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# Dimension of the column space or rank

Dimension of the Column Space or Rank. Created by Sal Khan.

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• So I conclude from this and the last video that for a Matrix A with the form (mxn), n=rank+nulity. Is this always true? •  That's exactly right, nullity is the number of redundant (free variable) columns, rank is the number of non-redundant (pivot) columns, so together they add up to the total number of columns. I was just going through those videos and looked it up to be sure I'd drawn the right conclusion too. Nice work.
• Isn't it clear that the vectors are linearly dependent? You have 5 vectors in R^4, it is obvious that (at least) one must be redundant. •  Yep. It's clear that at least one is redundant. So the rank of A must be 4 or less. But in order to find the rank, you need to figure out how many are redundant. Sal was kind enough to make this example with two columns redundant so we can see that the answer is not simply Rank(A) = 4.
• So every matrix in reduced row echelon form is linearly independent? Correct? •  Rodrigo,

No, this isn't the case. If the columns of a matrix are not linearly independent, then the columns of the reduced row echelon form of the matrix will ALSO not be linearly independent.

For example, let's look at a matrix whose columns are obviously not linearly independent, like:
|1 2|
|2 4|

Obviously, we can get the second column by multiplying the first column by 2, so they are linearly dependent, not independent. Now let's put the matrix into reduced row echelon form.

Step 1. Get all zeros in the 1st column except for the top term. I can do this by adding -2 times the first row to the second row, to eliminate the 2nd term of the first column:

|1 2|
|0 0|

But notice, that this is as far as I can go. This IS reduced row echelon form for this matrix, and notice also that the columns are not linearly independent. The second column is STILL 2 times the first column.

Hope that helps.
• I watched all the linear algebra video's up to this point. I can construct RREF, find the null space, column space, the nullity and the rank. etc... but I still miss the point of it all. What is the purpose of these definitions. I still have no clue about the context, what problems will this help us solve? • Great question! Linear Algebra has many applications - for instance almost anything in Physics is expressed as a vector of one form or another. Calculus is all about taking non-linear things and approximating them linearly, that was the whole point of derivatives. This is due to the fact that linear equations are relatively easy to solve, whereas non-linear ones can be very difficult. A standard technique in mathematics is looking at a non-linear system and finding a linear approximation. Often times in physics you have a taylor series expansion over differential pieces of length, area, volume, etc. so that the square and higher terms cancel. In Computer Science everything explicitly uses linear algebra. It's probably the most important mathematics, that and fourier analysis.
• Does the rank of A also equal to number of non-zero rows of RREF(A)? • Hi, i am kind's new to khan academy and stuff, is there any way i can practice the knowledge above. I mean i have watched a bunch of videos, but i think i will forget all the concepts unless i practice them. • is there any video talking a bout the rank of matrix ? • I think "" does not need a proof as they're just i j k l unit vectors. • Isn't 6-2(-3) =12 instead of zero? • No thesergcan's answer was incorrect.

The question was correct as it does compute to 12. However, Sal mentions at in the video that 6 + 2 times minus 3, which is 6+2(-3), does indeed equal 0.

However, "Here you multiply the numbers first: -2*(-3) = 6 ( you can think of the equation as 6+-2(-3) =12). So 6-6 = 0." will never result in 6-6 = 0.
(1 vote)
• At you state that r1, r2, and r4 are linearly independent. Are those columns (r1, r2, and r4 ) dependent variables though as they have a leading 1 value ? 