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Given m > 0, show that there exists a 2$\times$2 matrix T that $T^{-1}$$\pmatrix{n&1\\0&n}$T = $\pmatrix{n&m\\0&n}$.

Not really sure how to work it out. Thanks!

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    $\begingroup$ 1. What are the eigenvalues of the matrix on the right? 2.How many eigenvectors does each one have? 3. What must its Jordan Normal Form be? 4. How are the JNF and the matrix related? Answer each of these (by clicking "edit") below your question, and you'll be on the way to an answer. $\endgroup$ Mar 4, 2018 at 16:59
  • $\begingroup$ See also en.wikipedia.org/wiki/Jordan_normal_form $\endgroup$
    – Widawensen
    Mar 4, 2018 at 17:59

3 Answers 3

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Taking T as $\begin{bmatrix}a&b\\c&d\end{bmatrix}$,you know $T^{-1}$ is: $\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$. Substituting into your equation you get a relation between a,b,c,d and m which you can solve.

Hint: you can take $c=0$.

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    $\begingroup$ He can also use directly $JT=TM$ $\endgroup$
    – Widawensen
    Mar 4, 2018 at 18:08
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Note that existence of matrix T depends on Jordan decomposition of $A=\pmatrix{n&m\\0&n}$, this matrix have a Jordan decomposition if its characteristic polynomial splits into linear factors over R (its field).

$$p_A(x)=x^2-2nx+n^2$$ $$p_A(x)=(x-n)^2$$ Then, exists T invertible such that $T^{-1}\pmatrix{n&1\\0&n}T=\pmatrix{n&m\\0&n}$

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$$ \pmatrix{n&1\\0&n} = nI + \pmatrix{0&1\\0&0} $$ so $$ T^{-1}\pmatrix{n&1\\0&n}T = nI+T^{-1}\pmatrix{0&1\\0&0}T $$ you require: $$ T^{-1}\pmatrix{0&1\\0&0}T =\pmatrix{0&m\\0&0} $$ or, equivalently: $$ \pmatrix{0&1\\0&0}T =T\pmatrix{0&m\\0&0} $$ if we set $$ T = \pmatrix{a&b\\c&d} $$ this gives: $c=0$ and $am=d$

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