Give a basis to get the specific matrix M

[mathmari]

For the last map :

Let $b_2=\begin{pmatrix}0 \\ 1\end{pmatrix}$, then $\gamma_{\mathcal{B}_1}(\phi_3(b_2))=\gamma_{\mathcal{B}_1}\begin{pmatrix}1 \\ 0\end{pmatrix}=\begin{pmatrix}1 \\ 0\end{pmatrix}$.
So we get \begin{equation*}\mathcal{M}_{\mathcal{B}_1}(\phi_3)=\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}\end{equation*} which is an upper triangular matrix.There is no basis such that $\mathcal{M}_{\mathcal{B}_1}(\phi_3)$ is a diagonal matrix, because the element at the second column and first row has to be non zero to get linearly independent vectors, right?


:unsure:

 

[I like Serena]

\begin{align*}\phi_3:\mathbb{R}^2\rightarrow \mathbb{R}, \ \begin{pmatrix}x\\ y\end{pmatrix} \mapsto \begin{pmatrix}y\\ 0\end{pmatrix} \end{align*}
1. Give (if possible) for each $i\in \{1,2,3\}$ a Basis $B_i$ of $\mathbb{R}^2$ such that $M_{B_i}(\phi_i)$ an upper triangular matrix.
2. Give (if possible) for each $i\in \{1,2,3\}$ a Basis $B_i$ of $\mathbb{R}^2$ such that $M_{B_i}(\phi_i)$ an diagonal matrix.

Let $b_2=\begin{pmatrix}0 \\ 1\end{pmatrix}$, then $\gamma_{\mathcal{B}_3}(\phi_3(b_2))=\gamma_{\mathcal{B}_3}\begin{pmatrix}1 \\ 0\end{pmatrix}=\begin{pmatrix}1 \\ 0\end{pmatrix}$.
So we get \begin{equation*}\mathcal{M}_{\mathcal{B}_3}(\phi_3)=\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}\end{equation*} which is an upper triangular matrix.

I've taken the liberty to change $\mathcal{B}_1$ into $\mathcal{B}_3$ in the above quote.
That was what was intended wasn't it? :unsure:

The eigenvalue was $0$ wasn't it?
Shouldn't we have $u_{11}=0$ then?
I don't think we have the correct $\mathcal{M}_{\mathcal{B}_3}(\phi_3)$. (Shake)

There is no basis such that $\mathcal{M}_{\mathcal{B}_3}(\phi_3)$ is a diagonal matrix, because the element at the second column and first row has to be non zero to get linearly independent vectors, right?

Correct. (Nod)

 

[mathmari]

The eigenvalue was $0$ wasn't it?
Should we have $u_{11}=0$ then?
I don't think we have the correct $\mathcal{M}_{\mathcal{B}_3}(\phi_3)$. (Shake)

Ahh yes, we have \begin{equation*}\mathcal{M}_{\mathcal{B}_3}(\phi_3)=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\end{equation*} :unsure:

 

[I like Serena]

Ahh yes, we have \begin{equation*}\mathcal{M}_{\mathcal{B}_3}(\phi_3)=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\end{equation*}

Right. (Nod)

 

[mathmari]

Right. (Nod)

Thank you !