Show that there exists this matrix (Jordan decomposition)

Question

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!

Answer 1

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 $c=0$.

Answer 2

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 $T^{-1}\pmatrix{n&1\\0&n}T=\pmatrix{n&m\\0&n}$

Answer 3

$$ \pmatrix{n&1\\0&n} = nI + \pmatrix{0&1\\0&0} $$ 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 $am=d$