Eigenvalues, Eigenvectors and Diagonalisation of Matrix A

Question

I have the following question, from a Linear Algebra exam paper:

I have found the eigenvalues and vectors as well as the basis:

$ \lambda_1=2,\lambda_2=2,\lambda_3=1\\ \\ v_1=(-1,0,1)\\ v_2=(0,1,0)\\ v_3(0,-1,1)\\ \\ \\ \Gamma=\begin{pmatrix}-1&0&0\\0&1&-1\\1&0&1\end{pmatrix}$

And the consequently I can diagonalise A as:

$\Gamma^{-1}A\Gamma=\begin{pmatrix}2&0&0\\0&2&0\\0&0&1\end{pmatrix}$

But in question c) they ask for $\mathcal{M}(id,\mathcal{E},\Gamma)$ and $\mathcal{M}(id,\Gamma,\mathcal{E})$ and I haven't seen this notation before (neither in the notes for the course, nor the book). I honestly don't know what they mean by it.

Answer 1

I think $M(f, A, B) $ stands for the matrix representing the linear application $f$, with respect to the basis $A$ in the domain space and basis $B$ in the image space. In the case of $f=id$, i.e. the identity function, this matrix is just the matrix for change of basis from $A$ to $B$

Answer 2

In that case, the answer must be:

$\mathcal{M}(id,\mathcal{E},\Gamma)=\Gamma^{-1}=\begin{bmatrix}1&0&1\\1&1&1\\-1&0&0\end{bmatrix}$

Because $\Gamma^{-1}A\Gamma=\Lambda$ ie. changes from usual coordinates to the $\Gamma$ base (ie. diagonalises A). I would usually write this $_{\Gamma}M_{\mathcal{E}}$

$\mathcal{M}(id,\Gamma,\mathcal{E})=\Gamma=\begin{bmatrix}0&0&-1\\-1&1&0\\-1&0&1\end{bmatrix}$

Because $\Gamma\Lambda\Gamma^{-1}=A$ changes back to the usual cooordinates from the $\Gamma$ coordinates. I would usually write this as $_{\mathcal{E}}M_{\Gamma}$