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The answer is either 'True' or 'False'

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Student decides on four different examples whether two subspaces form a direct sum and whether the sum is the whole vectorspace.

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No randomisation, just the four examples. The question is set in explore mode, so that after deciding, students are asked to give reasons for their choices.

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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

For each of the following subspaces, determine whether the sum \\(U + W\\) is a direct sum, and whether the sum gives the whole vector space \\(V\\) (i.e. whether \\(U + W = V\\) ). The parts will talk you through some of the steps.

", "advice": "

a) Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\x \\end{pmatrix} \\middle| x \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} -y\\\\y \\end{pmatrix} \\middle| y \\in \\mathbb{R} \\right\\} \\) be subspaces of \\(\\mathbb{R}^2\\).

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We can see that if \\(v\\in U\\cap W\\), then we have to have \\(v=\\begin{pmatrix} x\\\\x \\end{pmatrix}=\\begin{pmatrix} -y\\\\y \\end{pmatrix}\\) for some \\(x\\) and \\(y\\). This means we need \\(x=-y\\) and \\(x=y\\), so the only option is \\(x=y=0\\). So \\(v=0\\) is the only vector in the intersection. So this is a direct sum.

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We can get any general vector \\(\\begin{pmatrix} a\\\\b\\end{pmatrix} \\in \\mathbb{R}^2\\) in this sum: \\(\\begin{pmatrix} a\\\\b\\end{pmatrix}=\\begin{pmatrix} \\frac{b+a}{2}\\\\\\frac{b+a}{2}\\end{pmatrix} +\\begin{pmatrix} -\\frac{b-a}{2}\\\\\\frac{b-a}{2}\\end{pmatrix}\\). I.e. \\(x=\\frac{b+a}{2}\\) and \\(y=\\frac{b-a}{2}\\). There is only one option here.

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b) Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\x\\\\-x \\end{pmatrix} \\middle| x \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} y\\\\z\\\\-z \\end{pmatrix} \\middle| y,z \\in \\mathbb{R} \\right\\} \\) be subspaces of \\( \\mathbb{R}^3 \\).

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Here we can see that there is some overlap: for example, the vector \\(\\begin{pmatrix}1\\\\1\\\\-1\\end{pmatrix}\\) is in \\(U\\cap W\\). In fact, \\(U\\subseteq W\\): any vector in \\(U\\) is also in \\(W\\). For any given \\(x\\), we just have to choose \\(y=x\\) and \\(z=x\\). So this is not a direct sum.

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It also does not give the whole space, as the second and third entry are very linked. So we cannot get the vector \\(\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}\\) or the vector \\(\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}\\) in the sum. (Or any vector where the third entry is different to minus the second entry.) In fact, since \\(U\\subseteq W\\), we get \\(U+W=W\\).

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c) Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\y\\\\-y \\end{pmatrix} \\middle| x, y \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} 0\\\\z\\\\-z \\end{pmatrix} \\middle| z \\in \\mathbb{R} \\right\\} \\) be subspaces of \\( \\mathbb{R}^3 \\).

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Again we see some overlap: \\( \\begin{pmatrix} 0\\\\z\\\\-z \\end{pmatrix} \\in U\\cap W\\) for any \\(z\\): we just put \\(x=0\\) and \\(y=z\\). So \\(W\\subseteq U\\). So it is not a direct sum.

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Therefore \\(U+W=U\\), which is not all of \\(\\mathbb{R}^3\\). As in b), we cannot get the vector \\(\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}\\) or the vector \\(\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}\\) in the sum.

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d) Let \\( P_2 \\) be the vectorspace of polynomials of degree at most \\( 2 \\) and consider the subspaces \\( U = \\{ p=a_2 X^2 + a_1 X \\vert a_1,a_2 \\in \\mathbb{R}\\} \\) and \\( W = P_1= \\{ q=b_1 X + b_0 \\vert b_0,b_1 \\in \\mathbb{R}\\}\\).

