Linear Algebra

Interactive Problem Sheet 3: Linear Space and Basis

- towards the Linear Vector Space concept Basis and Change of Basis
DEAR STUDENT
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Purpose of the Worksheet
The purpose of this Worksheet is to reinforce the understanding of the concepts
spanning vector sets
linear independence and
basis.

Contents

Brief Instructions
I Span
II Linear Independence
III Basis

IV Student Feedback

Brief Instructions

This example applet construction visualizes the standard vector addition +, multiplication by scalars (as the scaling function) and linear combinations of plane vectors.
This page requires a Java capable browser. In these JavaSketchpad applet constructions which we call sketches here) you usually can drag (move in the usual way using the mouse) the basic coloured points, which the constructions depend on.
The red point labeled u in the sketch beside corresponds to a plane vector starting from the origin of the two-dimensional Euclidean plane R2, while the red point c tied to a line, stands for a real scalar (number) on the real line R.
The result vector u+v is the image of the pair (u, v) with respect to the ordinary coordinatewise vector addition and the vector c∗u is the image of the pair (c, u) with respect to the ordinary coordinatewise multiplication by scalars.

A sketch is activated by a mouse click. To restart an active sketch from its original situation press the keybord key 'R'.

I Span of a vector set

In each of the following plane geometry figures you see scalars on the real line and vectors in the plane. You are supposed to solve some problems concerning the span of those vectors
Let us recall the definition of span:
DEFINITION OF SPAN
Let (V, +, ) be a real vector space and U a nonempty subset of V.

1. The span [U] of the set U is the set of all linear combinations of the elements of U, that is:

[U] := {c1u1 + c2u2 + c3u3 + ... + cnun | ci ∈ R, ui ∈ U, n ∈ N}.
2. U is a spanning set of V (and spans it) if every vector of V can be expressed as a linear combination of elements of U.

I Problem 1. In which Figures A-F the vectors span the whole plane V := R2 ?
Maximum 6 x 1 = 6 points.
Hint: To make things more exciting, the origin is not visible in most Figures. It may help if you place the draggable point a to the position where the origin should lay. Moreover, you can use the white segment as a "line" and place it wherever you see advantageous.

This page requires a Java capable browser. This page requires a Java capable browser. This page requires a Java capable browser.
Figure A
Figure B
Figure C
Is there a spanning set for R2 in Figure A? yes no I don't know
Is there a spanning set for R2 in Figure B? yes no I don't know
Is there a spanning set for R2 in Figure C? yes no I don't know
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Figure D
Figure E
Figure F
Is there a spanning set for R2 in Figure D? yes no I don't know
Is there a spanning set for R2 in Figure E? yes no I don't know
Is there a spanning set for R2 in Figure F? yes no I don't know
I Problem 2. In the case(s) above, where the span is not the whole plane R2, tell what the span is like geometrically and what is its dimension as a vector space (recall that the span is always a subspace, thus also a vector space on its own).
Maximum 4 points.
Your answer to I Problem 2:

I Problem 3. What is the span of the two vectors in the following Sketch?
Maximum 4 points.
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HINTS

For measurements, you can freely use the movable white point a and the line PQ, whose coordinates can be seen on the left side of the Sketch.

A sketch is activated by a mouse click.

To restart an active sketch from its original situation press the key 'R'.

Your answer to I Problem 3:

II Linear Independence

Also in this part we deal with linear combinations of vectors in the plane V : = R2, now in the sense of linear independence.

Problem type: Which vector sets of the plane R2 in the forthcoming Sketches are linearly independent?
In case they are linearly dependent, express graphically the null vector as their linear combination and give the values you found for the scalars, all of which are not zero.

