Linear Algebra. Please check my answers and help me find mistakes!

  • #1

Homework Statement


Give bases for row(A), col(A), and null (A). Also find bases for row(A) and col(A) using AT.
A=[1, 1, -3; 0, 2, 1; 1, -1, -4]
AT: [1, 0, 1; 1, 2, -1; -3, 1, -4]


Homework Equations





The Attempt at a Solution





For the first part I got the bases for row(A) to be [1, 0, -7/2] and [0, 1, 1/2].
For the col(A) I got [1;0;1] and [1;2;-1]

For the null(A) I got [7/2;1/2;1]


For the second part I got bases for row(A) to be [1, 0, 0] and [0, 1, 0]. For the basis for col(A) I got [1; 0; 1] and [0; 1; -1]

My teacher did not explain these concepts because he ran out of time, and it is due tomorrow. Please let me know where I messed up at, if anywhere. Thanks
 

Answers and Replies

  • #2
vela
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For the first part I got the bases for row(A) to be [1, 0, -7/2] and [0, 1, 1/2].
For the col(A) I got [1;0;1] and [1;2;-1]

For the null(A) I got [7/2;1/2;1]
The bases for row(A) and col(A) look fine, but you have a sign error for null(A).
For the second part I got bases for row(A) to be [1, 0, 0] and [0, 1, 0]. For the basis for col(A) I got [1; 0; 1] and [0; 1; -1]
Could you explain what you did to find these?
 
  • #3
I corrected the first null(A) to [7/2; -1/2; 1]

For the second part, I reduced the transposed matrix to [1, 0, 1; 0, 1, -1; 0, 0, 0]. I took the row(A) to be the first two columns of the transposed, reduced matrix, and I took the col(A) to be the first two rows of the transposed, reduced matrix.
 
  • #4
vela
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The first basis you found for the row space allows a non-zero z-component, but the second basis you found doesn't, which means their spans are different. So something went wrong there since the two bases are supposed to be spanning the same space. I think if you double-check your method carefully, you'll find your error. (Hint: Your reduced matrix is fine. That's not where the problem is.)

Can you think of a way to check if your bases for the column space are consistent with each other?
 
  • #5
I see what you are saying, but I don't see where I went wrong. My notes tell me that for the row(A) of a transposed matrix I have to look at the reduced form and use the columns as my answer. I am a bit confused.
 
  • #6
vela
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I see what you are saying, but I don't see where I went wrong. My notes tell me that for the row(A) of a transposed matrix I have to look at the reduced form and use the columns as my answer. I am a bit confused.
Just to make sure, when you say "row(A) of the transposed matrix," you mean the basis of the row space of the untransposed matrix A found via working with the transposed matrix, right? I think that's what you mean but what you wrote could be interpreted a bit differently.

When you found the column space of A, you looked at the reduced matrix, located which columns contained the pivots, and then selected the corresponding columns of the original matrix A. When you find the basis for the column space of AT, you have to do the same thing.
 
  • #7
Yes that is what I mean.

When you find the column space of the transposed matrix, don't you have to look at the rows rather than the columns since it is transposed?
 
  • #8
vela
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Yes, what you're saying is that row(A)=col(AT). You're just not finding the basis vectors correctly.

In the first part, you found a basis for col(A) to be {(1,0,1), (1,2,-1)}. From which matrix did you get those vectors? Now, when looking for a basis for the column space of AT, which is the row space of A, where are you grabbing the vectors from? You're not following the same process, which is why it's not working out.
 
  • #9
I got those vectors from the pivots of the reduced form. I took note of what columns those were, and took the original numbers from A.
I am getting the vectors for the second part from the reduced form of the transposed matrix.
 
  • #10
Would the col(A) be [1;0;1] and [1;2;-1] for the second method as well? If I look back at the way I solved the first one?
 
  • #11
vela
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Are you talking about col(A) or col(AT)?
 
  • #12
I meant col(A) using the AT method.
When I look at AT, I selected the top two rows, so that these correspond to the first two columns. The result this gave me is the same col(A) that I had for the first example that didn't require the use of AT
 
  • #13
vela
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I meant col(A) using the AT method.
In other words, row(AT).
When I look at AT, I selected the top two rows, so that these correspond to the first two columns. The result this gave me is the same col(A) that I had for the first example that didn't require the use of AT
Forget that AT is the transpose of A. Just think of it as another matrix. Find bases for AT's row and column space (making no reference to A).

Now how are row(A), col(A), row(AT), and col(AT) related?
 
  • #14
Right.

row(A)=col(AT)....col(A)=row(AT)

So row(AT)= [1,0,1] and [0,1,-1]
and col(AT)= [1;1;-3] and [0,2,1]
??
 
  • #15
vela
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Right.

row(A)=col(AT)....col(A)=row(AT)

So row(AT)= [1,0,1] and [0,1,-1]
and col(AT)= [1;1;-3] and [0,2,1]
??
Yup!
 
  • #16
Ok so in summary:

First part: row(A)=[1,0,-7/2] and [0,1,1/2]
col(A)= [1;0;1] and [1;2;-1]
null(A)= [7/2;-1/2;1]

Second part: row(A)= [1,1,-3] and [0,2,1]
col(A)= [1;0;1] and [0,1,-1]

All good?
 
  • #17
Is there any way I can ask you one more question? It shouldn't take as long as this last one took, but you seem helpful and I would really appreciate it.
 
  • #18
vela
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Looks good. You should check your answers are consistent. You should be able to write the vectors in one basis as a linear combination of the vectors in the other basis of that space.
 
  • #19
vela
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Is there any way I can ask you one more question? It shouldn't take as long as this last one took, but you seem helpful and I would really appreciate it.
Sure, just start up another thread with your new question. If I can't help, someone else likely can.
 
  • #20
I am a little confused about how to check if they are consistent. I set up a matrix to test if a linear combination is present, but it didn't work
 
  • #21
vela
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You can solve them by inspection. Take, for example,

(1,1,-3) = a (1,0,-7/2) + b (0,1,1/2)

You should be able to see what a and b have to equal for the first two slots to work out. All you have to do is verify that the third slot works out.

First part: row(A)=[1,0,-7/2] and [0,1,1/2]
Second part: row(A)= [1,1,-3] and [0,2,1]
 
  • #22
It does work out when a=b=1. What about when you use [0,2,1] instead of [1,1,-3]?
 
  • #23
I guess my questions would be better phrased as...
how did you know which combination to select, because it seems like other combos don't work.
 
  • #24
vela
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It has to work for both vectors, though you'll get different values for a and b when you have a different vector on the LHS. If it doesn't work, that means the vectors don't have the same span, so they can't both be a basis for row(A).
 
  • #25
It doesn't work for [0,2,1] though does it?
 

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