The linear transformation T:R^4 > R^4 is represented by the matrix A, where
A=
(1 2 0 3)
(2 7 11 38)
(3 9 3 21)
(5 14 4 31)
i)Find the rank of A.

Sometimes the question asks we to "show that the rank of A is 2", but in this question, it didn't ask you to show but find it. I am new to this topic, and I find it really bad to reduce this to the echelon form without knowing what the rank is. And actually I don't know what is "echelon form", I knew this term in a solution given to a question.
In short, I would like to know what's the method to do this question? And can we know what its rank is at first, so it's "easier" for us to show the rank?
Thank you.

Civ217
 Follow
 0 followers
 1 badge
 Send a private message to Civ217
 Thread Starter
Offline1ReputationRep: Follow
 1
 08102006 17:52

Juwel
 Follow
 3 followers
 18 badges
 Send a private message to Juwel
 Visit Juwel's homepage!
Offline18ReputationRep: Follow
 2
 08102006 20:51
Don't think there's a way to know the rank before finding it, but you absolutely have to learn the Gaussian elimination method to make a reduced row echelon matrix if you are to succeed in linear algebra. And when you perform the Gaussian elimination, the number of pivots is the rank of the matrix.

El Matematico
 Follow
 1 follower
 0 badges
 Send a private message to El Matematico
Offline0ReputationRep: Follow
 3
 08102006 21:01
You need to row reduce it. To do this, subtract 2 lots of the first row from the second, then add three lots and 5 lots of the first row onto the third and fourth respectively to get:
(1 2 0 3)
(0 11 11 44)
(0 3 3 12)
(0 4 4 16)
Then divide rows 2, 3, 4 by 11, 3, 4 respectively then add 1 lots of row 2 to rows three and four and you can see that the rank is 2 since there are 2 nonzero rows.
If T>V is a linear map and you know the Nullity and dimension of U then you can use the dimension theorem to calculate the rank. But generally you wont know this and you will have to row reduce the matrix. 
Civ217
 Follow
 0 followers
 1 badge
 Send a private message to Civ217
 Thread Starter
Offline1ReputationRep: Follow
 4
 09102006 17:45
Thank you Juwel and El Matematico.
(Original post by El Matematico)
You need to row reduce it. To do this, subtract 2 lots of the first row from the second, then add three lots and 5 lots of the first row onto the third and fourth respectively to get:
(1 2 0 3)
(0 11 11 44)
(0 3 3 12)
(0 4 4 16)
Then divide rows 2, 3, 4 by 11, 3, 4 respectively then add 1 lots of row 2 to rows three and four and you can see that the rank is 2 since there are 2 nonzero rows.
If T>V is a linear map and you know the Nullity and dimension of U then you can use the dimension theorem to calculate the rank. But generally you wont know this and you will have to row reduce the matrix.
Also, is there any difference if you perform the row operation or column operation?
What is a basis?
Thank you. 
El Matematico
 Follow
 1 follower
 0 badges
 Send a private message to El Matematico
Offline0ReputationRep: Follow
 5
 09102006 18:23
Column Rank = Row Rank = Rank so you could perform the same operations to the columns and see how many linearly independent columns there are which will also be the rank so theres no difference between row operations and column operations other than you apply them to the columns instead.
As long as you follow the algorithm, you should be able to reduce the matrix given so that:
(i) All non zero rows come above zero rows
(ii) The first non zero entry in each row is 1, call this C(i) for row i
(iii) c(1) < c(2) <...<c(S)
(iv) Entries below alpha_ic_(i) are all zero
(v) Entries above and below alpha_i(c_i) are 0
When your matrix has been reduced to satisfy these conditions it is rowreduced. You can then count the nonzero rows which will be the rank. You always follow the algorithm and it will always work. So just practice the algorithm and you will always be able to find the rank with little difficulty. Look at a few examples in a text book or something to see how to apply it if you are unsure or post another example and ill go through the stages.
v(1),...,v(n) form a basis if they are linearly independent and span V. So (1,1) and (1,1) form a basis of R^2. 
Civ217
 Follow
 0 followers
 1 badge
 Send a private message to Civ217
 Thread Starter
Offline1ReputationRep: Follow
 6
 10102006 15:39
(Original post by El Matematico)
As long as you follow the algorithm, you should be able to reduce the matrix given so that:
(i) All non zero rows come above zero rows
(ii) The first non zero entry in each row is 1, call this C(i) for row i
(iii) c(1) < c(2) <...<c(S)
(iv) Entries below alpha_ic_(i) are all zero
(v) Entries above and below alpha_i(c_i) are 0
1)The linear transformation T:R^4 >R^4 is represented by matrix A, where
A=
(1 1 2 3)
(2 1 1 11)
(3 2 3 14)
(4 3 5 17)
i)Find the rank of A and a basis for the null space of T.
ii)The vector
(1)
(2)
(1)
(1)
is denoted by e. Show that there is a solution of the equation Ax=Ae of the form
x=
(p)
(q)
(1)
(1)
, where p and q are to be found.

