &= -X^2(X-3) \end{align*} So 0 (with algebraic multiplicity 2) and 3 (with algebraic multiplicity 1) are the eigenvalues of $J$. To check whether $J$ is diagonaziable we will compute the geometric multiplicity of 0, that is $\dim \ker (J-0) = \dim\ker J$. We do Gaussian elimination: Subtrating the first row from the third and the second gives that $$ \ker J = \ker \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0&0\\ 0 & 0 & 0 \end{pmatrix}$$ This matrix has rank 1, so its kernel has dimension 2. So $\dim\ker J = 2$ and $J$ is diagonaziable.