How to Compute Coordinate Column Vectors in Different Bases?

[mathmari]

Hey!

We have the matrices $E_{k\ell}\in \mathbb{R}^{2\times 2}$ with $1$ iin the position $(k,\ell)$ and $0$ in the other positions and \begin{equation*}\sigma_0=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}, \ \sigma_1=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}, \ \sigma_2=\begin{pmatrix}0&-i\\ i&0\end{pmatrix}, \ \sigma_3=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix} \ \in \mathbb{C}^{2\times 2}\end{equation*}

We have the bases $\displaystyle{B=\left (E_{11}, \ E_{12}+E_{21}, \ i(E_{12}-E_{21}), \ E_{22}\right )}$ and $\displaystyle{C=\left (\sigma_0, \ldots , \sigma_3\right )}$ of $\mathbb{R}$-vector space \begin{equation*}H:=\left \{\begin{pmatrix}a & b+ic\\ b-ic & d\end{pmatrix}\in \mathbb{C}^{2\times 2}: a, b, c, d\in \mathbb{R}\right \}\end{equation*}

Compute for $v\in H$ the coordinate column vectors $v_B$ and $v_C$ and verify that $v_C=A_{\text{id}, B,C}\cdot v_B$. I have done the following:

Let $v\in H$. Then $v=\begin{pmatrix}a & b+ic\\ b-ic & d\end{pmatrix}$ for $a, b, c, d\in \mathbb{R}$.

Let $v_B=\begin{pmatrix}x \\ y \\ z \\ w \end{pmatrix}$ be the coordinate column vector. Since the elements of the basis $B$ are real matrices, we have that $x,y,z,w\in \mathbb{R}$.

Then \begin{equation*}v=x\cdot E_{11}+y\cdot \left ( E_{12}+E_{21} \right )+z \cdot \left [ i(E_{12}-E_{21})\right ]+ w \cdot E_{22}\end{equation*}

So, we have the following:
\begin{align*}\begin{pmatrix}a & b+ic\\ b-ic & d\end{pmatrix}&=x\cdot \begin{pmatrix}1& 0 \\ 0 & 0\end{pmatrix}+y\cdot \left [ \begin{pmatrix}0& 1 \\ 0 & 0\end{pmatrix}+\begin{pmatrix}0& 0 \\ 1 & 0\end{pmatrix} \right ]+z \cdot i\cdot \left [\begin{pmatrix}0& 1 \\ 0 & 0\end{pmatrix}-\begin{pmatrix}0& 0 \\ 1 & 0\end{pmatrix}\right ]+ w \cdot \begin{pmatrix}0& 0 \\ 0 & 1\end{pmatrix} \\ & = x\cdot \begin{pmatrix}1& 0 \\ 0 & 0\end{pmatrix}+y\cdot \begin{pmatrix}0& 1 \\ 1 & 0\end{pmatrix}+z \cdot i\cdot \begin{pmatrix}0& 1 \\ -1 & 0\end{pmatrix}+ w \cdot \begin{pmatrix}0& 0 \\ 0 & 1\end{pmatrix} \\ & = \begin{pmatrix}x& 0 \\ 0 & 0\end{pmatrix}+ \begin{pmatrix}0& y \\ y & 0\end{pmatrix}+ \begin{pmatrix}0& z \cdot i \\ -z \cdot i & 0\end{pmatrix}+ \begin{pmatrix}0& 0 \\ 0 & w\end{pmatrix} \\ & = \begin{pmatrix}x& y+z \cdot i \\ y-z \cdot i & w\end{pmatrix} \end{align*}

So, we get:
\begin{equation*}\begin{cases}a=x \\ b+ic=y+iz \\ b-ic = y-iz \\ d=w\end{cases}\ \Rightarrow \ \begin{cases}x=a \\ y=b \\ z=c \\ w=d\end{cases}\end{equation*}

The coordinate column vector of $v$ in respect to the basis $B$ is \begin{equation*}v_B=\begin{pmatrix}a \\ b \\ c \\ d \end{pmatrix}\end{equation*}



Let $v_C=\begin{pmatrix}x \\ y \\ z \\ w \end{pmatrix}$ be the coordinate column vector. Since the elements of the basis $C$ are complex matrices, we have that $x,y,z,w\in \mathbb{C}$.

