(28 Points.) The following are True/False questions. For this problem only, you do not have to show any work. There will be no partial credit given for this problem. For this problem:
F (a) Five vectors in R^{5 }always span R^{5}.
F (b) If T : R^{n }→ R^{m }and U : R^{m }→ R^{p }are both linear transformations, then the standard matrix for the linear transformation U ◦T is n × p.
T (c) If the Row Echelon form of a matrix A has a pivot in every column, then whenever A~x = ~b has a solution, it must be unique.
T (d) The sum of two solutions to the homogeneous problem A~x = ~0 is also a solution.
T (e) If a linear transformation T : R^{n }→ R^{m }is both one-to-one and onto, then n must equal m.
F (f) Two vectors whose entries are all positive must be linearly independent.
T (g) AssumeT : R^{n }→ R^{m }is a linear transformation and~v_{1},. . . ,~v_{p }are vectors in R^{n}. If {T(~v_{1}),. . . ,T(~v_{p})} are linearly independent, then {~v_{1},. . . ,~v_{p}} are also linearly independent.
Each statement below is either true (in all cases) or false (in at least one case).
If false, construct a counterexample. If true, give a justification that carefully notes everywhere you use the definition of “linear (in)dependence.”
(a) True. We want to show that ~v_{1 }can be written as a linear combination of v_{1 }and v_{2}:
~v_{1},~v_{2},~v_{3} being linearly dependent means that there are numbers a_{1}, a_{2}, and a_{3 }not all zero so that
a_{1}~v_{1}+ a_{2}~v_{2}+ a_{3}~v_{3 }= 0.
Note that a1 , 0 because otherwise we would have
a_{2}~v_{2}+ a_{3}~v_{3 }= 0,
where at least one of a_{2 }or a_{3 }is nonzero – which can’t happen because ~v_{1 }and ~v_{2 }are linearly independent.
So we’re safe to divide through by a_{1}, giving:
a2 a3
~v_{1}+ ~v_{2}+ ~v_{3 }= 0, a1 a1
and thus:
a2 a3
~v_{1 }=− ~v_{2}− ~v_{3}; a1 a1
and so ~v_{1 }is in Span ~v_{1},~v_{2} .
(b) True. Because u_{1 }and u_{2 }are linearly dependent, there are c_{1 }and c_{2 }not both zero so that
c_{1}u~_{1}+ c_{2}u~_{2 }= 0.
If c_{1 }, 0, then
c2 u~_{1 }=− u~_{2}.
c1
So u_{1 }is a constant multiple of u_{2}. If, on the other hand, c_{1 }= 0, then c_{2 }, 0, in which case:
c1
u~_{2 }=− u~_{1},
c2
and thus u_{2 }is a constant multiple of u_{1}.
Consider the Linear Transformation T : R^{4 }→ R^{3 }whose standard matrix is:
Reminder: you must show work to receive credit.
Let’s solve this entire problem in one computation (you don’t have to, but it’s more efficient to do so). In particular, we solve, for any ~b:
(24 Points.) Consider the following system of equations (where a and b are constants):
−2x_{1 }+ 4x_{2 }= 1
4x_{1 }− ax_{2 }= b
Let’s row reduce: −.
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