You should play with the Sketches and try to solve the problems, but let us first recall the definition:


DEFINITION OF LINEAR INDEPENDENCE
Let (V, +, ) be a real vector space, 0 its null vector (algebraically: neutral element) and U a nonempty subset of V. U is linearly independent if
c1u1 + c2u2 + c3u3 + ... + cnun = 0
ONLY IF ALL the scalars ci are equal to 0.
An alternative way is to say that the null vector 0 cannot be reached by linear combinations of U, unless all scalars in it are zero.

II Problem 4. a) Is the vector set {u, v} in the following Sketch a linearly independent subset of the plane R2?
b) In case they are linearly dependent, express graphically the null vector as their linear combination and give the values you found for the scalars, all of which are not zero.
Maximum 4 points.
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II Problem 4 Sketch

Hints
You only can drag the points (numbers) on the real axis.

A sketch is activated by a mouse click.
To restart an active sketch from its original situation press the key 'R'.

Answer area
Is {u,v} linearly independent? YES NO no idea

If the answer was NO, give your scalar values (not all 0):
c = d =
Your comments:

II Problem 5. a) Is the vector set {u,v} in the following Sketch a linearly independent subset of the plane R2?
b) In case they are linearly dependent, express graphically the null vector as their linear combination and give the values you found for the scalars, all of which are not zero.
Maximum 4 points.
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II Problem 5 Sketch

Hints
You only can drag the points (numbers) on the real axis.

A sketch is activated by a mouse click.
To restart an active sketch from its original situation press the key 'R'.

Answer area
Is {u,v} linearly independent? YES NO no idea

If the answer was NO, give your scalar values (not all 0):
c = d =
Your comments:

II Problem 6. a) Is the vector set {u, v, w} in the following Sketch a linearly independent subset of the plane R2?
b) In case they are linearly dependent, express graphically the null vector as their linear combination and give the values you found for the scalars, all of which are not zero.
Maximum 4 points.
This page requires a Java capable browser.

II Problem 6 Sketch

Hints
You only can drag the points (numbers) on the real axis.

A sketch is activated by a mouse click.
To restart an active sketch from its original situation press the key 'R'.

Answer area
Is {u,v,w} linearly independent? YES NO no idea

If the answer was NO, give your scalar values (not all 0):
c = d = e =
Your comments:

II Problem 7. a) Is the vector set {u,v,w} in the following Sketch a linearly independent subset of the plane R2?
b) In case they are linearly dependent, express graphically the null vector as their linear combination and give the values you found for the scalars, all of which are not zero.
Maximum 4 points.
This page requires a Java capable browser.

II Problem 7 Sketch

Hints
You only can drag the points (numbers) on the real axis.

A sketch is activated by a mouse click.
To restart an active sketch from its original situation press the key 'R'.

Answer area
Is {u,v,w} linearly independent? YES NO no idea

If the answer was NO, give your scalar values (not all 0):
c = d = e =
Your comments:

III Basis of A Linear Space

When we combine the properties of spanning and linear independence, we get an optimal set of vectors, a basis. The importance of basis lies in the property, that every vector of the space can be represented in exactly one way as linear combination of the given basis.

In the following Sketches you have to play with basis and/or coordinates and representations of vectors.
Let us recall the definitions:


DEFINITION OF BASIS AND COORDINATES
Let (V, +, ) be a real vector space. A subset E of V is a basis for V if it is both
a) a spanning set for V, ie. [E] = V, and
b) linearly independent.
If a vector set E = {e1, e2, ... , en} is a basis for V, then we obtain coordinates with respect to this basis:

For every vector u ∈ V there is exactly one sequence of scalars c1, c2, ... , cn, for which

u = c1u1 + c2u2 + c3u3 + ... + cnun.
We call these scalars coordinates of u with respect to the basis E. Sometimes this is denoted by
uE = (c1, c2, ... , cn)T.

III Problem 8. a) Determine whether the vector set E := {e, f} is a basis for the plane or not.
b) If it is, then find out the coordinates of the vectors u and v with respect to the basis E = {e, f}.
Maximum 5 points.
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III Problem 8 Sketch

You only can drag the points (numbers) on the real axis.