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There is an overlap here: the polynomial \\(X\\) is in \\(U\\cap W\\), because \\(X=0\\cdot X^2+1\\cdot X=1\\cdot X+0\\). So of course any \\(dX\\in U\\cap W\\). So this is not a direct sum. But we do not have one of the two spaces contained in the other: no polynomials in \\(U\\) have any non-zero constant terms, and no polynomials in \\(W\\) have any non-zero \\(X^2\\) term.

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However, we do get the whole space as the sum:

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\\(c_2X^2+c_1X+c_0= (c_2X^2+c_1X)  + (0X+c_0)\\), or \\(c_2X^2+c_1X+c_0= (c_2X^2+0X)  + (c_1X+c_0)\\), or \\(c_2X^2+c_1X+c_0= (c_2X^2+0.4c_1X)  + (0.6c_1X+c_0)\\), or \\(c_2X^2+c_1X+c_0= (c_2X^2+(c_1+14)X)  + ((c_1-14)X+c_0)\\), etc. As long as \\(a_1+b_1=c_1\\),  and \\(a_2=c_2\\), \\(b_0=c_0\\), we can make this sum. So you see there is some redundancy coming from the overlap.

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e) Let \\( P_2 \\) be the vectorspace of polynomials of degree at most \\( 2 \\) and consider the subspaces \\( U = \\{ p=a_2 X^2 \\vert a_2 \\in \\mathbb{R}\\} \\) and \\( W = P_0=\\{ q=b_0 \\vert b_0 \\in \\mathbb{R}\\} \\).

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We see here that there is no overlap: if \\(p\\in U\\cap W\\), then \\(p=a_2X^2 + 0X+0=0X^2+0X+b_0\\), so \\(a_2=0\\) and \\(b_0=0\\), so \\(0\\) is the only element in the intersection. So this is a direct sum.

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The sum does not give all of \\(P_2\\): we cannot get any non-zero \\(X\\) term. So for example \\(X\\notin U+W\\), or any \\(c_2X^2+c_1X+c_0\\) with \\(c_1\\neq 0\\).

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Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\x \\end{pmatrix} \\middle| x \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} -y\\\\y \\end{pmatrix} \\middle| y \\in \\mathbb{R} \\right\\} \\) be subspaces of \\( \\mathbb{R}^2 \\).

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Is \\(U + W \\) a direct sum?

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[[0]]

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Is \\(U + W = \\mathbb{R}^2\\) ?

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[[1]]

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Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\x \\end{pmatrix} \\middle| x \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} -y\\\\y \\end{pmatrix} \\middle| y \\in \\mathbb{R} \\right\\} \\) be subspaces of \\( \\mathbb{R}^2 \\).

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You correctly said that \\(U + W \\) is a direct sum.

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Check the statements which, after established true, prove your claim.

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Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\x \\end{pmatrix} \\middle| x \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} -y\\\\y \\end{pmatrix} \\middle| y \\in \\mathbb{R} \\right\\} \\) be subspaces of \\( \\mathbb{R}^2 \\).

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In order to show that \\(U + W = \\mathbb{R}^2\\), choose \\(x,y\\) such that \\( \\begin{pmatrix} x\\\\x \\end{pmatrix} + \\begin{pmatrix} -y\\\\y \\end{pmatrix} = \\begin{pmatrix} a\\\\b \\end{pmatrix} \\), for a general vector \\(\\begin{pmatrix}a\\\\b\\end{pmatrix}\\) in \\(\\mathbb{R}^2\\).

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\\(x = \\) [[0]]

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\\(y = \\) [[1]]

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Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\x\\\\-x \\end{pmatrix} \\middle| x \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} y\\\\z\\\\-z \\end{pmatrix} \\middle| y,z \\in \\mathbb{R} \\right\\} \\) be subspaces of \\( \\mathbb{R}^3 \\).

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Is \\(U + W \\) a direct sum?

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[[0]]

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Is \\(U + W = \\mathbb{R}^3 \\)?

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[[1]]

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Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\x\\\\-x \\end{pmatrix} \\middle| x \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} y\\\\z\\\\-z \\end{pmatrix} \\middle| y,z \\in \\mathbb{R} \\right\\} \\) be subspaces of \\( \\mathbb{R}^3 \\).

