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LINEAR ALGEBRA: Find vectors that span the image of A

  • Thread starter VinnyCee
  • Start date
  • #1
489
0
For each matrix A, find vectors that span the image of A. Give as few vectors as possible.

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


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


My work:

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

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

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

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

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

For the second part:

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

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

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

How do I show such?
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
955
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, any one of the rows will work.
 
  • #3
489
0
[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?
 
  • #4
4
0
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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