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I know of the $A = PDP^{-1}$ formula but it only works if P is invertible.

My example I've spent hours on is:

Find a 3x3 matrix where the only eigenvector is $ \begin{pmatrix} 1 \\ 1 \\ 0 \\ \end{pmatrix} $ for which the eigenvalue is 1. (Any constant multiplication of the eigenvector is naturally fine as well).

One example is \begin{bmatrix} 1 & 0 & 1 \\ 1/2 & 1/2 & 1/2 \\ -1/2 & 1/2 & 3/2 \end{bmatrix}.

But I cannot for the life of me figure out how to get to this answer.

Limit
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  • You have a solution..... to go backwards, name your matrix $M$ and find a matrix $Q$ such that $Q^{-1} MQ = J$ is in Jordan form. With eigenvalue $1,$ the right hand column of $Q$ is a column vector $q_3$ such that $(M_I)^3 q_3 = 0$ BUT $(M_I)^2 q_3 \neq 0.$ Next $q_2 = (M_I) q_3 $ and then $q_1 = (M_I) q_2 = (M_I)^2 q_3 $ That is your matrix $Q,$ columns $q_1, q_2, q_3$ If you switch the order to $q_3, q_2, q_1$ the outcome of $Q^{-1} MQ $ is the transpose of $J$ – Will Jagy Mar 28 '24 at 17:28

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in the U.S., Jordan form is upper triangular. The transformation with characteristic polynomial $(x-1)^3 $ but minimal polynomial the same is just the Jordan block

$$ J= \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) $$

That was the hard part. Now, you want to switch the eigenvector

$$ e_1 = \left( \begin{array}{r} 1 \\ 0 \\ 0 \\ \end{array} \right) $$
for the required

$$ v = \left( \begin{array}{r} 1 \\ 1 \\ 0 \\ \end{array} \right) $$

That is, you need to pull back $v$ to $e_1$ You require an (invertible) matrix $R$ such that

$$ \left( \begin{array}{r} 1 \\ 0 \\ 0 \\ \end{array} \right) \; = \; \; R \; \; \left( \begin{array}{r} 1 \\ 1 \\ 0 \\ \end{array} \right) $$
so write out the system. My first try was lower triangular so it is easy to confirm the nonzero determinant,

$$ R= \left( \begin{array}{rrr} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$

Note the eigenvectors of $R$ do not matter . Next

$$ R^{-1}= \left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ as we can confirm by calculating $R^{-1} R$

Finally, the matrix similar to $J,$ and an answer to your problem, is $$ R^{-1} J R $$

Things you should know: similar matrices have the same characteristic and minimal polynomials; I believe the proof for the characteristic polynomial has several steps. The minimal has a one line proof: for any one variable polynomial $f(x),$ we have $f(Q^{-1} M Q ) = Q^{-1} f(M) Q.$ Thus, if $f(M) = 0$ we also have $f(Q^{-1} M Q ) = 0$

Next, with just one eigenvalue, the degree of the minimal polynomial is the size of the largest Jordan block. see

Why does the largest Jordan block determine the degree for that factor in the minimal polynomial?

and https://en.wikipedia.org/wiki/Jordan_normal_form#Complex_matrices

ADDED I thought of a good display for the "largest block" idea.

$$ M = \left( \begin{array}{rrrrrr} 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 3 \\ \end{array} \right) $$

The characteristic polynomial is $(x-1)^2 (x-2)^2 (x-3)^2 .$ This is also the minimal polynomial; that says that each eigenvalue gets a Jordan block of size 2.

Meanwhile, $(x-1)(x-2)(x-3) = x^3 - 6 x^2 + 11 x - 6 $ and

$$ M^3 - 6 M^2 + 11 M - 6 I = \left( \begin{array}{rrrrrr} 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right) $$ which is not the zero matrix, indeed there is something nonzero in each of the three blocks

Will Jagy
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