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### Course: Linear algebra>Unit 2

Lesson 7: Transpose of a matrix

# Rowspace and left nullspace

Rowspace and Left Nullspace. Created by Sal Khan.

## Want to join the conversation?

• Its unclear to me why the dimC(A) = dimR(A)? Its clear that this is the case--but i've researched this a bit and haven't found the why this is the case. thanks
• When you row reduce A, all the rows left in the rref are linearly independent, otherwise one of the remaining rows would be a multiple of another row, and could be eliminated. But the number of rows in the rref is the same as the number of pivot columns, which is the same as the number of linearly independent column vectors.

The row vectors and column vectors of A are respectively the column vectors and row vectors of A transpose, so, as above, the number of linearly independent rows in A transpose is the same as the number of its linearly independent columns. Therefore Rank(A transpose) = Rank(A).
• I am having some trouble figuring out a couple proof type problems in this area of intro linear algebra. For example, right now I have to prove that for any matrix A if vector u is in RowA and vector v is in NullA, then product of u(transpose) and v is 0. Or that if u belongs to both RowA and NullA, then u equals the zero vector.
• This happens because RowA and NullA are orthogonal subspaces.
(1 vote)
• Is "nullspace" the same thing as the span of the "kernel"?
• Yes, and a better way to say it is that the kernel is the nullspace. The span of the kernel and the span of the nullspace are just themselves since they are already subspaces.
• Between and Sal explains how A^tx=0 is equivalent to x^TA=0^t. This sounds very confuse because the left part of the equivalence creates a 2x1 matrix and the second part creates a 1x3 matrix. The left part i just need to equal to zero, but what should I do with the right part?
Also, just to be sure, a transpose of zero matrix is zero right?
(1 vote)
• A is a n⨉m matrix, so Aᵀx⃗ is a m⨉1 column vector. (since Aᵀ is a m⨉n matrix and x⃗ is a m⨉1 column vector.) Therefore in the equation Aᵀx⃗ = 0⃗, the 0⃗ is the zero m⨉1 column vector.

Now, the equation x⃗ᵀA is a 1⨉m row vector (since x⃗ᵀ is a 1xm row vector), and therefore 0⃗ᵀ is also the zero 1⨉m row vector.
• At about , Sal says that x1 is the pivot entry. But couldn't we have solved for any of the other x's also, and then expressed them in the same way?
• Yes, he could. Depending on which row you put on top, and which column comes first and second, and how you do the elimination, you could make any of the 3 variables the pivot variable. Also notice that in this problem the span of any of the column vectors is the same as span of any other.
• For a system of equations Ax=b, how do you prove that the solution for x lies in the row space of A? I tried pondering over this, in vain.
• x doesn't necessarily lie in the row space. Consider:
[2 1][x1]=[3]
[x2]

x1 = 1.5
x2 = 0

The given values for x solve the single equation, but x is not a linear multiple of the [2 1] row.
• If Rank(A)=Rank(A Transpose), then that should mean dim( (C(A) )= dim ( C(A Transpose) ). If we have, say, 2 pivots then that means we have two Lin. Ind. rows. In that case dim ( C(A transpose) )= 2. The two independent rows form a basis for the column space of A transpose.
But if we have 2 pivots, then we also have two Lin. Ind. columns and dim ( C(A) )= 2.The two independent columns form a basis for the column space of A. So dim ( C(A) ) =2= dim ( C(A Transpose) ). Is this reasoning correct?
• It's absolutely correct.......The number of columns with pivots = the number of rows with pivots........v.i.z. .....Number of independent columns = Number of independent rows
So, C(A) = C(A Transpose) = Rank(A).
• I used x2 as the pivot variable and got the basis of c(a) right but not the null space. Can there not be different free variables in one matrix?
(1 vote)
• Maybe your null space has the same dimensions (R^2), and is even the same 2 dimensional subspace of R^3, but has different basis vectors. You could check this by taking the cross product of the basis vectors of both versions to see if they're parallel.