Also, the syllabus states that we should be able to
understand the terms "column space", "row space", "range space" and "null space", and determine the dimensions of, and bases for, these spaces in simple cases;
But I do not understand any of the terms. What are they actually? I know the steps to find the rank, but I don't understand why it is so and what I am actually doing.
Thank you. 
El Matematico
 Follow
 1 follower
 0 badges
 Send a private message to El Matematico
Offline0ReputationRep: Follow
 7
 10102006 19:48
First of all, condition (iv) in my above post is not needed because of condition (v) which is stronger (If we use condition (iv) instead of (v) the matrix will be in reduced echelon form after applying the algorithm). The rank is defined to be dim(Im(T)), which is equivalent to the number of nonzero rows in a matrix after it has been rowreduced.
To row reduce the matrix given I will apply the rowreduction algorithm systematically, so start at the top row. The first nonzero entry is a 1, so condition (ii) is satisfied so make it the pivot. If the first entry was not 1, then you could either swap it with another row which does have a 1 as its first nonzero entry or divide the row by the first entry (providing it isnt zero). Now, to satisfy condition (v), we need to add (2) lots of row 1 to row 2, (3) lots of row 1 to row 3 and then (4) lots of row 1 to row 4 to obtain:
(1 1 2 3)
(0 1 3 5)
(0 1 3 5)
(0 1 3 5)
So the first column now satisfies (v). We now need to repeat the process. Choose position (2,2) as the pivot ( i.e. ( 0 1 3 5) in the 2nd row in the matrix.
To satisfy property (v), we need to add 1 lot of row 2 to row 1, (1) lots of row 2 to row 3 and (1) lots of row 2 to row 4 to get:
(1 0 1 8)
(0 1 3 5)
(0 0 0 0)
(0 0 0 0)
We now note that the matrix satiffies all of the conditions required for it to be in row reduced form since all nonzero rows appear above zero rows, each nonzero row starts with a 1 and each element above and below it is 0 and condition (iii) is satisfied. We can now read off the rank as being 2. You should look at examples of matrices that are row reduced and others which arent. For example,
(1 2 1)
(0 0 1)
(0 0 0)
Is not row reduced since (v) fails.
Now for a basis of the nullspace (which we now know must be dimension 2),
apply A in row reduced form to the column vector (a b c d) to get:
(a c +8d, b + 3c +5d)
Now a  c + 8d = 0
and b + 3c + 5d = 0 since it maps to the kernel(T) (or nullspace(T))
Since we have 2 equations in 4 unknowns, let a = m and b = n where m,n constants. Solving these equations I get the following values:
a = m, b = n, c = (5m  8n)/29, d = (3m + n)/29
So Ker(T) = {(m, n, (5m  8n)/29, (3m + n)/29)  m,n belong to R} Any vectors with this property map to 0.
So a basis of Ker(T) = (1, 0, 5/29, 3/29) and (0, 1, 8/29, 1/29)
Hence the dimension is 2, as we would expect. The values may be wrong, but the method is fine.
Im not sure about the terminology used at the end, but Im assuming rowspace is a different way of saying rowrank. If it is then rowrank is the number of linearly independent rows of a matrix, and the column rank is the number of linearly independent columns in a matrix. The range space is probably dim(Im(T)), i.e. the Rank which is equal to rowrank and columnrank. Null space is the Kernel, Nullity is the Dim(Ker(T)). Hope that helps. 
