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Prove that the inverse of an invertible upper triangular matrix of order 3 is invertible and upper triangular.

I have checked all the similar questions but I couldn't understand any of them.

I supposed random 3x3 upper triangular matrix and tried to find its inverse, but it came out lower triangular matrix, not the upper triangular. Can anyone please give me a suggestion, how to prove it?

$$ A=\left(\begin{array}{rrr}% a&b&c\\% 0&d&e\\% 0&0&f\\% \end{array}\right)% $$ $$ x11=\left(\begin{array}{rrr}% d&e\\% 0&f\\% \end{array}\right)% =df, x12=-\left(\begin{array}{rrr}% 0&e\\% 0&f\\% \end{array}\right)% =0, x13=\left(\begin{array}{rrr}% 0&d\\% 0&0\\% \end{array}\right)% =0 $$ $$ x21=-\left(\begin{array}{rrr}% b&c\\% 0&f\\% \end{array}\right)% =-bf, X22=\left(\begin{array}{rrr}% a&c\\% 0&f\\% \end{array}\right)% =af, X23=-\left(\begin{array}{rrr}% a&b\\% 0&0\\% \end{array}\right)% =0 $$ $$ x31=\left(\begin{array}{rrr}% b&c\\% d&e\\% \end{array}\right)% =bc-cd, x32=-\left(\begin{array}{rrr}% a&c\\% 0&e\\% \end{array}\right)% =ac, x31=\left(\begin{array}{rrr}% a&b\\% 0&d\\% \end{array}\right)% =ad $$ $$ adjoint A = \left(\begin{array}{rrr}% df&0&0\\% -bf&af&0\\% bc-cd&-ac&ad\\% \end{array}\right)% $$ $$ det A = a\left(\begin{array}{rrr}% d&e\\% 0&f\\% \end{array}\right)% =adf $$ $$ Inverse-A =1/adf \left(\begin{array}{rrr}% df&0&0\\% -bf&af&0\\% bc-cd&-ac&ad\\% \end{array}\right)% $$ It came out lower triangular matrix. Is there any way to make it upper triangular matrix?

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  • $\begingroup$ If you show your steps when you tried it before and it came out lower triangular, we could probably diagnose the error. $\endgroup$ – Greg Martin Nov 3 '14 at 7:33
  • $\begingroup$ I added the steps which I did. $\endgroup$ – Hafiz Temuri Nov 3 '14 at 15:43
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    $\begingroup$ What you call the adjoint is in fact the cofactor matrix, you should take the transpose of this. See en.wikipedia.org/wiki/Adjugate_matrix. PS: there is a slight miscalculation in $x_{31}$ and $x_{32}$. $\endgroup$ – Marc Bogaerts Nov 3 '14 at 16:56
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    $\begingroup$ You need to take the transpose of the cofactor or adjoint matrix. See here: en.m.wikipedia.org/wiki/Invertible_matrix $\endgroup$ – Robert Lewis Nov 3 '14 at 17:16
  • $\begingroup$ I got it, thank guys :) $\endgroup$ – Hafiz Temuri Nov 3 '14 at 21:54
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It is not too difficult to solve directly $$ \left(\begin{array}{rrr}% a&b&c\\% 0&d&e\\% 0&0&f\\% \end{array}\right)% \left(\begin{array}{rrr}% x&y&z\\% 0&y&v\\% 0&0&w\\% \end{array}\right)% = \left(\begin{array}{rrr}% 1&0&0\\% 0&1&0\\% 0&0&1\\% \end{array}\right)% $$ giving

$$ \left(\begin{array}{rrr}% 1/a& -b/(ad)&(be-cd)/(afd)\\% 0&1/d&-e/(fd)\\% 0&0&1/f\\% \end{array}\right)% $$ from which we see directly that the matrix is invertible if all $a,d$ and $f$ are different from zero.

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  • $\begingroup$ That's my question, how you come up with that inverse? cos when I try to solve it, I got lower triangular matrix. Please check my steps which I just added to my main question, and let me know if you find any mistake in it. Thanks! $\endgroup$ – Hafiz Temuri Nov 3 '14 at 15:45
  • $\begingroup$ Note. A particular case appears in the study of the Heisenberg group, wherein $a = d = f = 1$. Then the inverse greatly simplifies. $\endgroup$ – dohmatob Nov 8 '18 at 19:23
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There is a nice trick for calculating the inverse of any invertible upper triangular matrix, one which avoids the computation of complicated determinants. Since it works for any such upper (or lower) triangular matrix $T$ of any size $n$, I'll explain it in that context.