A sketch is activated by a mouse click.
To restart an active sketch from its original situation press the key 'R'.

Answer area
Is {e,f} a basis for the plane? YES NO no idea
If the answer was YES, then give the
coordinates of u with respect to the basis {e,f}:
c = d =
coordinates of v with respect to the basis {e,f}:
c = d =
Your comments:

III Problem 9. a) Determine whether the vector set E := {e,f} is a basis for the plane or not.
b) If it is, then find out the coordinates of the vectors u and v with respect to the basis E = {e,f}.
Maximum 5 points.
This page requires a Java capable browser.

III Problem 9 Sketch

You only can drag the points (numbers) on the real axis.

A sketch is activated by a mouse click.
To restart an active sketch from its original situation press the key 'R'.

Answer area
Is {e,f} a basis for the plane? YES NO no idea
If the answer was YES, then give the
coordinates of u with respect to the basis {e,f}:
c = d =
coordinates of v with respect to the basis {e,f}:
c = d =
Your comments:

III Problem 10. a) Determine whether the vector set E := {e,f} is a basis for the plane or not.
b) If it is, then find out the coordinates of the vectors u and v with respect to the basis E = {e,f}.
Maximum 5 points.
This page requires a Java capable browser.

III Problem 10 Sketch

You only can drag the points (numbers) on the real axis.

A sketch is activated by a mouse click.
To restart an active sketch from its original situation press the key 'R'.

Answer area
Is {e,f} a basis for the plane? YES NO no idea
If the answer was YES, then give the
coordinates of u with respect to the basis {e,f}:
c = d =
coordinates of v with respect to the basis {e,f}:
c = d =
Your comments:

III Problem 11. Here we have again a fixed basis E := {e,f}. You have to find the coordinates of some vectors with respect to this and the standard basis.
Follow the instructions beside the Sketch.
Maximum 5 points.
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III Problem 11 Sketch

You can drag the points on the real axis and the vector u.
A sketch is activated by a mouse click.
To restart an active sketch from its original situation press the key 'R'.
Instructions: Write the two coordinates in the same input box.
By moving c and d
a) find the coordinates of e and f with respect to the basis {e,f}:
eE =
fE =
b) find the coordinates of 0 with respect to the basis {e,f}:
0E =

Using also the vector u, whose standard coordinates are shown,
c) find the standard basis coordinates of 2e
2e =
d) find the standard basis coordinates of e + f
e+f =

Your comments:

III Problem 12. Here we have two fixed bases, the standard basis J = {i, j} and another E := {e, f}. We shall try to obtain a formula that translates the standard coordinates to the coordinates with respect to the other basis.
Follow the instructions beside the Sketch.
Maximum 8 points.
You can dragg the points on the real axis and the vector u.
A sketch is activated by a mouse click.
To restart an active sketch from its original situation press the key 'R'.
This page requires a Java capable browser.

III Problem 12 Sketch Instructions

Here you should write each coordinate in its own box, and this time as column vectors.

a) By moving c and d, find the coordinates of i and j with respect to the new basis E = {e,f}. Write them below, in the appropriate column boxes iE and jE.

b) Write the standard coordinates of u (see the upper left corner of the Sketch) of your choice in the boxes columnwise.
Do not change u after that!

c) Find the coordinates of u with respect to the new basis E = {e,f} and write them in the last column.
iE jE u uE
d) Finally, multiply the matrix constituted by the two columns iE and jE by the column vector u.
The same result is (hopingly) seen in the last column uE!
Was the result correct? EXACTLY ALMOST NO

This was an introduction to the process change of basis.

Your comments:

IV Student Feedback

Problem 13.
Extra bonus 2 points.
Here we ask you to tell what do you think about this kind of Problems and working on this kind of worksheet.



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Updated 20.3.2013