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$$U(a,b)=\left(\begin{matrix}a&b&b&b\\b&a&b&b\\b&b&a&b\\b&b&b&a\end{matrix}\right)$$ Is there an easy way to find the inverse of $U(1,2)$ , a trick to solve this problem easy?(its part of an exam with answers)

Martin Sleziak
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Lola
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4 Answers4

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Note that $$U(a,b)U(c,d) = U(ac+3bd, bc+ad+2bd)$$ This can be seen because, if we let $A$ be the matrix each of whose entries are $1$'s, since $U(a,b)=(a-b)I+bA$, and since $A^2=4A$ $$((a-b)I+bA)((c-d)I+dA)$$ $$ = (a-b)(c-d)I+(b(c-d)+d(a-b)+4bd)A$$ $$= ((ac+3bd)-(bc+ad+2bd))I+(bc+ad+2bd)A$$ So, if $U(a,b)^{-1}$ exists and is equal to $U(c,d)$ for some $c,d$, then we would have $ac+3bd=1$ and $bc+ad+2bd=0$. The second equation gives us $$bc = -(a+2b)d$$ The determinate of $U(a,b)$ is actually $(a-b)^3(a+3b)$, so suppose $U(a,b)\neq 0$, so that neither of these factors are zero. Plugging the above into the first equation, $$-a(a+2b)d+3b^2d=b$$ $$\implies d = \frac{b}{-a^2-2ab+3b^2} = -\frac{b}{(a-b)(a+3b)}$$ Finally, $$c = \frac{a+2b}{(a-b)(a+3b)}$$ Therefore, assuming $a\neq b$ and $a\neq -3b$, $$U(a,b)^{-1}=U\left(\frac{a+2b}{(a-b)(a+3b)} , -\frac{b}{(a-b)(a+3b)}\right)$$

florence
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    To compute the determinant of $U(a,b)$, one can look here : https://math.stackexchange.com/questions/904926/determinant-of-a-rank-1-update-of-a-scalar-matrix-or-characteristic-polynomia – Arnaud D. Jun 12 '17 at 10:09
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this is $$ (a-b)I + b J, $$ where $I$ is the identity matrix and $J$ is the matrix of all ones.

The main observation is that $J^2 = n J,$ with square $n$ by $n$ matrices. Here, $$ J^2 = 4 J. $$

So, solve $$ \left( (a-b)I + b J\right) \left( xI + y J\right) = I $$ Which becomes the system of equations in $x,y$ $$ (a-b)x = 1, \; \; \; (a-b) y + bx + 4by = 0 $$ $$ x = \frac{1}{a-b}, \; \; \; (a+3b) y + \frac{b}{a-b} = 0, $$ $$ x = \frac{1}{a-b}, \; \; \; y = \frac{-b}{(a-b)(a+3b)}. $$ THE REQUESTED INVERSE IS $$ \frac{1}{a-b} I - \frac{b}{(a-b)(a+3b)} J $$ In particular, the resulting diagonal entries (which are $x+y$) are $$ \frac{a +2b}{(a-b)(a+3b)} $$

Note that this is sometimes impossible; it is easy to find the eigenvalues of the original matrix. That is, we can tell easily when it has no inverse. The eigenvalues are, simply, $(a+3b,a-b,a-b,a-b).$

The columns of the matrix below are pairwise perpendicular and are eigenvectors of the given matrix. This fails to be an orthogonal matrix in that the columns are of varying lengths, but that can be fixed by dividing the columns by four separate numbers, namely $2, \sqrt 2, \sqrt 6, \sqrt {12}$ $$ \left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 2 & -1 \\ 1 & 0 & 0 & 3 \end{array} \right). $$

Will Jagy
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    This does not appear to naswer the question. Where is the inverse? – Bill Dubuque Jun 12 '17 at 01:14
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    @BillDubuque I guess the point here is that we don't need to work in the 4-dimensional algebra of circulant matrices, but can work inside the simpler 2-dimensional algebra spanned by $I$ and $J$ instead. – Jyrki Lahtonen Jun 12 '17 at 08:14
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    @Jyrki My point is one should not have to "guess" what is intended to finish the solution. Reducing to some other problem P (which may not necessarily be easier for readers), then concluding by saying "so solve P" does not constitute an answer without also explaining how one intends to solve P. I've lost count of half-baked MSE "answers" that lead nowhere. It is essential to have higher standards for answers. – Bill Dubuque Jun 12 '17 at 14:23
  • I left finding the inverse of $a+b(x+x^2+x^3)$ modulo $x^4-1$ as an exercise for reader, so I will cheerfully let Will leave finding the inverse of $(a-b)+bx$ modulo $x^2-4x$ as an exercise for the reader as well. CRT applies there equally well AFAICT. – Jyrki Lahtonen Jun 12 '17 at 15:14
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    @Jyrki But you explained explicitly what needs to be done to complete the solution, and one way to do so.. Many readers will have no idea how to infer that simply from the above terse remark "So solve ...". As written, this is at most a comment, not an answer. – Bill Dubuque Jun 12 '17 at 15:46
  • @Will Kudos for adding the extra details (I had hoped my original comment would have been enough of a nudge). – Bill Dubuque Jun 12 '17 at 18:44
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Note $\, U = a + b x + b x^2 + b x^3\, $ for a circulant $x$ satisfying $\,0 = x^4 - 1 = (x\!-\!1)\ (x\!+\!1)(x^2\!+\!1)$. Thus it suffices to compute $U^{-1}\!\!\pmod{x^4\!-1}$ by CRT, working modulo the above listed factors.

