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.
\nIf you have not provided an answer to every input gap of a question or part of the question, and you try to submit your answers to the question or part, then you will see the message \"Can not submit answer - check for errors\". In reality your answer has been submitted, but the system is just concerned that you have not submitted an answer to every input gap. For this reason, please ensure that you provide an answer to every input gap in the question or part before submitting. Even if you are unsure of the answer, write down what you think is most likely to be correct; you can always change your answer or retry the question.
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Let $V$ be a vector space, and let $\\{ \\boldsymbol{v_1} , \\boldsymbol{v_2} , \\ldots , \\boldsymbol{v_k} \\} \\in V$.
\ni) $\\{ \\boldsymbol{v_1} , \\boldsymbol{v_2} , \\ldots , \\boldsymbol{v_k} \\}$ is a basis of $V$ if and only if which one of the following conditions if is true:
\n[[0]]
\nii) 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[[1]]
\nPlease press the \"Show feedback\" button once you have completed all the gaps in this part of the question.
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", "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 $\\lambda_1 , \\lambda_2, \\ldots , \\lambda_k = 0$.
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", "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:
\nLet $V$ 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 $\\begin{pmatrix} b_1 \\\\ b_2 \\end{pmatrix}_B$.
\nNow that we have covered these notational remarks, we proceed with the question.
\n\nConsider 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 $\\boldsymbol{f_1} = \\var{{ab}}, \\boldsymbol{f_1} = \\var{{cd}}$.
\ni) Explicitely state what the values of $(\\boldsymbol{e_1})_S$ and $(\\boldsymbol{e_2})_S$ are.
$(\\boldsymbol{e_1})_S =$ [[0]]$_S$
$(\\boldsymbol{e_2})_S =$ [[1]]$_S$
ii) Explicitely state what the values of $(f_1)_S$ and $(f_2)_S$ are.
$(\\boldsymbol{f_1})_S =$[[2]]$_S$
$(\\boldsymbol{f_2})_S =$[[3]]$_S$
ii) Explicitely state what the values of $(f_1)_B$ and $(f_2)_B$ are.
$(\\boldsymbol{f_1})_B =$[[4]]$_B$
$(\\boldsymbol{f_2})_B =$[[5]]$_B$
Continuing from part b, we will explicitely state the values of $(\\boldsymbol{e_1})_B$ and $(\\boldsymbol{e_2})_B$.
\nFirst 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 $(\\boldsymbol{e_1})_B = \\begin{pmatrix} x \\\\ y \\end{pmatrix}_B$.
\nNotice that we can express this system as $A \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}$, where $A=$ [[0]].
\nWe now find the reduced row echeleon form of $(A|\\boldsymbol{e_1})$ so that we can determine $x$ and $y$:
\n$(\\boldsymbol{e_1})_B =\\begin{pmatrix} x \\\\ y \\end{pmatrix}_B =$ [[1]]$_B$.
\nWe do the same as above, but for $\\boldsymbol{e_2}$ instead of $\\boldsymbol{e_1}$, and we get
\n$(\\boldsymbol{e_2})_B =$ [[2]]$_B$.
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\nConsider the subspace $V$ of $\\mathbb{R}^3$ defined by $V = \\Bigg\\{ \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\bigg| \\simplify{{f}x+{g}y+{h}z}=0 \\Bigg\\}$.
\nIn order to find an orthonormal basis, we will first find a basis and then apply the Gram-Schmidt process to orthonormalise it.
\ni) 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
$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} =$ [[0]]$s+$ [[1]]$t$.
Hence, we can see that $\\boldsymbol{b_1} :=$ [[0]] and $\\boldsymbol{b_2} :=$ [[1]] form a basis for $V$.
\nii) 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
$\\| \\boldsymbol{b_1} \\| =$ [[2]]
(If you wish to write something like $\\sqrt{2}$ then please write (2)^(1/2). )
Hence, our first vector in our orthonormal basis is $ \\boldsymbol{u_1} := \\frac{1}{\\| \\boldsymbol{b_1} \\|} \\boldsymbol{b_1} = \\Big(1 \\Big/ \\big($ [[2]]$\\big) \\Big)$ [[0]] .
\niii) 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, $\\boldsymbol{w_2} =$ [[3]] .
\nFinally, we need to normalise $\\boldsymbol{w_2} =$ [[3]]:
$\\| \\boldsymbol{w_1} \\| =$ [[4]].
Hence, our second vector in our orthonormal basis is $ \\boldsymbol{u_2} := \\frac{1}{\\| \\boldsymbol{w_2} \\|}\\boldsymbol{w_2} = \\Big(1 \\Big/ \\big($ [[4]] $\\big) \\Big)$ [[3]].
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.
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For all $\\boldsymbol{g_1} , \\boldsymbol{g_2} \\in V$, $\\langle \\boldsymbol{g_1} , \\boldsymbol{g_2} \\rangle := \\int_{-2}^{2} \\boldsymbol{g_1} (x) \\boldsymbol{g_2} (x) \\mathrm{d}x$.
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.
\ni) First, we calculate $\\| \\boldsymbol{f_1} \\| = \\Big(\\int_{-2}^{2} \\boldsymbol{f_1} (x) \\boldsymbol{f_1} (x) \\mathrm{d} x \\Big)^{\\frac{1}{2}} =$ [[0]].
\nHence, our first vector in our orthonormal basis is $\\boldsymbol{u_1}$ defined by $\\boldsymbol{u_1} (x) = \\frac{1}{\\| \\boldsymbol{f_1} \\|} \\boldsymbol{f_1} (x) = \\Big( 1 \\Big/$ [[0]]$\\Big)$, for all $x \\in [-2,2]$.
\nii) 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 $\\boldsymbol{w_2} = \\boldsymbol{f_2} - \\langle \\boldsymbol{f_2} , \\boldsymbol{u_1} \\rangle \\boldsymbol{u_1}$.
\nNow, $\\langle \\boldsymbol{f_2} , \\boldsymbol{u_1} \\rangle = \\int_{-2}^{2} \\boldsymbol{f_2} (x) \\boldsymbol{u_1} (x) \\mathrm{d} x = \\int_{-2}^{2} \\frac{x+1}{2} \\mathrm{d} x =$ [[1]].
\nHence, $\\boldsymbol{w_2}$ is defined by $\\boldsymbol{w_2} (x) =$ [[2]], for all $x \\in [-2,2]$.
\niii) 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) ).
$\\| \\boldsymbol{w_2} \\|=$ [[3]].
Hence, our second vector in our orthonormal basis is $\\boldsymbol{u_2}$ defined by $\\boldsymbol{u_1} (x) = \\Big(1 \\Big/$ [[3]]$\\Big)$ [[2]], for all $x \\in [-2,2]$.
\nWe have now completed the Gram-Schmidt process. You can confirm yourself that $\\boldsymbol{u_1}$ and $\\boldsymbol{u_2}$ are orthogonal to each other.
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