// Numbas version: exam_results_page_options {"name": "MA100 MT Week 10", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "MA100 MT Week 10", "statement": "

Week 10 (lectures 19 and 20): In this question, you will look at bases of vector spaces, the dimension of a vector space, orthonormal bases, and the Gram-Schmidt process.

\n

\n

\n

As with all questions, there may be parts where you can choose to \"Show steps\". This may give a hint, or it may present sub-parts which will help you to solve that part of the question. Furthermore, remember to always press the \"Show feedback\" button at the end of each part. Sometimes, helpful feedback will be given here, and often it will depend on how correctly you have answered and will link to other parts of the question. Hence, always retry the parts until you obtain full marks, and then look at the feedback again.

Keep in mind that in order to see the feedback for a particular part of a question, you must provide a full (but not necessarily correct) answer to that part. Do not worry though, as you can look at the feedback and then ammend your answer accordingly.

Furthermore, as with all questions, choosing to reveal the answers will only show you the answers which change every time the question is loaded (i.e. answers to randomised questions); the fixed answers will not be revealed.

", "parts": [{"type": "gapfill", "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "sortAnswers": false, "variableReplacements": [], "unitTests": [], "showFeedbackIcon": true, "scripts": {"mark": {"order": "after", "script": "this.markingFeedback = [];\n//This clears the feedback that would normally be given automatically by the built in script. We just want to show our own feedback.\n\n//For the code below: \"this.credit tells us the proportion of the marks that the student got for this part of the question. For example, if the student scored 1 mark out of three, then \"this.credit\" would equal to 1/3.\nif (this.credit < 1){\n this.markingComment('Not quite. Try again and have a look at this feedback section once each gap has been answered correctly as there will be some useful feedback.');\n}else if (this.credit == 1){\n this.markingComment('Good! Feedback for part i: A basis must be both linearly independent and span the vector space, like you said. One of the other two options simply described linear independence only, and the other simply described spanning of $V$ only. Any one of these conditions is a necessary condition for a basis, but it is not enough when taken alone: You need both for a sufficient condition. Feedback for part ii: As stated in lecture 19, if a vector space has a finite basis of size $k$, then all bases have size $k$. Hence, you only need to find one basis to determine the dimension of the vector space. In any case, vector spaces generally have an infinite number of bases, so it would be difficult to write them all down!');\n}"}}, "prompt": "

Let $V$ be a vector space, and let $\\{ \\boldsymbol{v_1} , \\boldsymbol{v_2} , \\ldots , \\boldsymbol{v_k} \\} \\in V$.

\n

i) $\\{ \\boldsymbol{v_1} , \\boldsymbol{v_2} , \\ldots , \\boldsymbol{v_k} \\}$ is a basis of $V$ if and only if which one of the following conditions if is true:

\n

[]

\n

ii) Now suppose $V$ is vector space with a basis of a finite size. Which one of the following best describes how we determine the dimension of $V$?

\n

[]

\n

Please press the \"Show feedback\" button once you have completed all the gaps in this part of the question.

", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "marks": 0, "gaps": [{"type": "1_n_2", "showCorrectAnswer": false, "variableReplacements": [], "choices": ["

$\\{ \\boldsymbol{v_1} , \\boldsymbol{v_2} , \\ldots , \\boldsymbol{v_k} \\}$ is a linearly independent set, and it spans $V$.

", "

Any $v \\in V$ can be written as a linear combination of the elements in $\\{ \\boldsymbol{v_1} , \\boldsymbol{v_2} , \\ldots , \\boldsymbol{v_k} \\}$.

", "

The only solution to $\\lambda_1 \\boldsymbol{v_1} + \\lambda_2 \\boldsymbol{v_2} + \\ldots + \\lambda_k \\boldsymbol{v_k} = 0$ is the trivial solution $\\lambda_1 , \\lambda_2, \\ldots , \\lambda_k = 0$.