Then \begin{equation*}v=x\cdot \sigma_0+y\cdot \sigma_1+z \cdot \sigma_2+ w \cdot \sigma_3\end{equation*}

So, we have the following:
\begin{align*}\begin{pmatrix}a & b+ic\\ b-ic & d\end{pmatrix}&=x\cdot \begin{pmatrix}1&0\\ 0&1\end{pmatrix}+y\cdot \begin{pmatrix}0&1\\ 1&0\end{pmatrix}+z \cdot \begin{pmatrix}0&-i\\ i&0\end{pmatrix}+ w \cdot \begin{pmatrix}1&0\\ 0&-1\end{pmatrix} \\ & = \begin{pmatrix}x&0\\ 0&x\end{pmatrix}+ \begin{pmatrix}0&y\\ y&0\end{pmatrix}+ \begin{pmatrix}0&-z \cdot i\\ z \cdot i&0\end{pmatrix}+ \begin{pmatrix}w&0\\ 0&-w\end{pmatrix} \\ & = \begin{pmatrix}x+w&y-z \cdot i\\ y+z \cdot i&x-w\end{pmatrix} \end{align*}

We get that
\begin{equation*}\begin{cases}a=x+w \\ b+ic=y-iz \\ b-ic = y+iz \\ d=x-w\end{cases}\end{equation*}

The coordinate column vector of $v$ in respect to the basis $C$ is \begin{equation*}v_C=\begin{pmatrix}\frac{a+d}{2} \\ b \\ -c \\ \frac{a-d}{2} \end{pmatrix}\end{equation*}
Is everything correct so far? (Wondering)

How could we compute $A_{\text{id}, B,C}$ ? (Wondering)

 

[I like Serena]

Is everything correct so far?

Hey mathmari!

It looks correct to me. (Nod)

How could we compute $A_{\text{id}, B,C}$ ?

What is $A_{\operatorname{id}, B,C}$ exactly?
Is it the transformation matrix for a vector with respect to B to a vector with respect to C?
Or the other way around? (Wondering)Anyway, let me write $A_{B\to C}$ for the matrix that transforms a vector with respect to basis B to a vector with respect to basis C.
Then we must have that:
$$v_C = A_{B\to C}\ v_B$$
yes? (Wondering)

We also have:
$$v_E= A_{B\to E}\ v_B = ( B_1 B_2 B_3 B_4)\ v_B$$
$$v_E= A_{C\to E}\ v_C = ( C_1 C_2 C_3 C_4)\ v_C$$
and:
$$A_{C\to E} = A_{E\to C}^{-1}$$

Can we find $A_{B\to C}$ (or $A_{C\to B}$ if that was intended) from that? (Wondering)Btw, you've effectively already (almost) found $A_{B\to H}$ and $A_{C\to H}$.
(What is the basis of $H$ anyway? And which are $A_{B\to H}$ and $A_{C\to H}$ exactly?)
So we can also find $A_{B\to C} = A_{H\to C}\ A_{B\to H}$ from that, can't we?
If it comes out the same, we have confirmation that you're previous calculations were correct. (Happy)

 

[mathmari]

We also have:
$$v_E= A_{B\to E}\ v_B = ( B_1 B_2 B_3 B_4)\ v_B$$
$$v_E= A_{C\to E}\ v_C = ( C_1 C_2 C_3 C_4)\ v_C$$
and:
$$A_{C\to E} = A_{E\to C}^{-1}$$

Can we find $A_{B\to C}$ (or $A_{C\to B}$ if that was intended) from that? (Wondering)

What is $E$ ? An other basis?

We have that $A_{B\to C}=A_{E\to C}\cdot A_{B\to E}$, right? (Wondering)

Btw, you've effectively already (almost) found $A_{B\to H}$ and $A_{C\to H}$.
(What is the basis of $H$ anyway? And which are $A_{B\to H}$ and $A_{C\to H}$ exactly?)
So we can also find $A_{B\to C} = A_{H\to C}\ A_{B\to H}$ from that, can't we?
If it comes out the same, we have confirmation that you're previous calculations were correct. (Happy)

We have that a basis of $H$ is $$\left \{\begin{pmatrix}1& 0\\ 0&0\end{pmatrix}, \begin{pmatrix}0& 1\\ 1&0\end{pmatrix}, \begin{pmatrix}0& i\\ -i&0\end{pmatrix} , \begin{pmatrix}0& 0\\ 0&1\end{pmatrix}\right \}$$ right? (Wondering)

$A_{B\to H}$ is the matrix of coefficients $a_{ij}$ of $T(b_j)=\sum a_{ij}h_i$, where $T=\text{id}$, right? (Wondering)

 

[I like Serena]

Let $v_C=\begin{pmatrix}x \\ y \\ z \\ w \end{pmatrix}$ be the coordinate column vector. Since the elements of the basis $C$ are complex matrices, we have that $x,y,z,w\in \mathbb{C}$.

