# Finding the range space of a matrix

1. Jan 3, 2012

### Samuelb88

1. The problem statement, all variables and given/known data
Represent the linear span of the four vectors x_1 = (-1,1,1,2), x_2 = (2,1,7,1), x_3 = (3,-2,0,5), and x_4 = (1,0,2,1) as the range space of some matrix.

3. The attempt at a solution
It's been along while since I've done any linear algebra and so I'm not sure what to do... According to wikipedia, if a matrix $A = [ x_1, ..., x_n$, where $x_i \in \mathbb{R}^n$, then the range space of A is $\{ y \in \mathbb{R}^n : y = c_1 x_1 + \cdots + c_n x_n, \, \, c_i \in \mathbb{R} \}$. So I want to say the answer is
$$\{ y \in \mathbb{R}^4 : y = c_1 x_1 + c_2 x_2 + c_3 x_3 + c_4 x_4, \, \, c_i \in \mathbb{R} \}$$

My question is, how should I express the answer? Is it standard to row reduce the matrix first to see if the vectors are linearly dependent or anything?

2. Jan 3, 2012

### Deveno

one thing to consider: you (should) know that {(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)} forms a basis for R4. suppose A is a 4x4 matrix. what would the image of A of these 4 basis elements (considered as column vectors) look like? why is the image of A often called "the column space" of A?

3. Jan 3, 2012

### Samuelb88

Let $e_1 = (1,0,0,0), ..., e_4 = (0,0,0,1)$. Then if A is a 4x4 matrix, then the vector $A e_1$ would be the first column of A, ..., $A e_4$ would be the fourth column of A. I'm not sure I'm understanding your post. What do you mean when you say the image of A?

Edit: I still don't understand what the question is asking me. I'm thinking I'm suppose to create a matrix A using the linear span of the four vectors. Could you/someone clarify for me what the question is asking?

4. Jan 3, 2012

### maCrobo

I think the exercise is asking you to find the range of the matrix and represent it with three appropriate vectors.
So first you should build up the matrix made of those four vectors, which is in R4. Then you know that a matrix has a range and a null space, the former being the space where AX=B, the latter being the space where AX=0. Here by A I mean the matrix you have built, by X a generic column vector and by B a defined column vector. So AX=B means that for every chosen B you have one and only one solution X (we are in linear spaces) while AX=0 takes all the X vectors in R4 that after the application of the matrix A becomes zero, so they go in a so called null space. Another way of seeing it is by imagining the matrix A as a transformation, an operator practically, that takes all the vectors in R4, which we call X, and send them to another space S that could be of dimension 4 or less; in the case the dimension is less than 4 you must have a null space.

Anyway, after you have built up your matrix you have to verify its range and, actually, the dimension of the range, I call n, and take n vectors that make a basis for the range. So finally you have a set of vectors that describe the linear span of the range of the matrix you started from.
I'm not very sure I made myself clear... If so ask :D

5. Jan 3, 2012

### Samuelb88

So I believe I understand what you're saying. Construct the matrix $A = [x_1 \, \, x_2 \, \, x_3 \, \, x_4 ]$. How do I verify it's range? Should I look at the transpose of the matrix A and row reduce, then the nonzero rows will be my linear span of the range of the matrix that I started from?

6. Jan 3, 2012

7. Jan 3, 2012

### maCrobo

Yes, check its range by manipulating the matrix until you get dependent rows or columns. Since it is a square matrix it will be enough to copy the, as you said, nonzero rows and you will have a basis.

P.S.:I got rank(A)=3.

8. Jan 3, 2012

### Samuelb88

Alright, so here's what I have. Let $A = \begin{bmatrix} 1 & -1&2& 3 \\ 0&1&1&-2\\2&1&7&0\\1&2&1&5 \end{bmatrix}$. Looking at $A^T$, and row reducing, I get that the independent columns of A (i.e. the independent rows of A^T) are (1,0,2,0), (0,1,0,0), and (0,0,0,1). Hence the range space of the matrix A is the span of those three vectors. Is this correct?

9. Jan 4, 2012

### maCrobo

I haven't found the basis, but to check it you can try to find the basis of the null space, create a determinant whose rows are the 3 basis vectors of the range + the basis vector of the null space and if you'll get a result different from zero, there's a very high probability that your answers are correct.
Another way to check, in this case not much longer, may be finding the the basis vector of the null space and a perpendicular vector to it and building a 2 rows matrix (where the rows are your vectors) M, then you should solve the system MX=0. This way you should find 2 other vectors perpendicular to that of the null space and since n orthogonal vectors in a n dimension space are independent, you will have your bases.