"], "shuffleChoices": true, "distractors": ["", "", ""], "minMarks": 0, "maxMarks": 0, "displayType": "radiogroup", "unitTests": [], "showFeedbackIcon": true, "scripts": {}, "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "displayColumns": "1", "marks": 0, "matrix": ["1", 0, 0], "extendBaseMarkingAlgorithm": true}, {"type": "1_n_2", "showCorrectAnswer": false, "variableReplacements": [], "choices": ["

We find any basis of $V$. Then, the dimension of $V$ is the number of elements in this basis.

", "

We find all bases of $V$. Then we find the one with the smallest number of elements. Then, the dimension of $V$ is the number of elements in this smallest basis.

", "

We find all bases of $V$. Then we find the one with the largest number of elements. Then, the dimension of $V$ is the number of elements inthis largest basis.

"], "shuffleChoices": true, "distractors": ["", "", ""], "minMarks": 0, "maxMarks": 0, "displayType": "radiogroup", "unitTests": [], "showFeedbackIcon": true, "scripts": {}, "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "displayColumns": "1", "marks": 0, "matrix": ["1", 0, 0], "extendBaseMarkingAlgorithm": true}]}, {"type": "gapfill", "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "sortAnswers": false, "variableReplacements": [], "unitTests": [], "showFeedbackIcon": true, "scripts": {}, "prompt": "

Before proceeding with this part of the question, we recall the following notational remarks:

\n

Let $V$ be a real vector space with basis $B = \\{ \\boldsymbol{v_1} , \\boldsymbol{v_2} \\}$. Also, let $a_1 ,a_2 \\in \\mathbb{R}$. We define $\\begin{pmatrix} a_1 \\\\ a_2 \\end{pmatrix}_B$ by $\\begin{pmatrix} a_1 \\\\ a_2 \\end{pmatrix}_B = a_1 \\boldsymbol{v_1} + a_2 \\boldsymbol{v_2}$. That is, $\\begin{pmatrix} a_1 \\\\ a_2 \\end{pmatrix}_B$ is a vector, but it is expressed in terms of the elements of the basis $B$. Now let $\\boldsymbol{v} \\in V$. Notice that we have yet to define $\\boldsymbol{v}$ in terms of a basis. We use the notation $(\\boldsymbol{v})_B$ to represent the coordinates of $\\boldsymbol{v}$ with respect to $B$. That is, if $\\boldsymbol{v} = b_1 \\boldsymbol{v_1} + b_2 \\boldsymbol{v_2}$ for some $b_1 , b_2 \\in \\mathbb{R}$, then $(\\boldsymbol{v})_B = \\begin{pmatrix} b_1 \\\\ b_2 \\end{pmatrix}_B$. Due to the current limitations of this system, we may occasionally need to write, for example, $\\begin{pmatrix} b_1 \\\\ b_2 \\end{pmatrix} {}_B$ instead of $\\begin{pmatrix} b_1 \\\\ b_2 \\end{pmatrix}_B$.

\n

Now that we have covered these notational remarks, we proceed with the question.

\n

\n

Consider the vector space $\\mathbb{R}^2$ and the bases $S = \\{ \\boldsymbol{e_1} , \\boldsymbol{e_1} \\}$ and $B = \\{ \\boldsymbol{f_1} , \\boldsymbol{f_2} \\}$, where $\\boldsymbol{e_1} = \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} , \\boldsymbol{e_1} = \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix}$ and $\\boldsymbol{f_1} = \\var{{ab}}, \\boldsymbol{f_1} = \\var{{cd}}$.

\n

i) Explicitely state what the values of $(\\boldsymbol{e_1})_S$ and $(\\boldsymbol{e_2})_S$ are.
$(\\boldsymbol{e_1})_S =$ []$_S$
$(\\boldsymbol{e_2})_S =$ []$_S$

\n

ii) Explicitely state what the values of $(f_1)_S$ and $(f_2)_S$ are.
$(\\boldsymbol{f_1})_S =$[]$_S$
$(\\boldsymbol{f_2})_S =$[]$_S$

ii) Explicitely state what the values of $(f_1)_B$ and $(f_2)_B$ are.
$(\\boldsymbol{f_1})_B =$[]$_B$
$(\\boldsymbol{f_2})_B =$[]$_B$

Continuing from part b, we will explicitely state the values of $(\\boldsymbol{e_1})_B$ and $(\\boldsymbol{e_2})_B$.

