Let $T(v_i)=v_i+v_{i-1}$. Find $[T_\beta]$ and $[T_\gamma]$.

Question

Let $\beta=(v_1,..v_n)$ Let $\gamma=(v_1,v_2-v_1,v_3-v_2,...v_n-v_{n-1})$ be basis of a finite dimensional vector space and $T:V\to V$ where $T(v_1)=v_1$ and $T(v_i)= v_i+v_{i-1}$

I believe the matrix of $Q$ here changes $\gamma$ coordinates to $\beta$ coordinates looks like this for example when $n=4$: $$Q=\left[ {\begin{array}{cccc} 1&-1&0&0\\ 0&1&-1&0\\ 0&0&1&-1\\ 0&0&0&1\\ \end{array} } \right]$$

So from my understanding to find $[T]_\gamma$ all I need to do is apply the transformation to this matrix, I get this new matrix.

$$[T]_\gamma=\left[ {\begin{array}{cccc} 1&0&-1&0\\ 0&1&0&-1\\ 0&0&1&0\\ 0&0&0&1\\ \end{array} } \right]$$

And then using this I would find $[T]_\beta$ using this formula $[T]_\beta=Q[T]_\gamma Q^{-1}$

I find $$Q^{-1}=\left[ {\begin{array}{cccc} 1&1&1&1\\ 0&1&1&1\\ 0&0&1&1\\ 0&0&0&1\\ \end{array} } \right]$$

And then using the formula shown above I find $$[T]_\beta=\left[ {\begin{array}{cccc} 1&0&-1&0\\ 0&1&0&-1\\ 0&0&1&0\\ 0&0&0&1\\ \end{array} } \right]$$

Is this correct?

I'm quite sure that I multiplied the matrices in the correct order and somehow I still end up with $[T]_\gamma=[T]_\beta$ So I'm pretty sure I've got the wrong idea how to to get $[T]_\gamma$ or my original matrix $Q$ is incorrect. Not homework just something I made for practice.

Answer 1

You should start by writing $[T]_\beta$ since your linear map $T$ is given in terms of the basis $\beta$:

$$[T]_\beta = \pmatrix{1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1}$$

Now $$[T]_\gamma = [I]_{\gamma, \beta} [T]_\beta [I]_{\beta, \gamma} = Q^{-1}[T]_\beta Q = \pmatrix{1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1} \pmatrix{1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1} \pmatrix{1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1}$$

We get

$$[T]_\gamma = \pmatrix{1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1}$$

which is the same as $[T]_\beta$ since

$$Tv_1 = v_1$$

$$T(v_i - v_{i-1}) = Tv_i - Tv_{i-1} = v_i + v_{i-1} - v_{i-1} - v_{i-2} = (v_i - v_{i-1}) + (v_{i-1} - v_{i-2})$$

so, interestingly, $T$ acts on $\gamma$ the same way it acts on $\beta$.