MHB How Do We Find the Kernel and Image Bases Using Matrix C?

[mathmari]

Hey!

Let $A\in \mathbb{C}^{2\times 2}$ and $L_A:\mathbb{C}^{2\times 2}\rightarrow \mathbb{C}^{2\times 2}, \ X\mapsto A\cdot X$.

We consider the matrix \begin{equation*}A=\begin{pmatrix}-1 & 2 \\ 2 & -4\end{pmatrix}\end{equation*} and the basis \begin{equation*}B=\left \{\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} , \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}, \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}\right \}\end{equation*}

I have calculated the transformation matrix of $L_A$ in respect to the basis, which is the following:
\begin{equation*}C=\begin{pmatrix}-1 & 0 &2 &0 \\ 0 &-1 & 0 & 2 \\ 2 & 0 & -4 & 0 \\ 0 & 2 & 0 & -4\end{pmatrix}\end{equation*}

I want to determine a basis of the kernel of $L_A$ and a basis of the image of $L_A$.

Can we determine these using the transformation matrix $C$ ? Or how can we calculated the bases? (Wondering)

 

[I like Serena]

Hey!

Let $A\in \mathbb{C}^{2\times 2}$ and $L_A:\mathbb{C}^{2\times 2}\rightarrow \mathbb{C}^{2\times 2}, \ X\mapsto A\cdot X$.

We consider the matrix \begin{equation*}A=\begin{pmatrix}-1 & 2 \\ 2 & -4\end{pmatrix}\end{equation*} and the basis \begin{equation*}B=\left \{\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} , \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}, \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}\right \}\end{equation*}

I have calculated the transformation matrix of $L_A$ in respect to the basis, which is the following:
\begin{equation*}C=\begin{pmatrix}-1 & 0 &2 &0 \\ 0 &-1 & 0 & 2 \\ 2 & 0 & -4 & 0 \\ 0 & 2 & 0 & -4\end{pmatrix}\end{equation*}

I want to determine a basis of the kernel of $L_A$ and a basis of the image of $L_A$.

Can we determine these using the transformation matrix $C$ ? Or how can we calculated the bases? (Wondering)

Hey mathmari!

Yes.

Let $X=\begin{pmatrix} w&x\\y&z\end{pmatrix}$. Then the representation of $X$ with respect to the basis $B$ is $\mathbf x = \begin{pmatrix}w\\x\\y\\z\end{pmatrix}$, isn't it?

So we can find the kernel by either solving $AX=0$ or by solving $C\mathbf x = \mathbf 0$. (Thinking)

To find the image, consider that each column in $C$ represents a matrix in the image.
Can we find a basis from the column vectors in $C$? (Wondering)

 

[mathmari]

Let $X=\begin{pmatrix} w&x\\y&z\end{pmatrix}$. Then the representation of $X$ with respect to the basis $B$ is $\mathbf x = \begin{pmatrix}w\\x\\y\\z\end{pmatrix}$, isn't it?

Why is that the representation? I got stuck right now. (Wondering)

So we can find the kernel by either solving $AX=0$ or by solving $C\mathbf x = \mathbf 0$. (Thinking)

Ah ok! So the kernel is $\ker (L_A)=\{X\in \mathbb{C}^{2\times 2} \mid L_A(X)=O \}=\{X\in \mathbb{C}^{2\times 2} \mid A\cdot X=O \}$.

So \begin{align*}A\cdot X=O\Rightarrow \begin{pmatrix}-1 & 2 \\ 2 & -4\end{pmatrix}\begin{pmatrix} w & x\\ y&z\end{pmatrix}=\begin{pmatrix} 0&0 \\ 0&0\end{pmatrix}\Rightarrow \begin{pmatrix}-w+2y & -x+2z \\ 2w-4y & 2x-4z\end{pmatrix}=\begin{pmatrix} 0&0 \\ 0&0\end{pmatrix}\Rightarrow \begin{cases} -w+2y=0 \\ -x+2z=0 \\ 2w-4y=0 \\ 2x-4z=0\end{cases}\Rightarrow \begin{cases} w=2y \\ x=2z\end{cases}\end{align*} So we get the kernel \begin{equation*}\left \{\begin{pmatrix} 2y & 2z\\ y&z\end{pmatrix}\middle |y,z\in \mathbb{C}\right \}=\left \{y\begin{pmatrix} 2 & 0\\ 1&0\end{pmatrix}+z\begin{pmatrix} 0 & 2\\ 0&1\end{pmatrix}\middle |y,z\in \mathbb{C}\right \}\end{equation*}
The basis is then \begin{equation*}\left \{\begin{pmatrix} 2 & 0\\ 1&0\end{pmatrix}, \begin{pmatrix} 0 & 2\\ 0&1\end{pmatrix}\right \}=\left \{2\begin{pmatrix} 1 & 0\\ 0&0\end{pmatrix}+\begin{pmatrix} 0 & 0\\ 1&0\end{pmatrix}, 2\begin{pmatrix} 0 & 1\\ 0&0\end{pmatrix}+\begin{pmatrix} 0 & 0\\ 0&1\end{pmatrix}\right \}=\left \{2b_1+b_3, 2b_2+b_4\right \}\end{equation*}
where $b_1=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}, \ b_2=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} , \ b_3=\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}, \ b_4=\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}$.

