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Right, I know how todoquestions on Jordan Normal Form and find a basis, but there is one part I don't understand.

Let's take for example this matrix, call it A.

\begin{bmatrix}

-3 & 1 & 0 \\

-1 & -1 & 0 \\

-1 & -2 & 1

\end{bmatrix}

We find the characteristic polynomial of A using

det(A-tI) = 0, where I is the identity matrix.

The roots of this equation are our eigenvalues, let's call them [itex]\lambda_{1}[/itex] and [itex]\lambda_{2}[/itex]

For each one we find A-[itex]\lambda[/itex]I.

Call this matrix B.

We find rk(B), and using the formula rk(B) + dimKer(B) = n, where n is the dimension of our vector space, find dimKer(B).

We find B^{2}, B^{3}, ..., B^{k-1}, B^{k}

until rk(B^{k-1}) = rk(B^{k})

We then have to find the vector(s) that spans the kernel of B^{k-1}and the first part of our basis is given by a vector which isn't in the kernel. This tends to change depending on various cases which we're working on.

Example, if rk(B) =/= rk(B^{2}) = 0 then we find the vectors that span the kernel of

B^{2}, bring it to reduced column echelon form and the vector we choose for part of our basis is the vector which makes up for the "missing leading one".

The last paragraph is basically what I don't understand.

Could someone please explain why this is done? (in general, not for any particular case)

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# Jordan Normal/Canonical basis

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