- Admin
- #26

- Mar 5, 2012

- 9,591

It should be:Using the equality $M^B_B(\phi_a)=M^E_B(\text{id}) \cdot a \cdot M^E_B(\text{id})^{-1}$ we get

\begin{align*}M^B_B(\phi_a)&=M^E_B(\text{id}) \cdot a \cdot M^E_B(\text{id})^{-1} \\ & =\begin{pmatrix}-\frac{1}{3} & \frac{4}{3} & -1 \\ \frac{5}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{2}{3} & -\frac{1}{3}\end{pmatrix}\cdot \begin{pmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}\cdot \begin{pmatrix}1 & 1 & -1 \\ 1 & 0 & 1 \\ 1 & -1 & 0\end{pmatrix}^{-1}\\ & =\begin{pmatrix}-\frac{1}{3} & \frac{4}{3} & -1 \\ \frac{5}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{2}{3} & -\frac{1}{3}\end{pmatrix}\cdot \begin{pmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}\cdot \begin{pmatrix}\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & -\frac{2}{3} \\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3}\end{pmatrix} \\ & = \begin{pmatrix}\frac{2}{9} & -\frac{1}{9} & \frac{11}{9} \\- \frac{2}{9} & \frac{13}{9} & -\frac{8}{9} \\ 0 & \frac{1}{3} & \frac{1}{3}\end{pmatrix} \end{align*}

$$M^B_B(\phi_a) = M^E_B(\text{id}) \cdot a\cdot M^E_B(\text{id})^{-1} = M^B_E(\text{id})^{-1} \cdot a\cdot M^B_E(\text{id})

=\begin{pmatrix}1 & 1 & -1 \\ 1 & 0 & 1 \\ 1 & -1 & 0\end{pmatrix}^{-1}\cdot \begin{pmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}\cdot \begin{pmatrix}1 & 1 & -1 \\ 1 & 0 & 1 \\ 1 & -1 & 0\end{pmatrix}

$$

For the record, we generally have for a vector $v$:

$$\gamma_B(\phi_a(v)) =M^B_B(\phi_a)\cdot \gamma_B(v) = M^E_B(\text{id}) \cdot a\cdot M^B_E(\text{id})\cdot \gamma_B(v)$$

That is, we start with the representation of $v$ with respect to $B$, which is $\gamma_B(v)$.

We convert it into a representation with respect to $E$ using the matrix $M^B_E(\text{id})$.

Now we can apply the matrix $a$.

The result is a vector with respect to $E$.

So we convert it back into a representation with respect to $B$ using the matrix $M^E_B(\text{id})$.

Correct.So $\mathcal{M}_{\mathcal{B}_1}(\phi_1)$ is a diagonal matrix and so also an upper triangular matrix if it is of the form $\begin{pmatrix}u_{11} & 0 \\ 0 & u_{22}\end{pmatrix}$.

Is that correct? Or can we not consider both cases (upper tridiagonal and diagonal) together?

And yes, we can consider both cases together.

Indeed.For map $\phi_2$ we cannot do that like that :

\begin{equation*}\phi_2 \begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}\end{equation*}

There are only complex eigenvalues

So what can we conclude?

What does that mean for the question in the OP?For the last map we have:

\begin{equation*}\phi_3 \begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}\end{equation*}

So there is only one eigenvalue $u_{11}=0$ and $b_1=\begin{pmatrix}1 \\ 0\end{pmatrix}$.