Linear Algebra - Basis and Kernel

succubus
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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,v3I'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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What are the vi here?
 
The v_{4} seems like

1
2
3
4

since the problem is asking for v_{4} to be written as a linear combination of \overline{v_{1}},\overline{v_{2}},\overline{v_{3}}
 
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succubus said:
The v_{4} 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.
succubus said:
since the problem is asking for v_{4} to be written as a linear combination of \overline{v_{1}},\overline{v_{2}},\overline{v_{3}}

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?
 
Since this is a column of only 1 vector representing a kernel, would this represent a plane? I have the equation as
\overline{v_{1}} + 2\overline{v_{2}} + 3\overline{v_{3}} + 4\overline{v_{4}} = 0


First, I solve for \overline{v_{4}}

So is it

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

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?
 
Can you post your question, precisely as it is stated, with no shorthand use of pronouns? Just straight from the text.
 
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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succubus said:
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.
succubus said:
I have the equation as
\overline{v_{1}} + 2\overline{v_{2}} + 3\overline{v_{3}} + 4\overline{v_{4}} = 0


First, I solve for \overline{v_{4}}

So is it

\overline{v_{4}} = c1(4\overline{v_{4}}) + c2(2\overline{v_{4}}) + c3(3/4\overline{v_{4}})
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.
succubus said:
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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