Find trace and determinant of matrix $A$ such that $A^2 = I$.

Question

I have a $ 2\times 2$ matrix $A$, where $A^2 = I$.

So the eigenvalues are $\lambda= \pm1$ .

I need to find its trace and determinant. There's no mention of upper or lower triangular matrix, therefore, the formulas for Trace and determinant := product of eigenvalues can't be used here.

The other part does give some specific information but I wonder is there any way to solve the first part because that's how it should be solved.

b) If the first row is (3,−1), what is the second row?

How can I solve this? Thanks in advance.

Answer 1

The fact that $A^2-I=0$ means precisely that $A$ is diagonalisable with its set of eigenvalues contained in the set $\{-1,1\}$ of (simple) roots of $X^2-1$. The multiset of roots for the characteristic polynomial $\chi_A$, which has size $2$, can then be one of $\{\!\{-1,-1\}\!\}$, $\{\!\{-1,1\}\!\}$, and $\{\!\{1,1\}\!\}$. The trace and determinant are respectively the sum and the product of that multiset, giving three possibilities $(-2,1)$, $(0,-1)$, $(2,1)$ for the pair $(\operatorname{tr}A,\det A)$.

Since being diagonalisable with a single eigenvalue$~\lambda$ means being equal to $\lambda I$, the outer two possibilities only occur for $-I$ respectively for $I$, and so they can be excluded for question b).

Answer 2

HINT

Here is a very direct approach. Let $$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ Then find $A^2$ and explicitly set to $I$ -- that gives you 4 equations in 4 unknowns...

After you discover the needed relationships between $a,b,c,d$, note that $\det A = ab - bc$ and $\mathrm{tr} A = a+d$...

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UPDATE

Many comments below claim it is a hard system to solve. Here is my approach: $$ A^2 = \begin{bmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I, $$ so $b(a+d) = 0 = c(a+d)$. Hence, either $b=c=0$ or $a=-d$.

In the first case, $a,d = \pm 1$ and $b,c = 0$, a fairly narrow set of choices.

In the second case, imposing the upper left and lower right equations, $$a^2 = 1 - bc = d^2 \iff a = -d = \pm \sqrt{1-bc},$$ which also leaves a very narrow margin of choices.

Answer 3

By Cayley-Hamilton, if $A\ne\pm I$, then we have that the characteristic polynomial of $A$ is $x^2-\mathrm{tr}(A)x+\det(A)=x^2-1$. Thus $\mathrm{tr}(A)=0$ and $\det(A)=-1$.

Answer 4

Do the two cases $A=\pm \,I$ separately. Henceforth, $A\ne\pm \,I.$ Let $f(x)=x^2-1.$ $f(A)$ is zero and neither proper divisor of $f(x)$ annihilates $A.$ Thus $f(x)$ is the minimal polynomial of $A$ and consequently the characteristic polynomial of $A.$ By the usual formula, the trace of $A$ is zero and $|A|=-1.$