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Answer: <\/strong>The only correct choice is c.<\/div>\n\n
Reasoning for a:<\/strong> \\( \\ker(L) \\) contains the constant polynomials, so \\( \\ker(L)\\neq\\{\\vec{0}\\} \\).<\/div>\n\n
Alternatively:<\/strong> