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Material Type: Assignment; Class: Introduction Linear Algebra; Subject: Mathematics & Statistics; University: California State University - Long Beach; Term: Spring 2003;
Typology: Assignments
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Newberger Math 247 Spring 03 Homework solutions: Section 2.8 #31-
Start your explanations with the assumptions (the suppose part). Since the column space of F is not equal to R^5 , F does not have a pivot in every row. Since F is square, we know F also does not have a pivot in every column. This means that the homogeneous equation F x = 0 has non-trivial solutions. Since Nul F is the solution set to the homogeneous equation, we can deduce that Nul F contains nonzero vectors, or in other words, Nul F 6 = { 0 }.
Start your explanation with the assumptions (the if part). Since Nul R is not the zero subspace, we know that the homogeneous equation Rx = 0 has nontrivial solutions. This means that R does not have a pivot in every column. Since R is square, this also means that R does not have a pivot in every row, so the columns of R do not span all of R^6. Thus Col R is not equal to R^6.
Start with the assumptions (the if part). Since Col Q = R^4 , the span of the columns of Q is all of R^4 , so Q has a pivot in every row. Furthermore, since Q is square, Q also has a pivot in every column. This means that Qx = b has a unique solution for all b in R^4. Keep straight in your mind that a pivot in every row means the solution always exists, and a pivot in every column means that when there is a solution, that solution is unique.
Start with the assumptions (the if part). Since Nul P is the zero sub- space, the homogeneous equation P x = 0 has only the trivial solution, which means that P has a pivot position in every column. Since P is square, P also has a pivot in every row. Thus P x = b has a unique solution for every b in R^5. Keep straight in your mind that a pivot in 1
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every row means the solution always exists, and a pivot in every column means when there is a solution, that solution is unique.
Since B has linearly independent columns, B has a pivot in every column. This means that the homogeneous equation Bx = 0 has only the trivial solution, so Nul B = { 0 } where 0 denotes the zero vector in R^4.
If the columns of A form a basis for Rm, the columns of A must span R^5 (so A must have a pivot in every row), and the columns must be linearly independent (so A must have a pivot in every column). Thus in this case m = n. Thus A is a square matrix.