Finding Jordan form Find Jordan form of the following matrix: $$\left(\begin{matrix}4&-5&2 \\ 5&-7&3\\ 6&-9&4 \end{matrix}\right)$$
So I got stuck pretty much trying to find the eigenvalues. 
Related question: Is the characteristic polynomial of the characteristic matrix, equals to the characteristic polynomial of the transpose of the characteristic matrix? Since their determinants are equal... 
Thanks! 
 A: The characteristic polynomial is given by $|A - \lambda I| = 0$, yielding:
$$\begin{vmatrix}4 - \lambda &-5 & 2 \\ 5 &-7 - \lambda & 3\\ 6 & -9 & 4 - \lambda \end{vmatrix} = 0$$
Using determinants, this reduces to the characteristic polynomial:
$$\lambda^2 - \lambda^3 = 0 \rightarrow -\lambda^2(\lambda - 1) \rightarrow \lambda_1 = 1, \lambda_{2,3} = 0, $$
To find the eigenvectors, we solve $[A-\lambda_i I]v_i = 0$
For the first eigenvalue, $\lambda_1 = 1$, we have:
$$v_1 = (1,1,1).$$
For the second eigenvalue, $\lambda_2 = 0$, we have a single eigenvector, which is:
$$v_2 = (1,2,3).$$
For the third, we need a generalized eigenvector and use $[A-\lambda I]v_3 = v_2$, yielding:
$$v_3 = (-1,-1,0).$$
I purposely left out the details so you can fill them in.
Also, you now have everything you need to find the Jordan Normal Form and there are other approaches to finding it.
A: Answer to related question: $$|tI-A|=|(tI-A)^T|=|(tI)^T-A^T|=|tI-A^T|$$
So yes, $A$ and $A^T$ have the same characteristic polynomials.
A: When you compute $\det(tI-A)$, you should find the characteristic polynomial to be $t^3 - t^2 = t^2(t-1)$.  So, your eigenvalues are $0$ and $1$.  Note that $0$ has one eigenvector and one generalized eigenvector, since $\dim \ker(A) = 1$.
Neat way to find your eigenvector for $1$: check the row sums.
As for your extra question: not only do $A$ and $A^T$ have the same characteristic polynomial, they actually have the same Jordan form as well!
