Find the characteristic polynomial $f(t)$ of $T$, where $T(a,b,c,d) = (a+b,b-c,a+c,a+d)$

Question

Find the characteristic polynomial $f(t)$ of $T$ and verify that the characteristic polynomial of $T_{W}$ divides $f(t)$. Where $T:\mathbb{R}^{4} \to \mathbb{R}^{4}$ and $T(a,b,c,d) = (a+b,b-c,a+c,a+d)$

This question is tied to another question where I found the basis of the $T$-cyclic subspace, $W$, generated by $e_{1} = (1,0,0,0)$. In that case the basis $\beta = \{e_{1}, T(e_{1}), T^{2}(e_{1})\} = \{(1,0,0,0), (1,0,1,1), (1,-1,2,2)\}$.

So now arriving at this question I am asked to find the characteristic polynomial of $T$ over $V = \mathbb{R}^{4}$.

The only way I know of finding the characteristic polynomial is by $f(t) = \det(T - tI)$. I do also have the following theorem:

Is it the case that I would need to try a variety of different vectors in $\mathbb{R}^{4}$ (most likely only the basis vectors) and find a $T$-cyclic subspace of dimension $4$ from which I could then apply the result of this theorem? Because the other idea I had was forming the matrix of the linear operator using the basis vectors and then calculating the determinant, which would be very labor intensive and I have a feeling they don't want me to do that in the problem.

Is the approach I have the right one? As a question of curiosity what do we do when we want to find eignevalues for larger dimension matrices? From what I've read calculating the determinant is not efficient.

Answer 1

The matrix for $T$ in the standard basis for $\mathbb R^4$ is $$ \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} $$ So you're looking for the determinant of $$ \begin{pmatrix} 1-t & 1 & 0 & 0 \\ 0 & 1-t & -1 & 0 \\ 1 & 0 & 1-t & 0 \\ 1 & 0 & 0 & 1-t \end{pmatrix} $$ This matrix has enough zeroes in it that we can simply wing it: $$ \left| \begin{array}{cccc} 1-t & 1 & 0 & 0 \\ 0 & 1-t & -1 & 0 \\ 1 & 0 & 1-t & 0 \\ 1 & 0 & 0 & 1-t \end{array} \right| = \left| \begin{array}{cccc} 1-t & 1 & 0 \\ 0 & 1-t & -1 \\ 1 & 0 & 1-t \end{array} \right| \cdot (1-t) $$ since there was only one nonzero element in the last column. And now we can just apply the well-known canned formula for a $3\times 3$ determinant -- four of its six terms vanish and we're left with $$ \cdots = \bigl((1-t)^3 + 1\cdot(-1)\cdot 1\bigr) \cdot (1-t) = \bigl((1-t)^3-1\bigr)(1-t) $$ You could multiply out the right-hand side here, but it's more enlightening to leave it in this shape because it's easier to read off its roots this way: $$ t=1 \qquad\text{or}\qquad (1-t)^3 = 1 \implies \left\{ \begin{array}{lll} 1-t = 1 & \implies & t = 0 \\ 1-t = e^{2\pi i/3} & \implies & t = \frac32 - \frac{\sqrt 3}2 i \\ 1-t = e^{-2\pi i/3} & \implies & t = \frac32 + \frac{\sqrt 3}2 i \end{array}\right . $$

Answer 2

You want to find the characteristic polynomial of

$I+ \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{pmatrix}$

The characteristic polynomial of the matrix on the right can be found to be $x(x^3+1)$ by expanding along the bottom row, and so it has roots $0,\omega^1,\omega^3,\omega^5$ where $\omega$ is a $6$'th root of unity.

It follows the characteristic polynomial of $T$ is $(x-1)(x-(\omega+1))(x-(\omega^5+1))(x-(-1+1)) = x(x-1)( (x-(\omega+1))(x-(\omega^5+1)) = (x^2-x)(x^2-3x+3)$