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In order to show that \\(U + V \\) is not a direct sum, you need to show that \\(U \\cap V \\) contains at least one vector which is not zero.

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Thus choose \\(x,y,z\\) such that \\( \\begin{pmatrix} x\\\\x\\\\-x \\end{pmatrix} = \\begin{pmatrix} y\\\\z\\\\-z \\end{pmatrix} \\neq \\begin{pmatrix} 0\\\\0\\\\0 \\end{pmatrix}.\\)

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\\( x = \\) [[0]]

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\\( y = \\) [[1]]

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\\( z = \\) [[2]]

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True or false: actually all vectors in \\(U\\) are also in \\(W\\), i.e. \\(U\\subseteq W\\).

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[[3]]

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Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\x\\\\-x \\end{pmatrix} \\middle| x \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} y\\\\z\\\\-z \\end{pmatrix} \\middle| y,z \\in \\mathbb{R} \\right\\} \\) be subspaces of \\( \\mathbb{R}^3 \\).

\n

You have seen that \\(U + W \\neq \\mathbb{R}^3\\).

\n

Give an example of a vector in \\(\\mathbb{R}^3\\) which is not contained in \\(U + W,\\) i.e. choose \\( v \\in \\mathbb{R}^3 \\) such that for all \\(x,y,z \\in \\mathbb{R} \\) \\( v \\neq \\begin{pmatrix} x\\\\x\\\\-x \\end{pmatrix} + \\begin{pmatrix} y\\\\z\\\\-z \\end{pmatrix} \\) 

\n

\\(v = \\) [[0]]

\n

True or false: in this case, \\(U+W\\) is one of the two original subspaces (\\(U+W=U\\) or \\(U+W=W\\)).

\n

[[1]]

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Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\y\\\\-y \\end{pmatrix} \\middle| x, y \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} 0\\\\z\\\\-z \\end{pmatrix} \\middle| z \\in \\mathbb{R} \\right\\} \\) be subspaces of \\( \\mathbb{R}^3 \\).

\n

Is \\(U + W \\) a direct sum?

\n

[[0]]

\n

Is \\(U + W = \\mathbb{R}^3 \\)?

\n

[[1]]

\n

", "gaps": [{"type": "1_n_2", "useCustomName": true, "customName": "direct sum?", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": "Decide if direct sum", "minMarks": 0, "maxMarks": 0, "shuffleChoices": false, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["yes", "no"], "matrix": ["-1", "1"], "distractors": ["", ""]}, {"type": "1_n_2", "useCustomName": true, "customName": "whole space?", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": "Decide if sum gives whole space", "minMarks": 0, "maxMarks": 0, "shuffleChoices": false, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["yes", "no"], "matrix": ["-1", "1"], "distractors": ["", ""]}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": true, "customName": "c) counterexample direct sum", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "mark:\ncorrectif(answers[0]=0 && answers[1] = answers[2] && answers[2] <> 0)", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [{"label": "Give the reaons for \"whole space\" answer", "rawLabel": "Give the reaons for \"whole space\" answer", "otherPart": 8, "variableReplacements": [], "availabilityCondition": "answered", "penalty": "", "penaltyAmount": 0, "lockAfterLeaving": false}], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": "Reasonings", "prompt": "

Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\y\\\\-y \\end{pmatrix} \\middle| x, y \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} 0\\\\z\\\\-z \\end{pmatrix} \\middle| z \\in \\mathbb{R} \\right\\} \\) be subspaces of \\( \\mathbb{R}^3\\).

\n

In order to show that \\(U + V \\) is not a direct sum you need to show that \\(U \\cap V \\) contains at least one vector which is not zero.

\n

Thus choose \\(x,y,z\\) such that \\( \\begin{pmatrix} x\\\\y\\\\-y \\end{pmatrix} = \\begin{pmatrix} 0\\\\z\\\\-z \\end{pmatrix} \\neq \\begin{pmatrix} 0\\\\0\\\\0 \\end{pmatrix}.\\)

\n

\\( x = \\) [[0]]

\n

\\( y = \\) [[1]]

\n

\\( z = \\) [[2]]

\n

True or false: actually all vectors in \\(W\\) are also in \\(U\\), i.e. \\(W\\subseteq U\\).

