Show that $T^3-2T^2-8T=0$

Question

Let $T : \mathbb{R}^3\rightarrow \mathbb{R}^3$ be $T(x, y, z) = (x + y - 2z, x + 2y - 3z, 3x + 4y - z)$. Show that $T^3 - 2T^2 - 8T = 0$.

I was able to write $T(x, y, z) = (x + y - 2z, x + 2y - 3z, 3x + 4y - z)$ as $T(x,y,z)=A[x,y,z]$, where A=[$(1,1,-2),(1,2,-3),(3,4,-1)$] and [$x,y,z$] are matrices.

I thought i can replace T with the matrix A in the equation $T^3 - 2T^2 - 8T = 0$, but this is not true. Please help, how to solve this? And also tell why can i not replace $T$ with $A$ in the equation. Thank you.

Any help appreciated

proof that the characteristic polynomial equals $t^3-2t^2-8t$ — John D, Sep 07 '17 at 16:32
@Magnusseen but the characteristic polynomial is $ t^3-2t^2+16t-6=0$ — Shobhit, Sep 07 '17 at 16:38
Why don't you try evaluating $T^3 - 2T^2 - 8T$ in the canonical base ? If it is zero for each element of it, then that holds — Max, Sep 07 '17 at 16:42
Did you check with a few values that what you are trying to prove is true? For instance, do you get $0$ when you evaluate $T^3-2T^2-8T$ at $(x,y,z)=(1,0,0)$? — Andrés E. Caicedo, Sep 07 '17 at 16:43
You have $$T^3-2T^2-8T= T^3-2T^2 +16^T-6I_3-8T+6I_3= 6I_3-8T=0$$ iff $T=\frac{3}{4}I_3,$ a contradiction. — John D, Sep 07 '17 at 16:44
@Magnusseen i get it that the statement is wrong. thank you. — Shobhit, Sep 07 '17 at 16:50

José Carlos Santos · Accepted Answer · 2017-09-07T16:47:22.183

3

You can do that. The problem is that the statement is false. In fact,$$A^2=\begin{pmatrix}-4 & -5 & -3 \\ -6 & -7 & -5 \\ 4 & 7 & -17\end{pmatrix}\text{ and }A^3=\begin{pmatrix}-18 & -26 & 26 \\ -28 & -40 & 38 \\ -40 & -50 & -12\end{pmatrix}.$$Therefore$$A^3-2A^2-8A=\begin{pmatrix}-18 & -24 & 48 \\ -24 & -42 & 72 \\ -72 & -96 & 30\end{pmatrix}\neq0.$$

edited Sep 07 '17 at 16:47

answered Sep 07 '17 at 16:43

José Carlos Santos

427,504

ok. got it. so i leave this statement, and i calculate the characteristic equation for T. It is a three degree equation, with only one real root, and i have to calculate the minimal polynomial. how can i proceed now? please help – Shobhit Sep 07 '17 at 16:46
@NV-US check my last comment – John D Sep 07 '17 at 16:48
@NV-US Since all roots of the charateristic polynomial are simple roots (in $\mathbb C$), the minimal polynomial is the characteristic polynomial. – José Carlos Santos Sep 07 '17 at 16:50

Show that $T^3-2T^2-8T=0$

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