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I am working on this practice problem, and I was wondering if I could get some help.

I have a $T$:$\mathbb{R^{2x2}}\to \mathbf{P_{2}}$, that is, from 2x2 matrices to polynomials of degree at most 2. The transformation is given as following: $$T\left(\begin{bmatrix} a & b\\ c & d \end{bmatrix}\right) = a-c+2d+(b+2c-d)t+(a-c+3d)t^{2}.$$

To get the basis of kernel of $T$, I solved a system of equations needed to get the 'O' element in the $\mathbf{P_{2}}$ -- $a-c+2d=0$, $b+2c-d=0$ and $a-c+3d=0$. As a result, I got the basis of the kernel equal to $$\begin{bmatrix} 1 & -2\\ 1 & 0\\ \end{bmatrix}.$$

When it comes to image, if I understand correctly, I need to factor out all the variables separately, to see what is it that they span. So I got $a(1+t^{2})+b(t)+c(-t^{2}+2t-1)+d(3t^{2}-t+2)$. So would I be correct in saying that these three polynomials (without the coefficients $a$, $b$, $c$, and $d$) form the basis for the image $T$? Thank you!

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To get the $\to$ for the functions, use \to. –  Arturo Magidin Feb 21 '11 at 3:03
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2 Answers

up vote 4 down vote accepted

Your answers are correct, but for the image part a) you need some additional reasoning and b) you can considerably simplify the result. a) These polynomials form a generating set for the image of $T$; to show that they form a basis, you have to check whether they're linearly independent (they are). b) You've correctly described the image space, but in quite a complicated way. What's the dimension of the space spanned by your basis of three polynomials? And what's the dimension of the entire codomain $\mathbf{P_2}$? What does that tell you about which space these elements actually span?

By the way, the basis element of the kernel should be written as a $2\times2$ matrix, not as a column vector, and the basis is the set containing that element, not the element itself.

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thanks, what do you mean by some additional reasoning? I did not write out all of my steps for part a, if that's what you mean. The dimension of my image is 3, and so is the dimension of the whole target-space, so I could say that $1, t, t^{2}$ is a basis, I assume. Thanks! –  LinAlgStudent Feb 21 '11 at 3:08
@LinAlgStudent: OK, if you didn't write it all out it's fine -- I thought you were saying that these three are a basis simply because they span the image. Yes, equal dimensions of $3$ and $1$, $t$, $t^2$ as a basis is what I meant. Seems like you've got this all figured out :-) –  joriki Feb 21 '11 at 3:11
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The simplest way to compute the image, abstractly, is to take a basis of the domain, and see what its image spans. So here, you could take the standard basis of $\mathbb{R}^{2\times 2}$, to wit, $$\left(\begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right),\quad \left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right),\quad \left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right),\quad \left(\begin{array}{cc} 0 & 0\\ 0 & 1 \end{array}\right).$$ Here, you have that the images of these basis vectors are, respectively, $$1+t^2,\quad t,\quad, -1+2t-t^2,\quad 2-t+3t^2.$$ So you would want to find out the span of these four polynomials in $\mathbb{P}^2$ (note that, of course, these four polynomials must be linearly dependent; after all, $\dim(\mathbf{P}_2) = 3$).

Alternatively, you can think about the matrix representation of $T$. Convince yourself that the image of $T$ corresponds to the columnspace of the matrix that represents $T$. One way to find the columnspace is to find a row-echelon form of the matrix, and then take the columns in the original matrix that correspond to the columns that contain the pivots of the row-echelon form. Since you may have already computed the row-echelon form of the matrix in order to find a basis for the null-space, it's likely you have already done all the needed work and can just exploit it.

Of course, as you noted in comments, if you already know that the dimension of the image is $3$ (by the Dimension Theorem, aka the Rank-Nullity Theorem, say), and you know that $\mathbf{P}_2$ has dimension $3$, then you know the image is all of $\mathbf{P}_2$ will do. The descriptions above are the general procedures you might use with any linear transformation.

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