\n

First let us look at $(\\boldsymbol{e_1})_B$. We need to find $x,y \\in \\mathbb{R}$ such that $\\boldsymbol{e_1} = x \\boldsymbol{f_1} + y \\boldsymbol{f_2}$. That is, $\\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}= x \\var{{ab}} + y \\var{{cd}}$. Then, we would have that $(\\boldsymbol{e_1})_B = \\begin{pmatrix} x \\\\ y \\end{pmatrix}_B$.

\n

Notice that we can express this system as $A \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}$, where $A=$ [].

\n

We now find the reduced row echeleon form of $(A|\\boldsymbol{e_1})$ so that we can determine $x$ and $y$:

\n

$(\\boldsymbol{e_1})_B =\\begin{pmatrix} x \\\\ y \\end{pmatrix}_B =$ []$_B$.

\n

We do the same as above, but for $\\boldsymbol{e_2}$ instead of $\\boldsymbol{e_1}$, and we get

\n

$(\\boldsymbol{e_2})_B =$ []$_B$.

\n

In this part of the question, we will find an orthonormal basis of a subspace of $\\mathbb{R}^3$. We define the inner product of two vectors $\\begin{pmatrix} x_1 \\\\ y_2 \\\\ z_2 \\end{pmatrix} , \\begin{pmatrix} x_2 \\\\ y_2 \\\\ z_2 \\end{pmatrix}$ by $\\Bigg\\langle \\begin{pmatrix} x_1 \\\\ y_2 \\\\ z_2 \\end{pmatrix} , \\begin{pmatrix} x_2 \\\\ y_2 \\\\ z_2 \\end{pmatrix} \\Bigg\\rangle = x_1 x_2 + y_1 y_2 + z_1 z_2$. It is with respect to this inner product that we discuss orthonormality.

\n

Consider the subspace $V$ of $\\mathbb{R}^3$ defined by $V = \\Bigg\\{ \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\bigg| \\simplify{{f}x+{g}y+{h}z}=0 \\Bigg\\}$.

\n

In order to find an orthonormal basis, we will first find a basis and then apply the Gram-Schmidt process to orthonormalise it.

\n

i) By definition, $V$ has one equation in its cartesian description: $\\simplify{{f}x+{g}y+{h}z}=0$. By letting $y=s$ and $z=t$, we can see that the general solution to this cartesian equation is
$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} =$ []$s+$ []$t$.

\n

Hence, we can see that $\\boldsymbol{b_1} :=$ [] and $\\boldsymbol{b_2} :=$ [] form a basis for $V$.

\n

ii) Now we normalise $\\boldsymbol{b_1}$. Recalling that the norm of a general vector $\\begin{pmatrix} a \\\\ b \\\\ c \\end{pmatrix}$ is $\\sqrt{a^2 + b^2 + c^2}$, we can see that
$\\| \\boldsymbol{b_1} \\| =$ []
(If you wish to write something like $\\sqrt{2}$ then please write (2)^(1/2). )

\n

Hence, our first vector in our orthonormal basis is $\\boldsymbol{u_1} := \\frac{1}{\\| \\boldsymbol{b_1} \\|} \\boldsymbol{b_1} = \\Big(1 \\Big/ \\big($ []$\\big) \\Big)$ [] .

\n

iii) Now we want to use $\\boldsymbol{b_2}$ to obtain a vector $\\boldsymbol{w_2}$ which is orthogonal to $\\boldsymbol{u_1}$. The Gram-Schmidt process tells us that this can be calculated explicitely by $\\boldsymbol{w_2} = \\boldsymbol{b_2} - \\langle \\boldsymbol{b_2} , \\boldsymbol{u_1} \\rangle \\boldsymbol{u_1}$. Hence, $\\boldsymbol{w_2} =$ [] .

