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Suppose columns $1,3,5,$ and 6 of a matrix $A$ are linearly independent (but are not necessarily pivot columns) and the rank of $A$ is $4 .$ Explain why the four columns mentioned must be a basis for the column space of $A .$

By the Basis Theorem, this set of four vectors is a basis for the column space.

Algebra

Chapter 2

Matrix Algebra

Section 9

Dimension and Rank

Introduction to Matrices

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supposed columns 135 and six of the matrix, a linearly independent and the rank of ace for Explain why these four columns must make a basis for the common space off, eh? Well, we know that the rank off a is equal before. So what this means is that the column space on a is equal to you are full. Now, what the basis do tells us is that any linearly independent sets off exactly p elements in hey H automatically is a basis for her age. So in this case, we have four literally independent sets which are in the which are in the column space off A and so therefore they must automatically make a basis for college face off.

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