The first thing one needs to remember is that the determinant of a triangular matrix is the product of its diagonal entries. This may easily be seen by induction on $n$. It is trivially true if $n = 1$; for $n = 2$, we have

$T= \begin{bmatrix} t_{11} & t_{12} \\ 0 & t_{22} \end{bmatrix}, \tag{1}$

so obviously

$\det(T) = t_{11} t_{22}. \tag{2}$

If we now formulate the inductive hypothesis that

$\det(T) = \prod_1^k t_{ii} \tag{3}$

for any upper triangular $T$ of size $k$,

$T = [t_{ij}], \; \; 1 \le i, j \le k, \tag{4}$

then for $T$ of size $k + 1$ we have that

$\det(T) = t_{11} \det(T_{11}), \tag{5}$

where $T_{11}$ is the $k \times k$ matrix formed by deleting the first row and comumn of $T$. (4) follows easily from the expansion of $\det(T)$ in terms of its first-column minors (see this wikipedia page), since $t_{i1} = 0$ for $i \ge 2$. From our inductive hypothesis,

$\det(T_{11}) = \prod_2^{k + 1} t_{ii}, \tag{6}$

whence from (5)

$\det(T) = t_{11} \det(T_{11}) = t_{11} \prod_2^{k + 1} t_{ii} = \prod_1^{k + 1} t_{ii}, \tag{7}$

proving our assertion.

It follows immediately from (7) that the characteristic polynomial $p_T(\lambda)$ of $T$ is

$p_T(\lambda) = \det(T - \lambda I) = \prod_1^n (t_{ii} - \lambda), \tag{8}$

and from (8) that the eigenvalues of $T$ are precisely its diagonal entries, i.e. the $t_{ii}$, $1 \le i \le n$; also follows from (7) the related fact that $T$ is nonsingular, that is, $\det(T) \ne 0$, precisely when its diagonal entries are all nonzero.

For non-singular $T$ we may compute $T^{-1}$ as follows: write

$T = \Lambda + T_u, \tag{9}$

where $\Lambda$ is the diagonal matrix formed from the diagonal of $T$; viz.,

$\Lambda = [\delta_{ij} t_{ij}]; \tag{10}$

then $\Lambda$ is nonsingular and $T_u = T - \Lambda$ is the strictly upper triangular matrix obtained by setting the diagonal of $T$ to zero, i.e. setting $t_{ii} = 0$ for $1 \le i \le n$. We may write

$T = \Lambda (I + \Lambda^{-1} T_u), \tag{11}$

whence

$T^{-1} = (I + \Lambda^{-1} T_u)^{-1} \Lambda^{-1}. \tag{12}$

The matrix $\Lambda^{-1} T_u$ occurring in (12) is itself in fact strictly upper triagnular as well as is $T_u$; indeed, for any diagonal $D$, $DT_u$ is strictly upper tirangular, an assertion which is easily validated by direct calculation. It follows that $\Lambda^{-1} T_u$ is in fact nilpotent; that is, $(\Lambda^{-1} T_u)^n = 0$. We may now use the well-known algebraic identity

$(1 + x)(\sum_0^m (-x)^j) = 1 - (-x)^{m + 1}, \tag{13}$

easily seen to hold in any unital ring, applied to the matrix $x =\Lambda^{-1} T_u$, yielding, with $m = n - 1$,

$(I + \Lambda^{-1}T_u)(\sum_0^m (-\Lambda^{-1}T_u)^j) = I - (-\Lambda^{-1}T_u)^{m + 1} = I - (-\Lambda^{-1}T_u)^n = I. \tag{13}$

(13) shows that the inverse of $I + \Lambda^{-1}T_u$ is given by

$(I + \Lambda^{-1} T_u)^{-1} = \sum_0^m (-\Lambda^{-1}T_u)^j. \tag{14}$

It follows from (14) that $(I + \Lambda T_u)^{-1}$ is upper triangular, since each of the matrices $(-\Lambda^{-1}T_u)^j$, $j \ge 1$, is strictly upper triangular, and $(-\Lambda^{-1}T_u)^0 = I$. It further follows then that $T^{-1} = (I + \Lambda T_u)^{-1}\Lambda^{-1}$ is also upper triangular, being the product of the upper triangular matrix $(I + \Lambda T_u)^{-1}$ and the diagonal matrix $\Lambda^{-1}$. We have thus shown that the inverse of any invertible upper triangular matrix, of any size $n$, is itself an upper triangular matrix.