\n

Finally, we need to normalise $\\boldsymbol{w_2} =$ []:
$\\| \\boldsymbol{w_1} \\| =$ [].
Hence, our second vector in our orthonormal basis is $\\boldsymbol{u_2} := \\frac{1}{\\| \\boldsymbol{w_2} \\|}\\boldsymbol{w_2} = \\Big(1 \\Big/ \\big($ [] $\\big) \\Big)$ [].

\n

We have now completed the Gram-Schmidt process. You can confirm yourself that $\\boldsymbol{u_1}$ and $\\boldsymbol{u_2}$ are orthogonal to each other.

In this part of the question we will do something similar to part d, but instead we will be working with the more abstract vector space $V$ defined by $V = \\{ f: [-2,2] \\longrightarrow \\mathbb{R} | \\text{ there exist } a,b \\in \\mathbb{R} \\text{ such that for all } x \\in [-2,2] \\text{ we have } f(x) = ax+b \\}$. This vector space just consist of functions with domain $[-2,2]$ and of the form $f(x)=ax+b$, for some $a,b \\in \\mathbb{R}$. We will refer to elements of $V$ as vectors. We can define the following inner product on $V$:
For all $\\boldsymbol{g_1} , \\boldsymbol{g_2} \\in V$, $\\langle \\boldsymbol{g_1} , \\boldsymbol{g_2} \\rangle := \\int_{-2}^{2} \\boldsymbol{g_1} (x) \\boldsymbol{g_2} (x) \\mathrm{d}x$.

\n

Now, it can be shown that $\\boldsymbol{f_1} , \\boldsymbol{f_2} \\in V$ defined by $\\boldsymbol{f_1} (x) = 1$ for all $x \\in [-2,2]$, and $\\boldsymbol{f_2} (x) = x+1$ for all $x \\in [-2,2]$, form a basis for $V$. We wish to apply the Gram-Schmidt process to this basis to obtain an orthonormal basis.

\n

i) First, we calculate $\\| \\boldsymbol{f_1} \\| = \\Big(\\int_{-2}^{2} \\boldsymbol{f_1} (x) \\boldsymbol{f_1} (x) \\mathrm{d} x \\Big)^{\\frac{1}{2}} =$ [].

\n

Hence, our first vector in our orthonormal basis is $\\boldsymbol{u_1}$ defined by $\\boldsymbol{u_1} (x) = \\frac{1}{\\| \\boldsymbol{f_1} \\|} \\boldsymbol{f_1} (x) = \\Big( 1 \\Big/$ []$\\Big)$, for all $x \\in [-2,2]$.

\n

ii) Now we want to use $\\boldsymbol{f_2}$ to obtain a vector $\\boldsymbol{w_2}$ which is orthogonal to $\\boldsymbol{u_1}$.The Gram-Schmidt process tells us that this can be calculated explicitely by $\\boldsymbol{w_2} = \\boldsymbol{f_2} - \\langle \\boldsymbol{f_2} , \\boldsymbol{u_1} \\rangle \\boldsymbol{u_1}$.

\n

Now, $\\langle \\boldsymbol{f_2} , \\boldsymbol{u_1} \\rangle = \\int_{-2}^{2} \\boldsymbol{f_2} (x) \\boldsymbol{u_1} (x) \\mathrm{d} x = \\int_{-2}^{2} \\frac{x+1}{2} \\mathrm{d} x =$ [].

\n

Hence, $\\boldsymbol{w_2}$ is defined by $\\boldsymbol{w_2} (x) =$ [], for all $x \\in [-2,2]$.

\n

iii) Finally, we need to normalise $\\boldsymbol{w_2}$:
(If you wish to write, for example, something like $\\frac{2}{\\sqrt{2}}$, then please write 2/sqrt(2) ).
$\\| \\boldsymbol{w_2} \\|=$ [].

\n

Hence, our second vector in our orthonormal basis is $\\boldsymbol{u_2}$ defined by $\\boldsymbol{u_1} (x) = \\Big(1 \\Big/$ []$\\Big)$ [], for all $x \\in [-2,2]$.

