Vector Spaces - Row Space and Rank

Chickentown

Hi,

I have attached a question I am working on and my solution.

I was wondering if someone could take a look and see if I am going about it right.

The matrix in question is the first one at the top of the page, I started row reduction notation straight away, sorry if this is misleading.

Thanks

TheBody

Hi there.

Please note that I have not checked your calculations to get the matrix to row echelon form.

The basis for the row space is incorrect. You read off the rows with the pivots (leading ones) in the reduced matrix. In other words the basis is:

(1,-2,0)
(0,1,1)
(0,0,1)


The basis for the column space is correct although you forgot a minus sign on the 2 in [latex]C_3[/latex].

Don't forget to write down the rank of the matrix!

Chickentown

TheBody wrote: »

Hi there.

Please note that I have not checked your calculations to get the matrix to row echelon form.

The basis for the row space is incorrect. You read off the rows with the pivots (leading ones) in the reduced matrix. In other words the basis is:

(1,-2,0)
(0,1,1)
(0,0,1)


The basis for the column space is correct although you forgot a minus sign on the 2 in [latex]C_3[/latex].

Don't forget to write down the rank of the matrix!

Ok thanks,

(1) Calculations aside you are saying the process is correct?

(2) To calculate the basis for a row space of a matrix all I have to do is; get the matrix in reduced row echelon form and then read off the rows that have '1' as a pivot - These are the basis for my row space?

(3) For column space I go back to the original matrix and use the columns that correspond with the pivots on the RREF matrix as the basis for the column space?

(4) I don't understand what the Rank a matrix actually is or how to find it, any advice on this would be greatly appreciated.

TheBody

Chickentown wrote: »

Ok thanks,

(1) Calculations aside you are saying the process is correct?

(2) To calculate the basis for a row space of a matrix all I have to do is; get the matrix in reduced row echelon form and then read off the rows that have '1' as a pivot - These are the basis for my row space?

(3) For column space I go back to the original matrix and use the columns that correspond with the pivots on the RREF matrix as the basis for the column space?

(4) I don't understand what the Rank a matrix actually is or how to find it, any advice on this would be greatly appreciated.

Everything is spot on for points 1-3.

To answer part 4, the rank of a matix is the dimension of the column space. As you have 3 vectors in the basis of the column space, the rank is 3.

Chickentown

TheBody wrote: »

Everything is spot on for points 1-3.

To answer part 4, the rank of a matix is the dimension of the column space. As you have 3 vectors in the basis of the column space, the rank is 3.

So whenever some guy asks me to get the rank of a matrix, immediately I must start thinking about getting the matrix in RREF then I just count the pivot columns and the result of this sum is the rank of the matrix.

Is it just that simple or are there some caveats attached, for example things like; in some arbitrary nXm matrix where n=m or n<m or n>m etc...

Thanks for that, you have no idea how much you have just helped me. I really appreciate it.

TheBody

Chickentown wrote: »

So whenever some guy asks me to get the rank of a matrix, immediately I must start thinking about getting the matrix in RREF then I just count the pivot columns and the result of this sum is the rank of the matrix.

Is it just that simple or are there some caveats attached, for example things like; in some arbitrary nXm matrix where n=m or n<m or n>m etc...

Thanks for that, you have no idea how much you have just helped me. I really appreciate it.

It is as simple as you said.

A good student friendly book to check out is Elementary Linear Algebra by Howard Anton and Chris Rorres. It's probably in your library.

Good luck with your studies!

Chickentown

Thanks for the advice, I will definitely look it up over the Christmas once I get my exams out of the way.

Thanks again

TheBody

No problem. Be sure to post again if you have any other questions.