Isn't C a basis for a $\mathbb R$-vector space?
Shouldn't we have $x,y,z,w\in \mathbb{R}$ then? (Wondering)

What is $E$ ? An other basis?

We have that $A_{B\to C}=A_{E\to C}\cdot A_{B\to E}$, right?

$E$ is the standard basis, which was also given in your problem statement.
And yes, that would hold.
But let's forget about that for now, since your problem statement only said to 'verify'.

We have that a basis of $H$ is $$\left \{\begin{pmatrix}1& 0\\ 0&0\end{pmatrix}, \begin{pmatrix}0& 1\\ 1&0\end{pmatrix}, \begin{pmatrix}0& i\\ -i&0\end{pmatrix} , \begin{pmatrix}0& 0\\ 0&1\end{pmatrix}\right \}$$ right?

Indeed.
Hey! Isn't that the same as the basis $B$? (Wondering)

$A_{B\to H}$ is the matrix of coefficients $a_{ij}$ of $T(b_j)=\sum a_{ij}h_i$, where $T=\text{id}$, right? (Wondering)

What are $a_{ij}$, $b_j$, and $h_i$? (Wondering)

$A_{B\to H}$ is the matrix such that when multiplying it with a vector wrt to B, we find the corresponding vector wrt to H, such that they describe the same actual vector. (Nerd)Anyway, we have:
$$
\begin{pmatrix}a\\b\\c\\d\end{pmatrix}
= v_H = A_{B\to H}\cdot v_B = A_{B\to H}\ \begin{pmatrix}x\\y\\z\\w\end{pmatrix}
$$
and you found that:
$$
\begin{pmatrix}a\\b\\c\\d\end{pmatrix}
= \begin{pmatrix}x\\y\\z\\w\end{pmatrix}
$$
So $A_{B\to H} = \operatorname{id}$ isn't it?

And if we can find $A_{C\to H}$ similarly, we have that $A_{B\to C} = A_{C\to H}^{-1}\cdot A_{B\to H}$ don't we? (Wondering)

 

[mathmari]

Can we do it also as follows?

We write the elements of $B$ as a linear combination of elements of $C$.

We have that
\begin{align*}&\text{id}(E_{11})=E_{11}=\frac{1}{2}\cdot \sigma_0+0\cdot \sigma_1+0\cdot \sigma_2+\frac{1}{2}\cdot \sigma_3\\ &\text{id}(E_{12}+E_{21})=E_{12}+E_{21}=0\cdot \sigma_0+1\cdot \sigma_1+0\cdot \sigma_2+0\cdot \sigma_3 \\ & \text{id}(i[E_{12}-E_{21}])=i[E_{12}-E_{21}]=0\cdot \sigma_0+0\cdot \sigma_1+(-1)\cdot \sigma_2+0\cdot \sigma_3 \\ & \text{id}(E_{22})=E_{22}=\frac{1}{2}\cdot \sigma_0+0\cdot \sigma_1+0\cdot \sigma_2+\left (-\frac{1}{2}\right )\cdot \sigma_3 \end{align*}

So, we get that \begin{equation*}A_{\text{id}, B,C}=\begin{pmatrix}\frac{1}{2} & 0 & 0 & \frac{1}{2} \\ 0& 1& 0 & 0 \\ 0 & 0 & -1 & 0 \\ \frac{1}{2} & 0 & 0 & -\frac{1}{2}\end{pmatrix}\end{equation*}

Therefore we have that
\begin{equation*}A_{\text{id}, B,C}\cdot v_B=\begin{pmatrix}\frac{1}{2} & 0 & 0 & \frac{1}{2} \\ 0& 1& 0 & 0 \\ 0 & 0 & -1 & 0 \\ \frac{1}{2} & 0 & 0 & -\frac{1}{2}\end{pmatrix}\cdot \begin{pmatrix}a \\ b \\ c \\ d \end{pmatrix}=\begin{pmatrix} \frac{a+d}{2} \\ b \\ -c \\ \frac{a-d}{2}\end{pmatrix}=v_C\end{equation*}

 

[I like Serena]

Yep. That works as well. (Nod)

 

[mathmari]

Yep. That works as well. (Nod)

Great! Thank you so much! (Mmm)