\n

We have now completed the Gram-Schmidt process. You can confirm yourself that $\\boldsymbol{u_1}$ and $\\boldsymbol{u_2}$ are orthogonal to each other.

This is the question for week 10 of the MA100 course at the LSE. It looks at material from chapters 19 and 20.

"}, "rulesets": {}, "variablesTest": {"maxRuns": 100, "condition": ""}, "functions": {}, "tags": [], "variables": {"g": {"definition": "random(-5..5 except(0))", "templateType": "anything", "name": "g", "description": "

This will be used in the cartesian description of the subspace in part d.

", "group": "Variables for part d"}, "cd": {"definition": "matrix([c],[d])", "templateType": "anything", "name": "cd", "description": "

This is the vector with entries c and d.

", "group": "Variables for parts b and c"}, "e1B": {"definition": "matrix([1/a +(b*c)/(a*(a*d - b*c))],[-b/(a*d - b*c)])", "templateType": "anything", "name": "e1B", "description": "

This is the vector e1 = (1,0)^T in terms of the the basis {(a,b)^T , (c,d)^T}

", "group": "Variables for parts b and c"}, "t": {"definition": "random(-3..3 except(0))", "templateType": "anything", "name": "t", "description": "

This is used in the definition of c and d.

", "group": "Variables for parts b and c"}, "b": {"definition": "random(-5..5 except(0) except(a) except(-a))", "templateType": "anything", "name": "b", "description": "

The variables a and b will form our first basis vector of R^2 in part b. It is necessary that b does not equal a or -a so that the vector (c,d)^T is linearly independent to (a,b)^T.

", "group": "Variables for parts b and c"}, "d": {"definition": "b+t*a", "templateType": "anything", "name": "d", "description": "

The vector (c,d)^T is defined by (c,d)^T = (a,b)^T + t (-b,a)^T . By definitions of a,b,t and the fact that (-b,a)^T is normal to (a,b)^T, we have that (a,b)^and (c,d)^T form a basis of R62

", "group": "Variables for parts b and c"}, "a": {"definition": "random(-5..5 except(0))", "templateType": "anything", "name": "a", "description": "

The variables a and b will form our first basis vector of R^2 in part b.

", "group": "Variables for parts b and c"}, "e2B": {"definition": "matrix([-c/(a*d - b*c)],[a/(a*d - b*c)])", "templateType": "anything", "name": "e2B", "description": "

This is the vector e2 = (0,1)^T in terms of the the basis {(a,b)^T , (c,d)^T}

", "group": "Variables for parts b and c"}, "f": {"definition": "random(-5..5 except(0))", "templateType": "anything", "name": "f", "description": "

This will be used in the cartesian description of the subspace in part d.

", "group": "Variables for part d"}, "h": {"definition": "random(-5..5 except(0))", "templateType": "anything", "name": "h", "description": "

This will be used in the cartesian description of the subspace in part d.

", "group": "Variables for part d"}, "c": {"definition": "a-t*b", "templateType": "anything", "name": "c", "description": "

The vector (c,d)^T is defined by (c,d)^T = (a,b)^T + t (-b,a)^T . By definitions of a,b,t and the fact that (-b,a)^T is normal to (a,b)^T, we have that (a,b)^and (c,d)^T form a basis of R62

", "group": "Variables for parts b and c"}, "ab": {"definition": "matrix([a],[b])", "templateType": "anything", "name": "ab", "description": "

This is the vector with entries a and b.

", "group": "Variables for parts b and c"}}, "variable_groups": [{"variables": ["a", "b", "t", "c", "d", "ab", "cd", "e1B", "e2B"], "name": "Variables for parts b and c"}, {"variables": ["f", "g", "h"], "name": "Variables for part d"}], "advice": "", "extensions": [], "type": "question", "contributors": [{"name": "Michael Yiasemides", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1440/"}]}]}], "contributors": [{"name": "Michael Yiasemides", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1440/"}]}