# Find the matrix $T$ in another basis

Let $$V=\mathbb R_{\leq 2}[x]$$ be the vector space of polynomials of degree at most $$2$$ over $$\mathbb R$$. Let $$e_1, e_2, e_3$$ and $$e_1',e_2',e_3'$$ be the bases $$1,x,x^2$$ and $$1+x^2,1+2x,(1-x)^2$$.

Firstly I'm asked to write down the matrices of linear maps $$e_i\mapsto e_i'$$ and $$e_i'\mapsto e_i$$ which I've calculated as $$P=\begin{pmatrix}1&0&1\\1&2&0\\1&-2&1\end{pmatrix}$$ for the first, and then for the second this is just $$P^{-1}=\begin{pmatrix}-1&1&1\\0.5&0&-0.5\\2&-1&-1\end{pmatrix}$$

Now I have to write the linear map $$T:V\rightarrow V$$ given by $$T(f(x))=f''(x)$$ in the basis $$e_1,e_2,e_3$$, for which I got $$A=\begin{pmatrix}0&0&2\\0&0&0\\0&0&0\end{pmatrix}$$.

Now the final part is what I'm struggling on. It says to now write $$T$$ in the basis $$e_1',e_2',e_3'$$ using the matrices calculated in the first part.

I tried using the equation $$[T]_{e_i'}=S^{-1}[T]_{e_i}S$$ where $$S$$ is the change of basis matrix from $$e_i'$$ to $$e_i$$, i.e $$S=P^{-1}$$ (Not sure if I've used the right matrices)

I calculated $$[T]_{e_i'}=\begin{pmatrix}4&-2&-2\\4&-2&-2\\4&-2&-2\end{pmatrix}$$ and then to see if it worked I considered the polynomial $$x^2+1$$, whose second derivative is $$2$$ and calculated $$\begin{pmatrix}4&-2&-2\\4&-2&-2\\4&-2&-2\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}$$ (in this basis $$1+x^2$$ corresponds to $$\begin{pmatrix}1\\0\\0\end{pmatrix}$$), but this gives $$\begin{pmatrix}4\\4\\4\end{pmatrix}$$ which clearly isn't equal to $$2$$ in this basis. Can somebody tell me where I'm going wrong? Or what I'm not understanding?

• the problem is that you constructed wrong your base change matrix, paulinho did set correctly Apr 8, 2020 at 20:14
• Can you explain how it should be constructed please? Apr 8, 2020 at 20:15
• you see the coefficients of the linear combinations and write them in the columns of the matrix, you are writing them in the rows and that causes the fail Apr 8, 2020 at 20:18

The matrix that transforms from the basis $$\alpha = \{e_1, e_2, e_3\}$$ to $$\beta = \{e'_1, e'_2, e'_3\}$$ is incorrectly transposed. So for example, the linear map that takes the vectors of $$\alpha$$ to the respective vectors in $$\beta$$ is $$\begin{bmatrix} 1 &1 &1 \\ 0 & 2 &-2 \\ 1 & 0 &1 \end{bmatrix}$$ (Alternatively, you can view this as the change of basis from a vector $$p_\beta$$ in the basis $$\beta$$ to a vector $$p_\alpha$$ in the basis $$\alpha$$.) I believe that is the only problem here. The rest of the work seems correct to me.
EDIT: To answer the question, the confusion stems from the fact that the polynomials $$e_1, e_2, e_3$$ are to be treated like vectors, not scalars. For example, if $$\{1, x, x^2\}$$ is the basis, then we should think of the polynomial $$f(x) = 1$$ as the vector $$(1, 0, 0)^T$$ and the polynomial $$g(x) = 1 + 5x + 2x^2$$ as the vector $$(1, 5, 2)$$. So when looking for the matrix that takes $$e_1 \to e'_1$$, $$e_2 \to e'_2$$, $$e_3 \to e'_3$$, you should think of finding the matrix that takes $$(1, 0, 0)^T \to (1, 0, 1)^T, (0,1,0)^T \to (1, 2, 0)^T$$, and $$(0, 0, 1)^T \to (1, -2, 1)^T$$.
• My reasoning was that $\begin{pmatrix}1&0&1\\1&2&0\\1&-2&1\end{pmatrix}\begin{pmatrix}e_1\\e_2\\e_3\end{pmatrix}=\begin{pmatrix}e_1'\\e_2'\\e_3'\end{pmatrix}$ Can you explain what I'm misunderstanding? Apr 8, 2020 at 20:12