# Find a basis for the range of linear operator and linear transformation.

A) Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be the linear operator given by the formula $T(x,y) = (2x-y, -8x+4y)$.

Find a basis for the range of the linear operator.

B) Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be the linear transformation given by the formula:

$T(x_1, x_2, x_3, x_4) = (4x_1+x_2-2x_3-3x_4, 2x_1+x_2+x_3-4x_4, 6x_1-9x_3+9x_4)$ Find a basis for the range of the linear transformation.

C) Let $T:P_2 \to P_3$ be the linear transformation defined by $T(p(x)) = xp(x)$.

Find the basis for the range of the linear transformation.

*Update 7/16/13: Working on part b:

I believe (correct me if I'm wrong) that the basis of the range of a linear transformation is just the column space of the linear transformation. If so, I should set the transformation up in a matrix and reduce to row echelon. Then, I think I'll use the columns in the reduced matrix that have pivots and correspond those columns to the original matrix - thus giving me my basis. However, I set up the following matrix.

B=\begin{bmatrix} 4 & 1 & -2 & -3 \\ 2 & 1 & 1 & -4 \\ 6 & 0 & -9 & 9 \end{bmatrix}

When I reduced this, I got

B=\begin{bmatrix} 1 & 0 & -(3/2) & 0 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}

From this, I would think that the column one, two and four contain my pivots. Therefore, (going back to the original matrix) my basis would be the column vectors <4,2,6>, <1,1,0> and <-3,-4,9>. I'm afraid I've gone wrong somewhere. Can someone help me with this?

• Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be closed, see meta. – Cameron Buie Jul 16 '13 at 6:09
• @icurays1: On part a, I tried to solve using a matrix, but my results were inconsistent. When I reduced, my second row was 0x+0y=5... – briteId Jul 16 '13 at 6:21
• For A) you will easily see that $T$ takes the canonical basis to two linearly dependent vectors. Throw one away. You are done. For C), the image of the canonical basis is linearly independent: you are done already. For B), what do yo know about rank and row/column operations? – Julien Jul 16 '13 at 6:39
• Ok, the canonical basis of $\mathbb{R}^2$ is $e_1=(1,0)$ (ie $x=1$ and $y=0$) and $e_2=(0,1)$. Then $T(e_1)=(2,-8)$ (using the formula defining $T$ with $x=1$ and $y=0$) and $T(e_2)=(-1,4)$ span the range of $T$. But $T(e_1)=-2T(e_2)$ (linear dependence) so we can throw, for instance, $T(e_1)$ away and keep $(-1,4)$. That's you basis. – Julien Jul 16 '13 at 6:42
• When things get more complicated (bigger dimensions), we have a standard procedure/algorithm for linear maps such as $T$ in B). Here is how to find a basis of the column space of a matrix by "elementary row operations". The range of $T$ is the column space of its representing matrix where the columns are the images of the vectors of the canonical basis. – Julien Jul 16 '13 at 6:47

## 2 Answers

In cases A and B, you can find the matrix of the linear transformation with respect to the canonical bases; in case A it is $$A=\begin{bmatrix} 2 & -1 \\ -8 & 4 \end{bmatrix}$$ and in case B it is $$B=\begin{bmatrix} 4 & 1 & -2 & -3 \\ 2 & 1 & 1 & -4 \\ 6 & 0 & -9 & 9 \end{bmatrix}$$ In general, when you have a linear transformation $T\colon\mathbb{R}^n\to\mathbb{R}^m$ and $\{e_1,e_2,\dots,e_n\}$ is the canonical basis of $\mathbb{R}^n$, you just write down (as columns), the vectors $T(e_1), T(e_2), \dots, T(e_n)$. A basis for the range can easily be computed by Gaussian elimination.

For case C, you don't have a "canonical basis", but you still can compute the matrix associated to the bases $\{1,x,x^2\}$ of $P_2$ (assuming it's the space of polynomials having degree at most 2) and $\{1,x,x^2,x^3\}$ of $P_3$. Since $T(1)=x=0\cdot1+1x+0x^2+0x^3$, $T(x)=x^2$, $T(x^2)=x^3$, the matrix is $$C=\begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ The rank of this matrix is? And what can you conclude from this?

• Let's work on the explanation of a and b before I tackle C. I see where you got the matrices A and B. However, I seem to be having trouble with finding the basis for the range of A and B. Can you help with this first? I know how to do Guassian elimination. The problem comes in when I need to decide whether to reduce the matrix B or the transpose of B. – briteId Jul 16 '13 at 20:56
• On part B, when I reduced, I got B=\begin{bmatrix} 1 & 0 & -(3/2) & 0 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} So, my thought was that since I have pivots in the first, second and fourth column - my basis would the the first, second and fourth column of the original matrix. That would make my basis equal to: Basis=\begin{bmatrix} 4\\ 2\\ 6 \end{bmatrix} \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix} \begin{bmatrix} -3\\ -4\\ 9 \end{bmatrix} – briteId Jul 16 '13 at 21:02
• @Cee I didn't do the calculation; if it's correct, then the answer is right. – egreg Jul 16 '13 at 21:12

Hints for you to sort and apply:

== If we have a linear map $\,T:V\to W\;,\;\;V,W\;$ finite dimensional vector spaces over the same field , then we have the dimension theorem:

$$\dim V=\dim\ker T+\dim\text{Im}\,T$$

== We clearly have $\;(-4)\cdot(2x,-y)=(-8x,4y)\implies\;$ the range of $\,T\,$ is $\,1-$dimensional

== In (B), what is the dimension of the range? Try reducing a coefficients matrix.

== In (C), the minimal degree of any non-zero element in the image is one...

• Thanks for your comment, but I'm not sure how to take the dimension of the range of T and turn that into the basis of the range of T. Please advise. – briteId Jul 16 '13 at 19:47
• In all the cases take the transpose matrix representing the operator, and reduce it. Delete any original vectors corresponding to rows of zeros after the reduction of the matrix. What is left gives you a basis for the range. For example, if you reduce $$\begin{pmatrix}4&2&6\\1&1&0\\-2&1&-9\\-3&-4&9\end{pmatrix}$$ say, first interchanging rows $\,1\leftrightarrow 2\,$ to make things easier, you get that the third row becomes all zeros and the other ones don't, so the vectors corresponding to the non-all-zero rows are a basis for the range... – DonAntonio Jul 16 '13 at 19:59
• You have to go back to the original vectors...if you want to. If you reduce a matrix by elementary operations you do not change the span of its rows. And yes: basis of range = basis of column space, and you were looking at the columns of the original matrix: I just took the transpose to reduce these columns (now rows!) since reducing by rows is way more usual than reducing by columns. If you feel comfortable with this last one do it with the original matrix: it's the same. – DonAntonio Jul 16 '13 at 20:26
• Well, if you reduce by rows in the original $\,3\times 4\,$ matrix then not sure what columns to choose to get your column-space basis. In this case, for example, the three rows are lin. independent since by rows and by columns you get the same rank. With the transpose you get a $\,4\times 3$ matrix, and I'm not sure why do you think you would have to move the all-zeros row to the bottom: the row already tells you which one of the four $\;3-$dimensional vectors is lin. dependent on the other ones, so just choose the other ones. – DonAntonio Jul 16 '13 at 20:46
• @Cee, then at least one of we two is wrong: if you get zeros in the fourth row and I in the third one then something seems to be wrong.... but in this particular case it doesn't matter since the operator's range is the whole $\,\Bbb R^3\;$ ! – DonAntonio Jul 16 '13 at 21:03