# why is the $\det T=\pm1$ for this linear transformation?

I am a little familiar with the linear transformation but unable to understand why $\det T = \pm1$ for $(Tp)(x)=a_n+ a_{n-1}x+...+a_0x^n$ where$T:V\to V$ a linear transformation ,V is a vector space of polynomials over R of degree $\le n$ and $p(x)=a_0+ a_1x +...+ a_nx^n$ in V? Why can't it be equal to other than $\pm1$?

• Hint: What happens if you apply $T$ twice? May 9, 2016 at 17:36
• @Crostul you misunderstood: the $a_i$ are the coefficients of the the polynomial $p$ (which is acted upon by $T$). May 9, 2016 at 17:39
• whether the basis elements are x^k s or $a_i x^k$s May 9, 2016 at 17:43
• @Crostul like any linear transformation, $T$ maps the zero vector (i.e. the zero oolynomial) to itself. May 9, 2016 at 18:35

If we apply $T$ twice on a polynomial $p$, then we end up where we started, in other words: $$T(Tp) = p \quad \Longrightarrow T^2=I$$ Taking determinants yields $$\left(\det T \right)^2 = 1$$ And hence $$\det T = 1 \quad \text{or} \quad \det T = -1$$
The vector space $V$ has the basis $<e_0, e_1, \ldots, e_n>, e_i=x^i$ and the operator $T$ acts as follows $T(e_i)=e_{n-i}.$ It is easy to see that the matrix of the operatot is an antidiagonal matrix with the entries equal to 1. The determinant of such matrix is $(-1)^{\dfrac{n(n+1)}{2}}$.