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012 Echelon 5 PointsLet Aand B be (m X n) matrices Suppose that BT is the row reduced echelon form of AT , then the nonzero columns of B form basis for the range of...

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012 Echelon 5 PointsLet Aand B be (m X n) matrices Suppose that BT is the row reduced echelon form of AT , then the nonzero columns of B form basis for the range of A_

012 Echelon 5 Points Let Aand B be (m X n) matrices Suppose that BT is the row reduced echelon form of AT , then the nonzero columns of B form basis for the range of A_



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A matrix in row echelon form is given. By inspection, find a basis for the row space and for the column space of that matrix. (a) $\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right]$ (b) $\left[\begin{array}{cccc}1 & -3 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]$

All right. Okay. Um so, given a matrix in row echelon form. Okay, so which looks like the following? Okay, 5310. These are Zeros, apologize for my uh poor handwriting on the tablet. Okay, so this note that this is a five by four uh matrix. So, uh if you just look at the roads long form, right? You see that one of the roses all zeros. Okay, So that means that if you look at the dimensions of the road space in the column space of this matrix, which we will call a for now. Okay, So the dimension of both the road space and the columns, column space is going to be four. Right? Because if you look at the columns, okay, there are four uh columns, none of which is all zeros. So that's fine. To the dimension of the column spaces for and for the road space there are 44 rows in which uh none of the entries are all, you know, None of the roads are all not all zero. Okay. And there's one row where it's all zeros. Okay, So five minus one gives you four. Okay? So in terms of finding the basis for these, all you have to do is to eyeball. It's okay. So for the road space because this is given in a row echelon form, we can just Go through the rose one x 1 and that would be the basis. So 1245 01 -30 001 -3 and 0001. Okay, so this would be our basis. Okay, for the road space for the columns based it's only slightly uh more difficult. All you have to do is to read through the columns. Okay, One by one. So it's 100021000 And then for negative 3100 And then last but not least. 50 negative 310 Okay, so these four vectors form the basis for the column space of this matrix. So let's look at another example. So suppose we have the following four by four uh 2100, and 5 3 -71. Okay, so again uh if you inspect the this matrix which is in row echelon form, you see that none of the rose nor the columns are all zeros. Therefore the road space and the column space are both going to be a full dimension for. Okay. So what does that mean? That means that we can just read off the rows and the columns of this matrix and get our basis for the road space and column space. So in particular for the road space, we just read it off. So it's 12 negative 15 01, 4, 3. Uh huh. 001 negative seven and 0001 Okay, so that's the basis for a row space. And for the column space, we just read off the column. So it's 10002100 negative 140 and 53 negative 71 Okay, so this is the basis for a road space.


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