Linear Algebra - Basis and Kernel

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Consider a 5 x 4 matrix....

We are told that the vector,

1
2
3
4



is in the kernel of A. Write
v4

as a linear combination of

v1,v2,v3


I'm a bit confused. Since this is a kernel of A, the kernel is a subset of R^m, therefore the other columns are linear combinations and therefore redundant. (since this is the only column represented) So, that means I can have the columns be anything I want, so why can't they just all be the same? Is this too easy or am I missing something?
 
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  • #2
Office_Shredder
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What are the vi here?
 
  • #3
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The v[tex]_{4}[/tex] seems like

1
2
3
4

since the problem is asking for v[tex]_{4}[/tex] to be written as a linear combination of [tex]\overline{v_{1}}[/tex],[tex]\overline{v_{2}}[/tex],[tex]\overline{v_{3}}[/tex]
 
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  • #4
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The v[tex]_{4}[/tex] seems like

1
2
3
4
What do you mean that v4 "seems like" the above? It either is or isn't a vector with coordinates 1, 2, 3, and 4.
since the problem is asking for v[tex]_{4}[/tex] to be written as a linear combination of [tex]\overline{v_{1}}[/tex],[tex]\overline{v_{2}}[/tex],[tex]\overline{v_{3}}[/tex]

I'm guessing that v1, v2, v3, and v4 are the columns of your matrix.

What does it mean that a vector x is in the nullspace of a matrix?
If you carry out the multiplication Ax, what do you get?
If you carry out the same multiplication, but using the vectors v1, v2, v3, and v4, what do you get?
 
  • #5
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Since this is a column of only 1 vector representing a kernel, would this represent a plane? I have the equation as
[tex]\overline{v_{1}}[/tex] + 2[tex]\overline{v_{2}}[/tex] + 3[tex]\overline{v_{3}}[/tex] + 4[tex]\overline{v_{4}}[/tex] = 0


First, I solve for [tex]\overline{v_{4}}[/tex]

So is it

[tex]\overline{v_{4}}[/tex] = c1(4[tex]\overline{v_{4}}[/tex]) + c2(2[tex]\overline{v_{4}}[/tex]) + c3(3/4[tex]\overline{v_{4}}[/tex])




Referring to above poster.

When I read the question, it basically seemed confusing at first. I was thinking of it in terms of a redundant column vector. Now I'm thinking each number is the coefficient to the column vector and it represents a plane. Am I correct in assuming this based on the information?
 
  • #6
Office_Shredder
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Can you post your question, precisely as it is stated, with no shorthand use of pronouns? Just straight from the text.
 
  • #7
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I thought I had it exact, but my formatting is all wrong. I kept copying and pasting and realized that I ws calling v's x's. I fixed it. :/ Sorry for the confusion.
 
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  • #8
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Since this is a column of only 1 vector representing a kernel, would this represent a plane?
The problem states that [1 2 3 4] is in the kernel, not that it represents the kernel. We know that the kernel is at least 1-dimensional (a line through the origin). It might be that the kernel is 2-dimensional (a plane in 4D) or higher.
I have the equation as
[tex]\overline{v_{1}}[/tex] + 2[tex]\overline{v_{2}}[/tex] + 3[tex]\overline{v_{3}}[/tex] + 4[tex]\overline{v_{4}}[/tex] = 0


First, I solve for [tex]\overline{v_{4}}[/tex]

So is it

[tex]\overline{v_{4}}[/tex] = c1(4[tex]\overline{v_{4}}[/tex]) + c2(2[tex]\overline{v_{4}}[/tex]) + c3(3/4[tex]\overline{v_{4}}[/tex])
No, but I think you have the right idea, assuming that my interpretation of this problem was correct; namely, that v1, v2, v3, and v4 are the columns of A.. Try again, and be more careful with your subscripts.
Referring to above poster.

When I read the question, it basically seemed confusing at first. I was thinking of it in terms of a redundant column vector. Now I'm thinking each number is the coefficient to the column vector and it represents a plane. Am I correct in assuming this based on the information?
 

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