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Give a basis to get the specific matrix M

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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*}
It should be:
$$M^B_B(\phi_a) = M^E_B(\text{id}) \cdot a\cdot M^E_B(\text{id})^{-1} = M^B_E(\text{id})^{-1} \cdot a\cdot M^B_E(\text{id})
=\begin{pmatrix}1 & 1 & -1 \\ 1 & 0 & 1 \\ 1 & -1 & 0\end{pmatrix}^{-1}\cdot \begin{pmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}\cdot \begin{pmatrix}1 & 1 & -1 \\ 1 & 0 & 1 \\ 1 & -1 & 0\end{pmatrix}
$$
🤔

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 $M^E_B(\text{id})$. 🧐

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?
Correct. (Nod)
And yes, we can consider both cases together.

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
Indeed.
So what can we conclude? 🤔

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 $b_1=\begin{pmatrix}1 \\ 0\end{pmatrix}$.
What does that mean for the question in the OP? 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
Correct. (Nod)
And yes, we can consider both cases together.
So we have a basis for both cases, for an upper tridiagonal matrix and a diagonal matrix.


Indeed.
So what can we conclude? 🤔
That means that there is no basis so that we get a diagonal matrix, right? But what about a tridiagonal matrix? :unsure:

What does that mean for the question in the OP? 🤔
That means that there is no basis so that we get a diagonal matrix, right? But what about a tridiagonal matrix? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
That means that there is no basis so that we get a diagonal matrix, right? But what about a tridiagonal matrix?
Did you you mean upper triangular instead of tridiagonal? 🤔

We needed a real eigenvalue for an upper triangular matrix, didn't we?
Or alternatively, we can find a solution for $-2x_1 y_1 -y_1^2+x_1^2=0$.
We can solve it as a quadratic equation with respect to $x_1$ can't we? 🤔


That means that there is no basis so that we get a diagonal matrix, right? But what about a tridiagonal matrix?
Upper triangular instead of tridiagonal?

We found that we needed a real eigenvalue and its eigenvector to ensure the lower left matrix entry becomes $0$.
What else do we need? 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
Did you you mean upper triangular instead of tridiagonal? 🤔

We needed a real eigenvalue for an upper triangular matrix, didn't we?
Or alternatively, we can find a solution for $-2x_1 y_1 -y_1^2+x_1^2=0$.
We can solve it as a quadratic equation with respect to $x_1$ can't we? 🤔
Oh yes, I mean upper triangular matrix not tridiagonal.

Since there only complex roots, there is no basis such that the matrix $M$ is a diagonal nor an upper triangular matrix, right?


We found that we needed a real eigenvalue and its eigenvector to ensure the lower left matrix entry becomes $0$.
What else do we need? 🤔
So can we consider any other vector as $b_2$, that is linearly independent to $b_1$, so that the matrix $M$ has the desired form?


:unsure:
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
Since there only complex roots, there is no basis such that the matrix $M$ is a diagonal nor an upper triangular matrix, right?

So can we consider any other vector as $b_2$, that is linearly independent to $b_1$, so that the matrix $M$ has the desired form?
Yes and yes. (Nod)

And we should verify that the matrix actually does get the desired form. (Sweating)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
\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)
 
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mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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)