${\rm mod}\ \color{#0a0}{(x\!+\!1)(x^2\!+\!1)}\!:\,\ U^{-1}\equiv \color{#c00}{\dfrac{1}{a\!-\!b}}\ $ by $\ x\equiv -1 $ or $\,x^2\equiv -1\,\Rightarrow\, U\equiv a\!-\!b\,$

${\rm mod}\ x\!-\!1\!:\, \ \dfrac{1}{a\!+\!3b}\equiv U^{-1}\equiv \color{#c00}{\dfrac{1}{a\!-\!b}} + c\:\color{#0a0}{(x\!+\!1)(x^2\!+\!1)}\equiv \dfrac{1}{a\!-\!b}+4c\ $ by $\ x\equiv 1$

hence $\ 4c \equiv \dfrac{1}{a\!+\!3b}-\dfrac{1}{a\!-\!b}\ $ therefore $\ c \equiv \dfrac{-b\ \ \ }{(a\!-\!b)(a\!+\!3b)}$

Thus $\ U^{-1} = \underbrace{\dfrac{1}{a\!-\!b} + \color{#0a0}{c}}_{\large d\,} +\color{#0a0}{c\,(x\!+\!x^2\!+\!x^3)}\, =\, \bbox[6px,border:1px solid red]{U(d,\,c)}\,,\ $ $\, d = \dfrac{a\!+\!2b}{(a\!-\!b)(a\!+\!3b)}$

Bill Dubuque
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    That is very nice. You get the $U^{-1}=1/((b-a)(a+3b))\cdot (-a-2b + bx +bx^2+bx^3)$ almost for free. – sharding4 Jun 12 '17 at 01:43
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    @sharding4 Your comment reminds me of some graffiti I once saw at MIT. $\begin{align}\ &\text{Now look at them yo-yo's that's the way you do it}\ &\text{You solve the congruences by the CRT}\ &\text{That ain't workin' that's the way you do it}\ &\text{Inverses for nothin' and tricks for free} \end{align}$ – Bill Dubuque Jun 12 '17 at 02:40
  • Nice! I only figured out a reasonable way to do this "in my sleep" - it was past midnight when I posted my answer :-) – Jyrki Lahtonen Jun 12 '17 at 05:11
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    @Jyrki Ah, dreamed proofs - probably every number theorist has dreamed elementary proofs of FLT. Btw, for readers too young to recognize the lyrical pun, it is morphed from the song Money for Nothing by Dire Straits. "I want my CRT, I want my CRT ..." – Bill Dubuque Jun 12 '17 at 16:06
  • LOL. My sister was somewhat of a fan of Dire Straits back in the day while I was doing a Sam the Eagle impersonation. – Jyrki Lahtonen Jun 12 '17 at 16:42
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I will use the structure of a circulant matrix.

Let $T$ be the matrix $$ T=\left(\begin{array}{cccc}0&1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0 \end{array}\right). $$ We have $T^4=I$, and your matrix is $$ U(a,b)=C(T):=aI+bT+bT^2+bT^3. $$ The matrix $T$ has eigenvalues $1,i,-1,-i$, so there is a complex matrix $P$ such that $P^{-1}TP=diag(1,i,-1,-i)$. Actually it is easy to see that (the discrete Fourier transform 4x4 matrix $$ P=\left(\begin{array}{cccc} 1&1&1&1\\ 1&i&-1&-i\\ 1&-1&1&-1\\ 1&-i&-1&i \end{array}\right) $$ works. The same matrix $P$ diagonalizes all the powers $T^j$ and $U(a,b)$ has eigenvalues $C(1)=a+3b$, $C(i)=a-b$, $C(-1)=a-b$ and $C(-i)=a-b$, so $$ P^{-1}U(a,b)P=diag(a+3b,a-b,a-b,a-b). $$ From this we see that $$ U(a,b)^{-1}=P\ diag(1/(a+3b),1/(a-b),1/(a-b),1/(a-b))\ P^{-1}. $$ If you find a cubic polynomial $Q(T)$ such that $Q(1)=1/(a+3b)$ and $Q(\pm i)=Q(-1)=1/(a-b)$, then you can deduce that $$ U(a,b)^{-1}=Q(T). $$

Jyrki Lahtonen
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