Is it possible to solve and find out the square of the matrix value answer without calculating?

Question

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 .

Answer 1

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) + ab\right) - a \left( (a-x) -b \right) + b \left(a + (-b-x) \right)\\ &= x \left( (b+x)(a-x) - ab\right) - 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) - ab\right) - a \left( 1-x \right) + b \left(1-x \right) \\ &= x \left( ba + (a-b)x - x^2 - ab\right) - 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 -1\left( 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. :)

Answer 2

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.

Answer 3

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

Answer 4

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 $rank(A-I)=rank\begin{pmatrix}-1&b+1&b\\1&-b-1&-b\\-1&b+1&b\end{pmatrix}=1$.