\n

[[3]]

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "x", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": false, "showFeedbackIcon": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "123", "maxValue": "123", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": true, "customName": "y", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": false, "showFeedbackIcon": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "123", "maxValue": "123", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": true, "customName": "z", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": false, "showFeedbackIcon": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "123", "maxValue": "123", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "true-false", "useCustomName": true, "customName": "True/false", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "settings": {"correct_answer_expr": "true"}}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": true, "customName": "c) counterexample whole space", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "mark:\n assert(all_valid or not settings[\"sortAnswers\"], fail(translate(\"question.can not submit\")));\n apply(answers);\n apply(gap_feedback);\n correctif(rank(matrix([1,0,0],[0,1,-1],transpose(answers[0])[0])) = 3)", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [{"label": "Try part d)", "rawLabel": "", "otherPart": 9, "variableReplacements": [], "availabilityCondition": "", "penalty": "", "penaltyAmount": 0, "lockAfterLeaving": false}, {"label": "Try part e)", "rawLabel": "", "otherPart": 12, "variableReplacements": [], "availabilityCondition": "", "penalty": "", "penaltyAmount": 0, "lockAfterLeaving": false}, {"label": "Give the reaons for \"direct sum\" answer if you haven't done it yet", "rawLabel": "Give the reaons for \"direct sum\" answer if you haven't done it yet", "otherPart": 7, "variableReplacements": [], "availabilityCondition": "answered", "penalty": "", "penaltyAmount": 0, "lockAfterLeaving": false}], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": "Reasonings", "prompt": "

Let \\( U = \\left\\{ \\begin{pmatrix} x\\\\y\\\\-y \\end{pmatrix} \\middle| x, y \\in \\mathbb{R} \\right\\} \\) and \\( W = \\left\\{ \\begin{pmatrix} 0\\\\z\\\\-z \\end{pmatrix} \\middle| z \\in \\mathbb{R} \\right\\} \\) be subspaces of \\( \\mathbb{R}^3\\).

\n

You have seen that \\(U + W \\neq \\mathbb{R}^3\\).

\n

Give an example of a vector in \\(\\mathbb{R}^3\\) which is not contained in \\(U + W,\\) i.e. choose \\( v \\in \\mathbb{R}^3 \\) such that for all \\(x,y,z \\in \\mathbb{R} \\) \\( v \\neq \\begin{pmatrix} x\\\\y\\\\-y \\end{pmatrix} + \\begin{pmatrix} 0\\\\z\\\\-z \\end{pmatrix} \\) 

\n

\\(v = \\) [[0]]

\n

True or false: in this case, \\(U+W\\) is one of the two original subspaces (\\(U+W=U\\) or \\(U+W=W\\)).

\n

[[1]]

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Let \\( P_2 \\) be the vectorspace of polynomials of degree at most \\( 2 \\) and consider the subspaces \\( U = \\{ p=a_2 X^2 + a_1 X \\vert a_1,a_2 \\in \\mathbb{R}\\} \\) and \\( W = P_1= \\{ q=b_1 X + b_0 \\vert b_0,b_1 \\in \\mathbb{R}\\}\\).

\n

Is \\(U+W\\) a direct sum?

\n

[[0]]

\n

Is \\( U+W=P_2\\)?

\n

[[1]]

", "gaps": [{"type": "1_n_2", "useCustomName": true, "customName": "direct sum?", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": "Decide if direct sum", "minMarks": 0, "maxMarks": 0, "shuffleChoices": false, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["yes", "no"], "matrix": [0, "1"], "distractors": ["", ""]}, {"type": "1_n_2", "useCustomName": true, "customName": "whole space?", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": "Decide if sum gives whole space", "minMarks": 0, "maxMarks": 0, "shuffleChoices": false, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["yes", "no"], "matrix": ["1", "0"], "distractors": ["", ""]}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": true, "customName": "d) counterexample direct sum", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "mark:\n assert(all_valid or not settings[\"sortAnswers\"], fail(translate(\"question.can not submit\")));\n apply(answers);\n correctif(answers[1] = answers[2] && answers[1] <> 0 && answers[0] = answers[3] = 0)", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [{"label": "Give the reaons for \"whole space\" answer", "rawLabel": "Give the reaons for \"whole space\" answer", "otherPart": 11, "variableReplacements": [], "availabilityCondition": "answered", "penalty": "", "penaltyAmount": 0, "lockAfterLeaving": false}], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": "Reasonings", "prompt": "

