Cfrgtkky

There is a matrix $A =$

begin{pmatrix} 0 & a & b \ 1 & -b & -b\ -1 & a & a end{pmatrix}

where $a-b = 1$ . $A^2$ is an Identity matrix of order $3$ . Is it possible to find $A^2$ without even calculating $A^2$ ? Please help .

Yes, there's another way. It's certainly not easier, but it works. I suspect it uses a number of ideas that you have not yet encountered as well, but that's how life is sometimes. Anyhow, here goes:

First, you can compute the characteristic polynomial, $c(x) = det(A - xI)$, which is
begin{align}
c(x)
&= det pmatrix{-x & a & b \ 1 & -b - x & -b \ -1 & a & a - x}\
&= (-x) left( (-b-x)(a-x) + abright) - a left( (a-x) -b right) + b left(a + (-b-x) right)\
&= x left( (b+x)(a-x) - abright) - a left( a-x -b right) + b left(a -b-x right) & text{substitute $a-b = 1$ to get}\
&= x left( (b+x)(a-x) - abright) - a left( 1-x right) + b left(1-x right) \
&= x left( ba + (a-b)x - x^2 - abright) - a left( 1-x right) + b left(1-x right) \
&= x left(x - x^2 right) - a left( 1-x right) + b left(1-x right) \
&= x^2 - x^3 + (b-a)left( 1-x right) \
&= x^2 - x^3 -1left( 1-x right) \
&= x^2 - x^3 -1 + x \
&= -x^3 + x^2 + x -1 \
&= -(x-1)^2(x+1)
end{align}
That means that the eigenvalues of $A$ are $1, 1, -1$. Hence the Jordan normal form of $A$ is either
$$
A= P^{-1} pmatrix{1& 0& 0 \ 0 & 1 & 0 \ 0 & 0 & -1} P
$$
or
$$
A= P^{-1} pmatrix{1& 1& 0 \ 0 & 1 & 0 \ 0 & 0 & -1} P
$$
for some invertible matrix $P$.

In the first case, there are two eigenvectors corresponding to the eigenvalue $1$; in the second there's only one. We can check which of these happens by looking at the nullspace of $A - 1 cdot I$, whose dimension is the number of eigenvectors for $+1$:

begin{align}
A - 1 cdot I =
pmatrix{-1 & a & b \
1 & -b-1 & -b \
-1 & a & a-1} & text{substitute $a = b+1; a-1 = b$}\
=pmatrix{-1 & b+1 & b \
1 & -(b+1) & -b \
-1 & b+1 & b}
end{align}
The second and third columns are obviously multiples of the first, so $A - 1 cdot I$ has rank $1$, i.e., there are two eigenvectors for the eigenvalue $1$. That means that for some matrix $P$, we have
$$
A= P^{-1} pmatrix{1& 0& 0 \ 0 & 1 & 0 \ 0 & 0 & -1} P
$$
whence

begin{align}
A^2
&=
(P^{-1} pmatrix{1& 0& 0 \ 0 & 1 & 0 \ 0 & 0 & -1} P)
(P^{-1} pmatrix{1& 0& 0 \ 0 & 1 & 0 \ 0 & 0 & -1} P) \
&=
P^{-1} pmatrix{1& 0& 0 \ 0 & 1 & 0 \ 0 & 0 & -1} (P
P^{-1}) pmatrix{1& 0& 0 \ 0 & 1 & 0 \ 0 & 0 & -1} P \
&=
P^{-1} pmatrix{1& 0& 0 \ 0 & 1 & 0 \ 0 & 0 & -1}^2 P \
&=
P^{-1} pmatrix{1& 0& 0 \ 0 & 1 & 0 \ 0 & 0 & 1} P \
&=
P^{-1} I P \
&=
P^{-1} P \
&=
I.
end{align}

It seems to me that it would have been a great deal easier to just compute $A^2$ directly. :)

Here is another approach. It is not as nice as the other answer, but it is shorter. Perform the elementary row operations (in C programming language's notation):
row 3 += row 2; col 2 -= col 3. We see that $A$ is similar to
$$
B=pmatrix{
0&1&b\
1&0&-b\
0&0&1}.
$$
One can verify that $pmatrix{1\ 1\ 0},pmatrix{1\ -1\ 0}$ and $pmatrix{b\ -b\ 2}$ are eigenvectors of $B$ corresponding to the eigenvalues $1,-1$ and $1$ respectively. When the underlying field is not of characteristic $2$, these eigenvectors are linearly independent. Therefore $B$ and in turn $A$ are similar to $operatorname{diag}(1,-1,1)$. Hence their squares are equal to the identity matrix.

What?? If "$A^2$ is the identity matrix" then $A^3= A(A^2)= A$. No "calculation" at all required!

A solution without calculation.

from $tr(A)=a-b=1$, we deduce that

$A^2=I$ IFF $A$ is diagonalizable and ${1,1}subset spectrum(A)$

IFF $rank(A-I)=1$.

Clearly, we see that $rank(A-I)=rankbegin{pmatrix}-1&b+1&b\1&-b-1&-b\-1&b+1&bend{pmatrix}=1$.