# Lin. algebra

1. May 17, 2007

### 413

Let b be the vector such that B = [A b] , and let a1, a2, a3 and a4 be the columns of A.

Let m be the number of linearly independent columns of A, let k be the number of parameters (free variables), and let n be the total number of columns in A. In our example above, n = 4.

Do you suppose that this relationship m + k = n will be true for all systems of linear equations?

2. May 17, 2007

### HallsofIvy

I'm afraid I don't recognize the notation "the vector such that B = [A b]". What does it mean?

3. May 17, 2007

### 413

um, well A is the co-efficient matrix of the system, and B is the augmented matrix of the system, b is just the answer column of the linear equations.

4. May 17, 2007

Yes, m + k = n should hold, since m = r(A).

5. May 19, 2007

### fopc

"Do you suppose that this relationship m + k = n will be true for all systems of linear equations?"

No.

Edit:
In the problem as stated there are n unknowns.
In the context of Gaussian elimination, they fall into
two groups, free and basic. In the stated problem
they are k and m, respectively. Their sum equals n,
i.e., k+m=n. This is a trivial result. I suppose that's
what fooled me. I guess I was expecting a "trick" question, or maybe something with a little more substance.

I might add that in the case ~(b=0), the solution *set* is not a "space" (in the vector space sense). It should properly be called a linear manifold.

Last edited: May 20, 2007
6. May 19, 2007

### daniel_i_l

Yes, if P is the solution space and r(A) is the rank then:
dimP = n - r(A)
outline of proof: (r=r(A))
First we can put the matrix A into row-echelon form, now the number or non-zero rows is the rank.
If v = (t_1, ... , t_r, ... , t_n) is a solution then the vectors {t_1, ... , t_r} are all linear combinations of {t_r+1, ... , t_n} . If
t_1 = C_11*t_r+1 + ... + C_1n*t_n then it's possible to write the general solution as:
t_r+1 (C_11, C_21, ... , 1, 0, ... , 0)
:
:
t_n (C_1(n-r) , C_2(n-r), ... , 0 , ... ,0 ,1)
since all these n-r vectors are also linearly independent they are a base to the solution and so dimP = n - r(A)
In this case dimP = k, r(A) = m so m+k = n