LINEAR ALGEBRA: Find vectors that span the image of A

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Homework Help Overview

The discussion revolves around finding vectors that span the image of given matrices A_1 and A_2 in the context of linear algebra. Participants are exploring the concepts of image, row space, and basis vectors.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to express the image of the matrices using linear combinations of their columns and seeks clarification on how to identify the spanning vectors for A_2. Some participants suggest reducing the rows to find a basis for the row space and question the relationship between the number of rows and the number of spanning vectors.

Discussion Status

Participants are actively engaging with the problem, offering different perspectives on how to approach finding the spanning vectors. There is a mix of interpretations regarding the dimensions of the image and the relationship between rows and vectors. Some guidance has been provided regarding the use of leading ones in row reduction.

Contextual Notes

There is a mention of the dimensionality of the image being limited by the number of columns, and the discussion includes references to specific vectors and their roles in spanning the image.

VinnyCee
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For each matrix A, find vectors that span the image of A. Give as few vectors as possible.

[tex]\mathbf{A_1} =<br /> \left[ \begin{array}{cc}<br /> 1 & 1 \\<br /> 1 & 2 \\<br /> 1 & 3 \\<br /> 1 & 4<br /> \end{array} \right][/tex]


[tex]\mathbf{A_2} =<br /> \left[ \begin{array}{cccc}<br /> 1 & 1 & 1 & 1 \\<br /> 1 & 2 & 3 & 4 \\<br /> \end{array} \right][/tex]


My work:

[tex]T(\overrightarrow{x})\,=\,A_1\,\overrightarrow{x}[/tex]

[tex]A_1\,\left[ \begin{array}{c}<br /> x_1 \\<br /> x_2 \\<br /> x_3 \\<br /> x_4 <br /> \end{array} \right]\,=\,\left[ \begin{array}{cc}<br /> 1 & 1 \\<br /> 1 & 2 \\<br /> 1 & 3 \\<br /> 1 & 4<br /> \end{array} \right]\,\left[ \begin{array}{c}<br /> x_1 \\<br /> x_2<br /> \end{array} \right][/tex]

[tex]x_1\,\left[ \begin{array}{c}<br /> 1 \\<br /> 1 \\<br /> 1 \\<br /> 1 <br /> \end{array} \right]\,+\,x_2\,\left[ \begin{array}{c}<br /> 1 \\<br /> 2 \\<br /> 3 \\<br /> 4 <br /> \end{array} \right][/tex]

The image of [tex]A_1[/tex] is the space spanned by

[tex]v_1\,=\,\left[ \begin{array}{c}<br /> 1 \\<br /> 1 \\<br /> 1 \\<br /> 1 <br /> \end{array} \right]\,\,and\,\,v_2\,=\,\left[ \begin{array}{c}<br /> 1 \\<br /> 2 \\<br /> 3 \\<br /> 4 <br /> \end{array} \right][/tex]

For the second part:

[tex]A_2\,\overrightarrow{x}[/tex]

[tex]\left[ \begin{array}{cccc}<br /> 1 & 1 & 1 & 1 \\<br /> 1 & 2 & 3 & 4 \\<br /> \end{array} \right]\,\left[ \begin{array}{c}<br /> x_1 \\<br /> x_2 \\<br /> x_3 \\<br /> x_4 <br /> \end{array} \right][/tex]

[tex]x_1\,\left[\begin{array}{c}<br /> 1 \\<br /> 1 \\<br /> \end{array}\right]\,+\,x_2\,\left[\begin{array}{c}<br /> 1 \\<br /> 2 \\<br /> \end{array}\right]\,+\,x_3\,\left[\begin{array}{c}<br /> 1 \\<br /> 3 \\<br /> \end{array}\right]\,+\,x_4\,\left[\begin{array}{c}<br /> 1 \\<br /> 4 \\<br /> \end{array}\right][/tex]

But now I don't know where to proceed to solve the second part. The B.O.B. says that the answer is

[tex]\left[\begin{array}{c}<br /> 1 \\<br /> 1 \\<br /> \end{array}\right],\,\left[\begin{array}{c}<br /> 1 \\<br /> 2 \\<br /> \end{array}\right][/tex]

How do I show such?
 
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You are trying to find a basis for the "row space". Treat each row as a vector and reduce. Obviously you have 4 rows, so 4 vectors, but the image is subset of R2 and so can have dimension no larger than 2.

Actually, if the dimension is 2, then (1,0), (0, 1) will work. If 1, anyone of the rows will work.
 
[tex]A_1[/tex] has four rows, but only two vectors. I am assuming that the number of columns should equal the number of vectors?

Where are you getting the (1, 0) and (0, 1 ) from?

By "dimension", do you mean the number of columns in a matrix, or the number of rows?
 
Leading ones.

Solve for leading ones. you will get something like: [(1,0,-1,-2);(0,1,2,3)]. You can see that the columns with leading ones refer you back to the original matrix, which gives you the answer that the BOB had. Hope that helps.
 

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