Let \\( P_2 \\) be the vectorspace of polynomials of degree at most \\( 2 \\) and consider the subspaces \\( U = \\{ p=a_2 X^2 + a_1 X \\vert a_1,a_2 \\in \\mathbb{R}\\} \\) and \\( W = P_1= \\{ q=b_1 X + b_0 \\vert b_0,b_1 \\in \\mathbb{R}\\}\\).

\n

You have seen that \\(U + W\\) is not a direct sum.

\n

Choose \\(a_2,a_1,b_1,b_0\\) such that \\( a_2 X^2 + a_1 X = b_1 X + b_0 \\neq 0 \\).

\n

\\(a_2 = \\) [[0]]

\n

\\(a_1 = \\) [[1]]

\n

\\(b_1 = \\) [[2]]

\n

\\(b_0 = \\) [[3]]

\n

True or false: actually all vectors in \\(U\\) are also in \\(W\\), i.e. \\(U\\subseteq W\\).

\n

[[4]]

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Let \\( P_2 \\) be the vectorspace of polynomials of degree at most \\( 2 \\) and consider the subspaces \\( U = \\{ p=a_2 X^2 + a_1 X \\vert a_1,a_2 \\in \\mathbb{R}\\} \\) and \\( W = P_1= \\{ q=b_1 X + b_0 \\vert b_0,b_1 \\in \\mathbb{R}\\}\\).

\n

In order to show that \\(U + W = P_2,\\) choose \\(a_2,a_1,b_1,b_0\\) such that \\( a_2 X^2 + a_1 X + b_1 X + b_0 = c_2 X^2 + c_1 X + c_0 \\). I.e. show that any general polynomial of degree at most two can be written as the sum of a poly in \\(U\\) and a poly in \\(W\\).

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Enter subscripts as c_2 or c_1 etc. Please use decimals instead of fractions: the marking algorithm will not understand fractions in this question.

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\\(a_2 = \\) [[0]]

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\\(a_1 = \\) [[1]]

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\\(b_1 = \\) [[2]]

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\\(b_0 = \\) [[3]]

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Let \\( P_2 \\) be the vectorspace of polynomials of degree at most \\( 2 \\) and consider the subspaces \\( U = \\{ p=a_2 X^2 \\vert a_2 \\in \\mathbb{R}\\} \\) and \\( W = P_0=\\{ q=b_0 \\vert b_0 \\in \\mathbb{R}\\} \\).

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Is \\(U+W\\) a direct sum?

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[[0]]

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Is \\( U+W=P_2\\)?

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[[1]]

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Let \\( P_2 \\) be the vectorspace of polynomials of degree at most \\( 2 \\) and consider the subspaces \\( U = \\{ p=a_2 X^2 \\vert a_2 \\in \\mathbb{R}\\} \\) and \\( W = P_0=\\{ q=b_0 \\vert b_0 \\in \\mathbb{R}\\} \\).

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You have seen that \\(U + W \\neq P_2\\).

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Give an example of a Polynomial in \\(P_2,\\) which is not contained in \\(U + W,\\) i.e. choose \\(c_0,c_1,c_2\\) such that \\( c_2 X^2 + c_1 X + c_0 \\notin U + W. \\)

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\\(c_0 = \\) [[0]]

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\\(c_1 = \\) [[1]]

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\\(c_2 = \\) [[2]]

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True or false: in this case, \\(U+W\\) is one of the two original subspaces (\\(U+W=U\\) or \\(U+W=W\\)).

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[[3]]

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