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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. –  David Mitra May 28 '12 at 14:32
    
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? –  Soham May 28 '12 at 14:44
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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. –  Jyrki Lahtonen May 28 '12 at 15:15
    
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. –  Soham May 28 '12 at 15:41
    
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. –  Jyrki Lahtonen May 28 '12 at 20:43

1 Answer 1

Please note that it makes no sense in general to "give" a linear transformation between linear spaces that are not the standard spaces (the spaces $F^n$; in this case, $\mathbb{R}^n$) simply by giving a matrix. Vectors in general vector spaces are not tuples.

In order to be able to interpret the matrix as a linear transformation, there has to be a specified basis for both the source and target spaces. In the case of $\mathbb{R}^n$, these bases are usually assumed to be the standard bases, but for other vector spaces there is usually no clear choice. Even for the vector spaces of polynomials, there are several bases that may be in use depending the setting; for instance, the basis $\{1,x,x^2\}$ for $P_2$ may seem "obvious", but then so does $\{x^2,x,1\}$, and this will give you a different transformation; and if you are working with $P_2$ as an inner product space with inner product given by $$\langle p,q\rangle = \int_a^b p(t)q(t)\,dt,$$ then neither of those bases makes good sense.

So when you start your question, you need to specify the bases you are using. Otherwise, giving the matrix does not really tell you what the linear transformation "really" is.

That said, there are some things that you can compute even without knowing what the bases are: the rank of the matrix will be the dimension of the image of the linear transformation (once you specify the bases you will be able to actually describe the image, not only say what its dimension is; it's possible you can say what the image is anyway, e.g., if you know the dimension is $0$ or the dimension is the dimension of the target space), and the nullity of the matrix will be the dimension of the nullspace of the linear transformation (but without knowing what the basis of the domain is, you may not be able to actually describe that nullspace, only say what its dimension is).

The nullspace is never empty: the nullspace is the collection of all vectors that map to $0$. Since the zero vector always maps to zero, the zero vector is always in the nullspace. In the case of your first map, since the nullity is zero, the dimension of the nullspace is $0$. The only subspace of dimension $0$ is the zero subspace, so the nullspace of your first linear transformation is $\{0\}$, regardless of what the basis is.

In your second example, the nullity of the matrix is spanned by the vector $(-10,3,4)^t$. That means that, if your basis consists of the polynomials $p_1(x)$, $p_2(x)$, and $p_3(x)$, then the nullspace will consist exactly of all vectors of the form $$\alpha\Bigl( -10p_1(x) + 3p_2(x) + 4p_3(x)\Bigr),\qquad \alpha\in\mathbb{R};$$ for example, if your ordered basis is $[1,x,x^2]$, then they are all scalar multiples of $-10 + 3x + 4x^2$. If your ordered basis is $[x^2,x,1]$, then they are all scalar multiples of $-10x^2 + 3x + 4$. If your ordered basis is $[1,1+x,1+x+x^2]$, then the nullspace consists of all multiples of $$-10(1) + 3(1+x) + 4(1+x+x^2) = -3 + 7x + 4x^2.$$ (Can you see why specifying the basis is important?)

Note that this is not the same as saying $-10x^2+3x+4=0$; that equation makes no sense in the context of these vector spaces and linear transformations.

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