I want to calculate the Kernel of a function (similar to the one shown below but not exactly) that goes through a linear transformation (I'm not sure if $L_1$ does go through a linear transformation but for the sake of understandign let's assume that it does), how would you calculate the Kernel?
If $L_1:\mathbb{R_{≤3}[x]} \to \mathbb{R^{2,2}} \ ; \ ax^3+bx^2+cx+d \to \left[ \begin{array}{cc} -2c & a-2c \\ c+a & b+3a \\ \end{array}\right]$ then how can I calculate the kernel of the function? The formula for calculating the kernel is $Ker(L_1)=[v∈L_1 \ ; \ L_1(\vec{v})=\vec{0}]$ and I think you would rewrite the formula as :
$$Ker(L_1)=[ax^3+bx^2+cx+d∈L_1 \ ; \ L_1(ax^3+bx+cx+d)=\vec{0}]$$
Since $L_1(ax^3+bx^2+cx+d)=\left[ \begin{array}{cc} -2c & a-2c \\ c+a & b+3a \\ \end{array}\right]$, I would try to solve for the zero vector, making it
$$\left[ \begin{array}{cc} -2c & a-2c \\ c+a & b+3a \\ \end{array}\right]=\left[ \begin{array}{c} 0\\0 \end{array}\right]$$
But how can I go on from here?
P.S. - I'M not sure if this example does go through a linear transformation, as I thought of it off the top of my head in an attempt to understand the concept of Kernel through examples, so I apologize in advance if it turns out that it doesn't go through a linear transformation.
I have no idea why you didn't get answers before. This looks like a perfectly reasonable question to me.
Correct. But the zero vector inside $\mathbb{R}^{2,2}$ is $\begin{pmatrix} 0&0\\0&0\end{pmatrix}$. Now both are $2\times 2$ matrices, so just set their corresponding entries equal.
Don't be confused by the fact that this is called a "vector" - it's just the zero element of the vector space.