Is everything correct? (Wondering)

To find the image, consider that each column in $C$ represents a matrix in the image.
Can we find a basis from the column vectors in $C$? (Wondering)

The first two vectors of $C$ form the basis, or not? (Wondering)

 

[I like Serena]

Let $X=\begin{pmatrix} w&x\\y&z\end{pmatrix}$. Then the representation of $X$ with respect to the basis $B$ is $\mathbf x = \begin{pmatrix}w\\x\\y\\z\end{pmatrix}$, isn't it?

Why is that the representation? I got stuck right now.

The meaning of the representation $\mathbf x = \begin{pmatrix}w\\x\\y\\z\end{pmatrix}$ with respect to a basis $\{\mathbf b_1, \mathbf b_2, \mathbf b_3, \mathbf b_4\}$ is that $\mathbf x = w\mathbf b_1 + x\mathbf b_2 + y\mathbf b_3 + z\mathbf b_4$.
It's just that if we don't mention a basis, it is assumed that we mean the standard basis $\{\mathbf e_1, \mathbf e_2, \mathbf e_3, \mathbf e_4\}$.

Ah ok! So the kernel is $\ker (L_A)=\{X\in \mathbb{C}^{2\times 2} \mid L_A(X)=O \}=\{X\in \mathbb{C}^{2\times 2} \mid A\cdot X=O \}$.

So \begin{align*}A\cdot X=O\Rightarrow \begin{pmatrix}-1 & 2 \\ 2 & -4\end{pmatrix}\begin{pmatrix} w & x\\ y&z\end{pmatrix}=\begin{pmatrix} 0&0 \\ 0&0\end{pmatrix}\Rightarrow \begin{pmatrix}-w+2y & -x+2z \\ 2w-4y & 2x-4z\end{pmatrix}=\begin{pmatrix} 0&0 \\ 0&0\end{pmatrix}\Rightarrow \begin{cases} -w+2y=0 \\ -x+2z=0 \\ 2w-4y=0 \\ 2x-4z=0\end{cases}\Rightarrow \begin{cases} w=2y \\ x=2z\end{cases}\end{align*} So we get the kernel \begin{equation*}\left \{\begin{pmatrix} 2y & 2z\\ y&z\end{pmatrix}\middle |y,z\in \mathbb{C}\right \}=\left \{y\begin{pmatrix} 2 & 0\\ 1&0\end{pmatrix}+z\begin{pmatrix} 0 & 2\\ 0&1\end{pmatrix}\middle |y,z\in \mathbb{C}\right \}\end{equation*}
The basis is then \begin{equation*}\left \{\begin{pmatrix} 2 & 0\\ 1&0\end{pmatrix}, \begin{pmatrix} 0 & 2\\ 0&1\end{pmatrix}\right \}=\left \{2\begin{pmatrix} 1 & 0\\ 0&0\end{pmatrix}+\begin{pmatrix} 0 & 0\\ 1&0\end{pmatrix}, 2\begin{pmatrix} 0 & 1\\ 0&0\end{pmatrix}+\begin{pmatrix} 0 & 0\\ 0&1\end{pmatrix}\right \}=\left \{2b_1+b_3, 2b_2+b_4\right \}\end{equation*}
where $b_1=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}, \ b_2=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} , \ b_3=\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}, \ b_4=\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}$.

Is everything correct?

Yep. All correct.
And we'll find the same result if we solve $C\mathbf x = \mathbf 0$.

The first two vectors of $C$ form the basis, or not?

Yep. Can we write those 2 vectors in $C$ as a basis? (Wondering)

 

[mathmari]

The meaning of the representation $\mathbf x = \begin{pmatrix}w\\x\\y\\z\end{pmatrix}$ with respect to a basis $\{\mathbf b_1, \mathbf b_2, \mathbf b_3, \mathbf b_4\}$ is that $\mathbf x = w\mathbf b_1 + x\mathbf b_2 + y\mathbf b_3 + z\mathbf b_4$.
It's just that if we don't mention a basis, it is assumed that we mean the standard basis $\{\mathbf e_1, \mathbf e_2, \mathbf e_3, \mathbf e_4\}$.

Ah ok!

Yep. Can we write those 2 vectors in $C$ as a basis? (Wondering)

We have that the basis is \begin{equation*}\left \{\begin{pmatrix}-1 \\ 0 \\ 2 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ -1 \\ 0 \\ 2\end{pmatrix}\right \}=\left \{-b_1+2b_3, -b_2+2b_4\right \}\end{equation*} right? (Wondering)

 

[I like Serena]

Ah ok!

We have that the basis is \begin{equation*}\left \{\begin{pmatrix}-1 \\ 0 \\ 2 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ -1 \\ 0 \\ 2\end{pmatrix}\right \}=\left \{-b_1+2b_3, -b_2+2b_4\right \}\end{equation*} right? (Wondering)

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Great! Thank you very much! (Mmm)