This is inspired from this post as I was mentally playing with the concepts. The statement is the same just the transformation different, though for the benefit of everybody, I am repeating it, with a slight change:
What is the null space of this transformation: $A:P_{2} \to P_{2} $ where $P_{2}$ is the space of all polynomials of degree ⩽2 over the real numbers. Here $A= \begin{bmatrix} 1 & 2 & 1\\ 2 & 0 & 5\\ 4 & 1 & 3 \end{bmatrix}$
- A Simple check with echelon form, shows that rank=3 for this matrix, which implies that $AX=0$ doesnt have a non trivial solution, and thus has a unique solution. So will it be fair to say, that the null space is empty?
- If instead $A= \begin{bmatrix} 1 & 2 & 1\\ 2 & 0 & 5\\ 4 & 0 & 10 \end{bmatrix}$ where I have intentionally made $R_{3}$ a multiple of $R_{2}$, upon solving for the same which leads to the "scalars" of {$x^{2},x,1$} as [-10,3,4]. So by definition it effectively means this polynomial $-10x^{2}+3x+4=0$ has a unique solution (which is true!); so what has this transformation effectively achieved?