Practice Questions+Answers for test 2.
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Recall the following strategies.

If a problem is asking for the basis/dimension of the nullspace of A, write
out solutions for AX=0, separate the variables. The basis is the
vectors you come up with.

If a problem is asking for the basis/dimension of the column space of
A, compute rref(A) and pick the columns of A corresponding to the
columns in rref(A) that have a leading 1.

If a problem is asking for the basis/dimension of the row space of A,
compute rref(A transpose), pick the rows in A (=columns in A
transpose) corresponding to columns in rref(A transpose) with leading
1's.

If a problem gives you W as the collection of all vectors of "the
following type where a,b,... are any real numbers", take what is
given, separate variables, and take the resulting vectors as
generators of W. 

If W is given as the collection of "all vectors satisfying the
following conditions in a,b,..." then write out solutions for the
linear system that these conditions form to get a basis as in the
first point up on top.

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1.:
Find a vector of length 5 perpendicular to (1,0,-1,2).

first find some verctor perpendicular, like (0,1,0,0). Then make it
length 5 by multiplying by 5.

2.: 

Find the standard matrix form of the linear transformation L: R^3 ->
R^2 which takes a vector to its projection onto the x-z-plane. 

(1,0,0) -> (1,0)
(0,1,0) -> (0,1)
(0,0,1) -> (0,0)

so A=[1 0 0]
     [0 1 0]

3.:

Consider the 2 systems of parametric equations:

(a) x=1+2t, y=2-3t, z=-1+t;

(b) x=3+4s, y=-1-6s, z=2s.

Do they describe the same line? Explain!

The first has tangent direction (2,-3,1) and the second
(4,-6,2). These are parallel, so the lines might be the same. To check
this we must check whether they have apoint in common. Question: is
(1,2,-1) on the second line? If yes, then s must by -1/2 (from the
x-component) and if you plug this s in you do indeed get the point
(1,2,-1). So the lines are identical.

4.: 

Find the parametric equation of the line that passes through
P=(2,1,-1) and is perpendicular to the plane E: x+2y+3z-4=0.

The normal (1,2,3) must be the line vector. So the line is
(2+t,1+2t,-1+3t).

5.:

We define a plane by 
E: (x,y,z) = (1,2,-1) + t(2,1,1) + s(1,0,2). 
(s and t are arbitrary real numbers)

a) Find parametric equations of 2 lines that lie in the plane (you
choose which lines!).

First line: (1,2,-1)+t(2,1,1)
second: (1,2,-1)+s(1,0,2)


b) Find a vector perpendicular to the 2 lines you chose.

take the cross product between the two line vectors. (2,1,1) x (1,0,2)
= (2,-3,-1). so the new line is (1,2,-1)+u(2,-3,-1).

6.: 

A= [1 2 -1 3]
   [1 3  2 4]
   [3 1 -2 4]

A has rref equal to 
 [1 0 0 1]
 [0 1 0 1]
 [0 0 1 0]

a) Find a basis for R^3 from the columns of A.

the first, second and third column of A. (because, the same is tru in
the rref)

b) Express the last column of A as a linear combination of the other
columns of A.

(4) = (1) + (2) 
(also from rref)

7.: 

Find the dimension of V and W and write down a basis for each.

V= { (x,y,z,w) with 2x+y=0 }

W= { all vectors in R^4 perpendicular to (1,0,-1,0) }

[2 1 0 0|0] has rref with one leading 1. so the solution space has
dimension 4-1=3. 
to find a basis, write down the solutions: (-2y,y,z,w). (there are
three free variables!)
so if you take this apart into what belongs to y,z and w you get basis
elements
(-2,1,0,0) , (0,0,1,0) and (0,0,0,1).

for the second part, the defining linear equation of W is 
x-z=0.
rref is [1 0 -1 0|0]
so again dim=3, and the basis here is 
(1,0,1,0), (0,1,0,0), (0,0,0,1)

8.:

L: R^2 -> R^2 is a linear transformation such that L((1,1)) = (1,2)
and L((1,-1)) = (3,0).

a) Find L((1,0)) and L((0,1)).

(1,0)=1/2 (1,1) + 1/2 (1,-1). (solve a little linear system to figure
that out)
L is linear, so 
L(1,0)=1/2 L(1,1) + 1/2 L(1,-1)=(1/2,1)+(3/2,0)=(2,1).

also, (0,1)=1/2 (1,1)-1/2 (1,-1) so 
L(0,1)=1/2 L(1,1) - 1/2 L(1,-1)=(1/2,1)-(3/2,0)=(-1,1).

b) Write L in standard matrix form.

