$U_n=$$\left( \begin{array}{rrrr} 1 & * & \cdots & * \\ 0 & \ddots & * & \vdots \\ \vdots & 0 & \ddots & * \\ 0 & \cdots & 0 & 1 \\ \end{array}\right) $$T_n=$$\left( \begin{array}{rrrr} * & * & \cdots & * \\ 0 & \ddots & * & \vdots \\ \vdots & 0 & \ddots & * \\ 0 & \cdots & 0 & * \\ \end{array}\right) $

I have to Show that $U_n$ is a normal subgroup of $T_n$

To this end I have to show (after I have verified that $U_n$ is a subgroup of $T_n$):

$$\forall_{x\in T_n}\forall_{a\in U_n}\exists_{b\in U_n}xa=ba$$

I have looked at the case n=2

the results are

$\begin{pmatrix} a & b \\ 0 & c \\ \end{pmatrix}$ $\begin{pmatrix} 1 & d \\ 0 & 1 \\ \end{pmatrix}$ $=\begin{pmatrix} a & ad+b \\ 0 & c \\ \end{pmatrix}$ $=\begin{pmatrix} 1 & adc^{-1} \\ 0 & 1 \\ \end{pmatrix}$ $\begin{pmatrix} a & b \\ 0 & c \\ \end{pmatrix}$

Now the Question is how can I choose the coefficients for my Matrix $b$ in Dependance to $a$?

I suspect that the coefficients of $b$ can be determined this way

(Note that I use the notation $b_{kj}$ which translates to: I look at the coefficient in the $k$-th row at the $j$-th column of the Matrix $b$)

$$b_{kj}\begin{cases}1 &\mbox{if } j=k \\ 0 & \mbox{if } j>k\\a_{(k-1)j}a_{kj}a_{k(j+1)} & \mbox{otherwise} \end{cases}$$

How can I prove this idea?


Lauds to our colleague studiosus for a concise and elegant answer; myself, I have reverted to the somewhat computationally-intensive approach which at least has the advantage of exposing some of the detailed structure of the situation with $T_n$ and its subgroup $U_n$:

Try it like this:


$M \in U_n \tag 1$

may be written in the form

$M = I + N_1, \tag 2$

where $N_1$ is strictly upper triangular, that is, upper triangular with vanishing main diagonal: likewise, every

$L \in T_n \tag 3$

takes the form

$L = D + N_2, \tag 4$

where $N_2$, like $N_1$, is strictly upper triangular; however,

$D = \text{diag}(d_1, d_2, \ldots, d_n) \tag 5$

is a diagonal matrix having every diagonal entry $d_i \ne 0$; this assumption is necessary lest $L$ be singular, since here

$\det L = \det D = \displaystyle \prod_1^n d_i; \tag 6$

scrutinizing (4), we see that it may be written

$L = D(I + D^{-1}N_2), \tag 7$

and thus

$L^{-1} = (I + D^{-1}N_2)^{-1}D^{-1}; \tag 8$


$D^{-1} = \text{diag}(d_1^{-1} , d_2^{-1} , \ldots, d_n^{-1} ) \tag 9$

is also a diagonal matrix with $d_i^{-1} \ne 0$, $1 \le i \le n$; furthermore, the matrix $D^{-1}N_2$ occurring in (7)-(8) is also easily seen to be strictly upper triangular, being the product of a strictly upper triangular matrix with a diagonal matrix; as such, $D^{-1}N_2$ is nilpotent:

$(D^{-1}N_2)^n = 0, \tag{10}$

which in turn implies that the matrix $I + D^{-1}N_2$ is invertible, since

$(I + D^{-1}N_2)^{-1} = \displaystyle \sum_0^{n - 1} (-D^{-1}N_2)^k$ $= I - D^{-1}N_2 + (-D^{-1}N_2)^2 + \ldots + (-D^{-1}N_2)^{n - 1}, \tag{11}$

as is easily seen by using (10) to work out the product

$(I + D^{-1}N_2) \displaystyle \sum_0^{n - 1} (-D^{-1}N_2)^k  = (I - (-D^{-1}N_2)) \displaystyle \sum_0^{n - 1} (-D^{-1}N_2)^k$ $= I + (-D^{-1}N_2)^n = I. \tag{12}$

We next observe that $D^{-1} N_2$ and all its powers are strictly upper triangular, and from thus via (11) we infer that that the diagonals of both $I + D^{-1}N_2$ and $(I + D^{-1}N_2)^{-1}$ are comprised solely of $1$s; thus we may in fact write

$(I + D^{-1}N_2)^{-1} = I + N_3, \tag{13}$

where $N_3$ is also strictly upper triangular, and we have, for any $L$, $M$ as in (2), (4):

$LML^{-1} = (D + N_2)(I + N_1)(D + N_2)^{-1}$ $= D(I + D^{-1}N_2)(I + N_1)( (I + D^{-1}N_2)^{-1}D^{-1}$ $= D(I + D^{-1}N_2)(I + N_1)(I + N_3)D^{-1}; \tag{14}$

we turn now to the product

$(I + D^{-1}N_2)(I + N_1)(I + N_3) \tag{15}$

occurring in (14); we have

$(I + N_1)(I + N_3) = I + N_1 + N_3 + N_1N_3, \tag{16}$

and each of the matrices $N_1$, $N_3$, $N_1N_3$ is strictly upper triangular, hence is their sum (it is easy to show the product of two such matrices is the same); thus we see that the product (16) is in $U_n$; and essentially the same argument gives us

$(I + D^{-1}N_2)(I + N_1)(I + N_3) \in U_n \tag{17}$

as well; finally, we observe that for any matrix such as $M$ (cf. (2)), and invertible $D$ as in (5),

$D(I + N_1)D^{-1} = DID^{-1} + DN_1D^{-1} = I + DN_1D^{-1}, \tag{18}$

where it is easy to affirm $DN_1D^{-1}$ is strictly upper triangular; now turning again to (15) we see that we may infer that

$LML^{-1} \in U_n; \tag{14}$

but $L$ may be any element of $T_n$; thus our proof that $U_n$ is normal in $T_n$ is complete.

In closing we observe that this argument largely hinges on two easy facts: first, that the product of two strictly upper triangular matrices is again strictly upper triangular; and second, the product of a strictly upper triangular matrix and a diagonal matrix is again strictly upper triangular. From these two easily established facts our demonstration flows.

Show that a certain set of invertible matrices is a normal subgroup of another set of invertible matrices (triangular)


I want to propose an alternative proof without computations. Note that $U_{n}$ consists exactly of those matrices in $T_{n}$ whose eigenvalues are all equal to $1$. Given $A\in U_{n}$ and $X\in T_{n}$, we immediately get that $$XAX^{-1}\in U_{n},$$ since eigenvalues are invariant under conjugation.

  • $\begingroup$ Very nice answer, as opposed to my tour de grunge; endorsed; + 1!!! $\endgroup$ – Robert Lewis Jan 30 '19 at 2:25

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