Regarding the diagonalizibility of matrices.

My matrix $A$ is $$ \begin{bmatrix} 2 & -1 \\ -4 & 2 \\ \end{bmatrix} $$

I know that A is diagonalizable if the geometric multiplicity and algebraic multiplicities ( of its eigenvalues, which are 4 and 0 ) are equal, and that the geometric multiplicity is less then or equal to the algebraic multiplicity. I've marked

$Mg ( E ) = Ma ( E )$ where $E$ = eigenvalue

If the algebraic multiplicity is the multiplicity of as a root of the characteristic polynomial, then Ma must be 1 because the characteristic polynomial is $(E-2)*(E-2)-4=0$

For the geometric multiplicity, the eigenvector of 4 is one dimensional, so Mg is 1 as well.

Thus the matrix is diagonilizable

Is this correct?


Any time an $n\times n$ matrix has $n$ distinct eigenvalues, it is diagonalizable, since each eigenvalue must have (at least) one eigenvector. Your matrix falls into this category. So you are correct.

  • $\begingroup$ Interesting, what theorem states that fact? $\endgroup$ – user140878 May 7 '14 at 19:22

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