[ 2 -1]
[ 1  1]

c) Describe the range of L.

all of R^2 because the two columns are linearly independent (check!),
so the image is of dimension 2 and so must fill all of R^2.

9.:

Which are subspaces?

a) U= { (2x+y,y+z,2) }

no. try adding (2,0,2) to itself.

b) V= { (x,y,z) with x^2+z=0 }

no, try adding (1,0,-1) to itself.

c) W= { (x,y,z) with x>0, z>0 }

no, try scaling.

d) X= Span{ (1,2,3), (2,0,1) }

yes (any span is a subspace)

Explain your answers.

10.: 

Two planes are given,

E1: x+2y+2z-5=0

E2: 2x-y+z-2=0.

a) Find a vector perpendicular to E1, and another vector perpendicular
to E2.

(1,2,2) and (2,-1,1)

b) Find a point in the intersection of the planes.

solve rref of

[1  2 2 -5]
[2 -1 1 -2] any solution is a point on both planes.

c) Find a parametrization of the entire intersection of the planes E1
and E2. Parts a) and b) could be useful...

pick a point P of part b).
take the cross product v of the vectors in a).

the line of intersection is P + tv.

11.: 

A= [1 1 4 1  2]
   [0 1 2 1  1]
   [1 -1 0 0 2]
   [2 1 6 0  1]

has rref equal to
 [1 0 2 0  1]
 [0 1 2 0 -1]
 [0 0 0 1  2]
 [0 0 0 0  0]
  x y z w  v
a) Find a basis for null(A).

there are 3 leading 1's. the ambient space is 5-dim'l. so the
solutions are 2-dim'l.

so we need 2 independent solutions. z and v are free.
write down the solutions:
(-v-2z,v-2z,z,-2v,v)=v(-1,1,0,2,1)+z(-2,-2,1,0,0).
there is the basis.

b) Are the first three columns of A linearly independent?

nope. they would need each a leading 1. a relation is 
(3) = 2(1) + 2(2)

12.:

a) Determine a basis of R^3 from among the following four vectors.

 v1=(1,1,0)
 v2=(2,1,2)
 v3=(3,2,2)
 v4=(1,0,1)

make A with columns equal to the vectors above. run rref. pick the
columns in which leading 1's show. they are a basis.

             ->
b) Express   i as a linear combination of the elements in that basis.

13.:

Make a linear system with matrix

A=(1 2 3 4)
  (2 3 4 5)

and write out solutions for AX=0. The use Gram Schmidt to make the
basis orthonormal.

14.:

First get orthonormal basis to W=span( (1 1 1 1)
                                       (2 4 2 6)
                                       (0 2 0 4) ).
Then project v=(1,2,3,4) onto W. Then find v - proj_W(v).

15.: 

A=(3 1) and b=( 2)
  (4 1)       ( 4)
  (7 1)       ( 5)
 (17 1)       ( 10)

Solve A^T*A*X=A^T*b. 

That gives slope and y-intercept of the line.

To get the parabola,

A=(9   3 1) and b=( 2)
  (16  4 1)       ( 4)
  (49  7 1)       ( 5)
 (289 17 1)       ( 10)

proceed as above to get quadratic, linear and constant coefficient.

16.:

17.:

First get orthonormal basis for span(1,x). 
Then project.

18.: 

Get orthonormal basis for W. Then project v into W. Then project v
into W'. These are the 2 components that v splits into.

19.:

Separate variables, use Gram-Schmidt.

20.:

Length.

21.:

Plug in.

22.:

Need (v,B^T*B*v) positive. By the Note, 
(v,B^T*B*v) = v^T*B^T*B*v.
But v^T*B^T*B*v = (B*v)^T * (B*v) = (B*v, B*v)

We know that the only vector that gives inner product with itself
equal to zero is the zerovector, and no vector has inner product with
itself negative. So the only case where (v,B^T*B*v) is not positive is
when B*v=0. 
But B is invertible, so B*v =0 has only one solution, namely
v=0. Hence (v,B^T*B*v) is always positive for nonzero v. Hence B^T*B
is positive definite.

23.:

First get basis for W={(a+b-c,a-c,0,a+b)}. Then make A with rows from
that basis. Then solve A*v=0 for d.