The inverse of any invertible matrix is invertible, the inverse of the inverse being the original matrix.

We can apply these considerations to the calculation of $A^{-1}$, where

$A = \begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{bmatrix}; \tag{14}$

here we have

$\Lambda = \begin{bmatrix} a & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & f \end{bmatrix} \tag{15}$

and

$T_u = \begin{bmatrix} 0 & b & c \\ 0 & 0 & e \\ 0 & 0 & 0 \end{bmatrix}; \tag{16}$

then

$\Lambda^{-1} T_u = \begin{bmatrix} a^{-1} & 0 & 0 \\ 0 & d^{-1} & 0 \\ 0 & 0 & f^{-1} \end{bmatrix} \begin{bmatrix} 0 & b & c \\ 0 & 0 & e \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix}; \tag{17}$

$(\Lambda^{-1} T_u)^2 = \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & bea^{-1}d^{-1} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}; \tag{18}$

$(\Lambda^{-1} T_u)^3 = 0; \tag{19}$

$\sum_0^2 (-\Lambda^{-1} T_u)^j = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & bea^{-1}d^{-1} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ $= \begin{bmatrix} 1 & -ba^{-1} & (be - cd)a^{-1}d^{-1} \\ 0 & 1 &- ed^{-1} \\ 0 & 0 & 1 \end{bmatrix}; \tag{20}$

finally,

$T^{-1} = (I + \Lambda^{-1} T_u)^{-1} \Lambda^{-1} = (\sum_0^2 (-\Lambda^{-1} T_u)^j) \Lambda^{-1}$ $= \begin{bmatrix} 1 & -ba^{-1} & (be - cd)a^{-1}d^{-1} \\ 0 & 1 &- ed^{-1} \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} a^{-1} & 0 & 0 \\ 0 & d^{-1} & 0 \\ 0 & 0 & f^{-1} \end{bmatrix}$ $= \begin{bmatrix} a^{-1} & -ba^{-1}d^{-1} & (be - cd)a^{-1}d^{-1}f^{-1} \\ 0 & d^{-1} &- ed^{-1}f^{-1} \\ 0 & 0 & f^{-1} \end{bmatrix}, \tag{21}$

this in agreement with Nimda's calculations. Indeed, we have

$\begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{bmatrix}\begin{bmatrix} a^{-1} & -ba^{-1}d^{-1} & (be - cd)a^{-1}d^{-1}f^{-1} \\ 0 & d^{-1} &- ed^{-1}f^{-1} \\ 0 & 0 & f^{-1} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \tag{22}$

as some simple algebra reveals.

Of course all this stuff applies to lower triangular matrices as well, and the demonstrations are similar and analogous, that is, essentially the same.

Hope this helps! Cheers,

and as always,

Fiat Lux!!!

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    $\begingroup$ Great answer! Can you please take a look at my question too? $\endgroup$ – Henry Apr 6 '18 at 19:33
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    $\begingroup$ Thanks. Yeah, give me a few minutes. And thanks for the bounty-boost! Cheers! $\endgroup$ – Robert Lewis Apr 6 '18 at 19:45
  • $\begingroup$ @Henry: I took a look at your problem but have nothing quite certain to add yet. Still thinking . . . $\endgroup$ – Robert Lewis Apr 6 '18 at 22:21
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As the question was made once more alive I will give for it the answer in a much more general sense, valid not only for the upper-triangularity property of matrices, but also for other properties if they are present in the described below circumstances.

Now suppose that for some matrices $A,B$ you consider a pattern of entries, say it could be mentioned "upper-triangularity" (UT) and you have proved that for any matrices with UT property the sum $A+B$ and the product $AB$ preserves UT (what is easy to prove).

If so then also powers $A^k$ preserve UT.

Consequently since any inverse can be expressed as polynomial $p(A)$ of $A$ directly calculated from Cayley-Hamilton theorem then also $ A^{-1} $ has the UT property.

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