Midterm 2 Solutions for Matrix Methods (APPM 3310) - Part 1, Exams of Mathematics

Solutions to midterm 2 questions for the matrix methods course (appm 3310). It includes finding the row space, column space, null space, and left null space of a matrix, determining when a set of vectors spans a certain dimension, and describing linear transformations of r2 performed by matrices. It also includes problems from the previous midterm revisited, such as proving that the column space of a symmetric matrix is orthogonal to its null space and finding the determinant of a skew-symmetric matrix.

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2012/2013

Uploaded on 02/23/2013

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Midterm 2 Solutions APPM 3310: Matrix Methods Page 1
Midterm 2 Solutions
1. Consider A=
1 5 0 3
2 10 1 4
15 1 5
. Use this definition of Afor all parts of this problem.
Give a basis for (a) the row space of A, (b) the column space of A, (c) the null space of A, (d)
the left null space of A.
Row reduction gives AU=
1503
0 0 1 2
0 0 0 0
.A basis for the row space is the first two
nonzero rows of U(or A), that is,
1
5
0
3
,
0
0
1
2
. For the column space, use the pivot columns
(the first and third columns) of A:
1
2
1
,
0
1
1
. (Note: zero points for using the first and
third columns of U.) For the null space, the columns 2 and 4 have free variables, leading to a basis
set
5
1
0
0
,
3
0
2
1
. To find the left null space, row reduce AT
1 2 1
0 1 1
0 0 0
0 0 0
.The third
column corresponds to a free variable, so the basis vector is
3
1
1
.
(e) What is the rank of A? (f) What is the rank of AT?
Ahas two pivots, so its rank = 2. Similarly, AThas two pivots, so its rank is 2. The purpose of
this question was to help you find errors: you should remember that rank A= rank ATalways! So
if you made an arithmetic error and claimed the ranks were different, you did not receive full credit.
2. You are given 4 vectors, x1,x2,x3,x4R4. This set spans a d-dimensional subspace of R4.
(a) When, if ever, is d < 1?
When all 4 vectors are the zero vector; when the matrix A= [x1x2x3x4] is the zero matrix.
(b) When, if ever, is d= 1?
When all 4 vectors point along the same line; when all 4 vectors are multiples of each other; when
the set has only one linearly independent vector; when the matrix A= [x1x2x3x4] is rank 1 (one
pivot).
(c) When, if ever, is d= 2?
When all 4 vectors lie in a same plane; when the set has only two linearly independent vectors;
when the matrix A= [x1x2x3x4] is rank 2 (two pivots).
(d) When, if ever, is d= 3?
When the set has only three linearly independent vectors; when the matrix A= [x1x2x3x4] is
rank 3 (three pivots).
(e) When, if ever, is d= 4?
When the vectors are linearly independent; when the matrix A= [x1x2x3x4] is full rank (four
pivots).
(f) When, if ever, is d > 4?
Never: it is impossible for vectors R4to span a space with dimension >4.
pf3

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Midterm 2 Solutions

  1. Consider A =

 

 . Use this definition of A for all parts of this problem.

Give a basis for (a) the row space of A, (b) the column space of A, (c) the null space of A, (d) the left null space of A.

Row reduction gives A ∼ U =

  

  .^ A basis for the row space is the first two

nonzero rows of U (or A), that is,

    

   

   

   

   

    

. For the column space, use the pivot columns

(the first and third columns) of A:

  

  

   ,

  

  

  .^ (Note: zero points for using the first and

third columns of U .) For the null space, the columns 2 and 4 have free variables, leading to a basis

set

    

   

   

   

   

    

. To find the left null space, row reduce AT^ ∼

   

   

. The third

column corresponds to a free variable, so the basis vector is

  

  

  

  .

(e) What is the rank of A? (f) What is the rank of AT^? A has two pivots, so its rank = 2. Similarly, AT^ has two pivots, so its rank is 2. The purpose of this question was to help you find errors: you should remember that rank A = rank AT^ always! So if you made an arithmetic error and claimed the ranks were different, you did not receive full credit.

  1. You are given 4 vectors, x 1 , x 2 , x 3 , x 4 ∈ R^4. This set spans a d-dimensional subspace of R^4. (a) When, if ever, is d < 1? When all 4 vectors are the zero vector; when the matrix A = [x 1 x 2 x 3 x 4 ] is the zero matrix. (b) When, if ever, is d = 1? When all 4 vectors point along the same line; when all 4 vectors are multiples of each other; when the set has only one linearly independent vector; when the matrix A = [x 1 x 2 x 3 x 4 ] is rank 1 (one pivot). (c) When, if ever, is d = 2? When all 4 vectors lie in a same plane; when the set has only two linearly independent vectors; when the matrix A = [x 1 x 2 x 3 x 4 ] is rank 2 (two pivots). (d) When, if ever, is d = 3? When the set has only three linearly independent vectors; when the matrix A = [x 1 x 2 x 3 x 4 ] is rank 3 (three pivots). (e) When, if ever, is d = 4? When the vectors are linearly independent; when the matrix A = [x 1 x 2 x 3 x 4 ] is full rank (four pivots). (f) When, if ever, is d > 4? Never: it is impossible for vectors ∈ R^4 to span a space with dimension > 4.

(g) In which case is the set {x 1 , x 2 , x 3 , x 4 } a basis, and for what space? In case (e), when the vectors are linearly independent, they span R^4 and are thus a basis for R^4.

  1. Describe the linear transformations of R^2 performed by the following matrices. (By “describe” I mean explain how the space is geometrically transformed. A sketch is recommended.)

(a) A =

[ 1 0 0 − 1

] (b) B =

[ 0 1 − 1 0

] (c) C =

[ 1 3 0 1

]

The transformation A reflects about the x axis. The transformation B rotates clockwise by 90 degrees. The transformation C shears in the x-direction. (Note: many people had difficulty recognizing the shear transformation, and wrote things like the transformation “rotates and stretches the y-axis” or “stretches the x-axis.” I was lenient in grading wording like this if the sketch was correct, but make sure you understand what shear means for the final.) (d) Describe the composite linear transformations performed by AB and by BA. Are they the same? Why or why not?

First, AB =

[ 0 1 1 0

]

. This transformation first rotates clockwise by 90 degrees, then reflects

through the x-axis. The result, AB, is a reflection through the line y = x. But BA =

[ 0 − 1 − 1 0

] .

When the order is reversed (first reflect, then rotate), the result is a reflection through the origin—you switch the x and y coordinates, then make both negative. Clearly AB 6 = BA, which demonstrates that reflections and rotations do not commute with each other. (Note: zero points if you got the order of the transformations wrong in the composition.)

  1. Problems from the previous midterm, revisited. (a) Prove that if a matrix is symmetric, its column space is orthogonal to its null space. First proof: the fundamental theorem of linear algebra tells us that the row space of a matrix is orthogonal to its null space. For a symmetric matrix, the row space and column space are the same. Therefore the column space is also orthogonal to the null space. Second proof: A vector x in the null space satisfies Ax = 0 , while a vector b in the column space can be written Ay = b. Take the inner product

bT^ x = (Ay)T^ x, = yT^ AT^ x, = yT^ Ax, = yT^0 , = 0.

Where we have used the fact that AT^ = A for a symmetric matrix and that Ax = 0.

(b) If you know the determinant of an n × n matrix A, what is det(−A)? First (more elegant) proof: We can write

| − A| = | − IA|, = | − I||A|.

But | − I| = (−1)n, because the determinant of a diagonal matrix is the product of the diagonal elements, and −I has n diagonal elements, each = −1. Therefore | − A| = (−1)n|A|.