Compute rank(T) by reducing matrix to echelon form

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Homework Statement


Let T: R4 --> R4 be defined by T(a,b,c,d) = (4a-b+c, b+3d, 3a+c+d, c-d, b+c+2d)
where a,b,c,d are in R. Compute rank(T).


Homework Equations


The Attempt at a Solution


Writing the transformation as a matrix,

T[a] = [4 -1 1 0][a]
[0 1 0 3]
[c] [3 0 1 1][c]
[d] [0 0 1 -1][d]
[0 1 1 2]

So that big matrix on top, call it A, is the standard matrix. We'll need to reduce A to echelon form in order to find the rank(T).
So this is how far I got:
A = [4 -1 1 0]
[0 1 3 0]
[3 0 2 0]
[0 0 -1 1]
[0 1 3 0]
I can't seem to reduce this any further without replacing the zero elements with nonzero elements, which is not the point. Rank(T) should be the number of nonzero rows in the matrix reduced to echelon form, right? So does that mean that rank(T) = 5? Although, I don't think A is completely reduced. Maybe my arithmetic does not work.
 

Answers and Replies

  • #2
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Perhaps you can try computing the dimension of the null space of T and then using the dimension theorem to find the rank of T.
 
  • #3
HallsofIvy
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Homework Statement


Let T: R4 --> R4 be defined by T(a,b,c,d) = (4a-b+c, b+3d, 3a+c+d, c-d, b+c+2d)
where a,b,c,d are in R. Compute rank(T).


Homework Equations


The Attempt at a Solution


Writing the transformation as a matrix,

T[a] = [4 -1 1 0][a]
[0 1 0 3]
[c] [3 0 1 1][c]
[d] [0 0 1 -1][d]
[0 1 1 2]

So that big matrix on top, call it A, is the standard matrix. We'll need to reduce A to echelon form in order to find the rank(T).
So this is how far I got:
A = [4 -1 1 0]
[0 1 3 0]
[3 0 2 0]
[0 0 -1 1]
[0 1 3 0]
I can't seem to reduce this any further without replacing the zero elements with nonzero elements, which is not the point. Rank(T) should be the number of nonzero rows in the matrix reduced to echelon form, right? So does that mean that rank(T) = 5? Although, I don't think A is completely reduced. Maybe my arithmetic does not work.

You need more practice on row reducing! Always work on one column at a time from left to right and top to bottom. Starting with
[tex]\left[\begin{array}{cccc}4 & -1 & 1 & 0 \\ 0 & 1 & 0 & 3 \\ 3 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \\ 0 & 1 & 1 & 2\end{array}\right][/tex]
The left column has 4 at the top and 3 in the third row- 0s else where. Subract 3/4 the first row from the third: it becomes 3-3= 0, 0- (-3/4)= 3/4, 1- 3/4= 1/4, 1- 0= 0. The matrix is now
[tex]\left[\begin{array}{cccc}4 & -1 & 1 & 0 \\ 0 & 1 & 0 & 3 \\ 0 & 3/4 & 1/4 & 1 \\ 0 & 0 & 1 & -1 \\ 0 & 1 & 1 & 2\end{array}\right][/tex]
Yes, some 0s were filled but the first column is "1 0 0 0" and THAT is the point.
Now look at the second column. There is a 1 in the second row and by subtracting 3/4 of the second row from the third, I can't get a 0 in the third row, second column while subracting the second row from the fourth gives me a 0 there:
[tex]\left[\begin{array}{cccc}4 & -1 & 1 & 0 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1/4 & -5/4 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & -1\end{array}\right][/tex]
Now do the third column. There is a 1/4 in "third row third column. I can get a 0 in the fourth row by subtracting 4 times the third row and a 0 in the fifth row by doing the same:
[tex]\left[\begin{array}{cccc}4 & -1 & 1 & 0 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1/4 & -5/4 \\ 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & 4\end{array}\right][/tex]
The last step should be obvious: subtract the fourth row from the fifth:
[tex]\left[\begin{array}{cccc}4 & -1 & 1 & 0 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1/4 & -5/4 \\ 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & 0\end{array}\right][/tex]k

By the way- it should have been obvious that the rank could NOT be 5. This transformation maps R4 to a subspace of R5. The dimension of that subspace is the "rank" and it cannot be higher than 4.
 
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