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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?
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    The null space is never empty. It might be trivial in the sense that it only contains the zero vector, though.2012-05-28
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    Oh okay, I guess, thats what it is- in the first case, the solution is unique as well as trivial, with [$a_{1},a_{2},a_{3}$] =[0,0,0]. That explains quite a bit, thanks. But David, what has the second transformation achieved,effectively?2012-05-28
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    Assuming that the matrix is w.r.t. the basis $\{x^2,x,1\}$, then the second transformation is simply the mapping $$ A:(ax^2+bx+c)\mapsto (a+2b+c)x^2+(2a+5c)x+(4a+10c). $$ The nullspace consists of those polynomials $p(x)$ that are mapped to the zero polynomial. There are infinitely many of them, all scalar multiples of $-10x^2+3x+4$ the one you found.2012-05-28
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    Aah, I see. Thanks, you answered my question. I was thinking, though there should be infinite of such polynomials, then why is it showing only 1.2012-05-28
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    Who says it's showing only one? $10x^2-3x-4$ will also do. And $70x^2-21x-28$. And any other scalar multiple. Also, read Arturo's answer carefully. You need to specify a basis, when you describe a linear transformation be giving its matrix. That's what my *Assuming that...* was also all about.2012-05-28

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