Chapter 4: Q8SE (page 191)
Let H be an n-dimensional subspace of an n-dimensional vector space V. Explain why \(H = V\).
H is a subspace of V, and B is also in V. As B is a linearly independent set in V, B must also be a basis for V.
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(M) Let \({{\mathop{\rm a}\nolimits} _1},...,{{\mathop{\rm a}\nolimits} _5}\) denote the columns of the matrix \(A\), where \(A = \left( {\begin{array}{*{20}{c}}5&1&2&2&0\\3&3&2&{ - 1}&{ - 12}\\8&4&4&{ - 5}&{12}\\2&1&1&0&{ - 2}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm a}\nolimits} _1}}&{{{\mathop{\rm a}\nolimits} _2}}&{{{\mathop{\rm a}\nolimits} _4}}\end{array}} \right)\)
Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).
In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.
\({\left( {{\bf{2}} - t} \right)^{\bf{3}}}\), \({\left( {{\bf{3}} - t} \right)^2}\), \({\bf{1}} + {\bf{6}}t - {\bf{5}}{t^{\bf{2}}} + {t^{\bf{3}}}\)
Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).
16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).
(Hint: Write \(A + B\) as the product of two partitioned matrices.)
A scientist solves a nonhomogeneous system of ten linear equations in twelve unknowns and finds that three of the unknowns are free variables. Can the scientist be certain that, if the right sides of the equations are changed, the new nonhomogeneous system will have a solution? Discuss.
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