Civ217
 Follow
 0 followers
 1 badge
 Send a private message to Civ217
 Thread Starter
Offline1ReputationRep: Follow
 8
 11102006 17:06
Thank you. But I still don't really understand, mainly is because of the algorithm
(i) All non zero rows come above zero rows
(ii) The first non zero entry in each row is 1, call this C(i) for row i
(iii) c(1) < c(2) <...<c(S)
(iv) Entries below alpha_ic_(i) are all zero
(v) Entries above and below alpha_i(c_i) are 0
First of all, condition (iv) in my above post is not needed because of condition (v) which is stronger.
To row reduce the matrix given I will apply the rowreduction algorithm systematically, so start at the top row. The first nonzero entry is a 1, so condition (ii) is satisfied so make it the pivot. If the first entry was not 1, then you could either swap it with another row which does have a 1 as its first nonzero entry or divide the row by the first entry (providing it isnt zero). Now, to satisfy condition (v), we need to add (2) lots of row 1 to row 2, (3) lots of row 1 to row 3 and then (4) lots of row 1 to row 4 to obtain:
A=
(1 1 2 3)
(2 1 1 11)
(3 2 3 14)
(4 3 5 17)
I don't understand the condition v, what is alpha_i(c_i)? Let's say then it's not one in a case, what do you mean by " swap it with another row which does have a 1 as its first nonzero entry or divide the row by the first entry (providing it isnt zero)."? Can you show me because I don't understand it.
Thank you. 
El Matematico
 Follow
 1 follower
 0 badges
 Send a private message to El Matematico
Offline0ReputationRep: Follow
 9
 11102006 17:16
You can get rid of condition (iv), my mistake. We only need (v). Basically, you make the pivot 1. The column which contains the first nonzero entry of a row is denoted C_(i) for row i. So,
(1 4 6)
(0 0 7)
(0 4 0)
In this case C(1) = 1, C(2) = 3 (Since the first nonzero entry is a 7, in column 3 so C(2) = 3) C(3) = 2
The algorithm requires that C(i) = 1 then make this the pivot (It also makes the arithmetic easier). Every entry above and below C(i) must be 0 (This is what condition (v) is saying). In my example of where (v) fails,
(1 2 1)
(0 0 1)
(0 0 0)
It fails because there is a 1 above the first non zero entry in row 2.
Related discussions
Related university courses

Cymraeg/Mathematics
Aberystwyth University

Economics and Mathematics
University of Derby

Mathematics
Queen Mary University of London

Mathematics (3 years)
Durham University

Mathematics (3 years)
Durham University

Mathematics (major with a minor)
University of Leicester

Mathematics with Business and Management
Aberystwyth University

Mathematics with Ocean and Climate Sciences
University of Liverpool

Mathematics with Physics (Including Placement Year)
University of Essex

Modern Languages with Mathematics
Royal Holloway, University of London
see more
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
 Notnek
 charco
 Mr M
 Changing Skies
 F1's Finest
 RDKGames
 davros
 Gingerbread101
 Kvothe the Arcane
 TeeEff
 Protostar
 TheConfusedMedic
 nisha.sri
 claireestelle
 Doonesbury
 furryface12
 Amefish
 Lemur14
 brainzistheword
 Quirky Object
 TheAnxiousSloth
 EstelOfTheEyrie
 CoffeeAndPolitics
 Labrador99
 EmilySarah00
 thekidwhogames
 entertainmyfaith
 Eimmanuel
 Toastiekid
 CinnamonSmol
 Qer
 The Empire Odyssey
 RedGiant
 Sinnoh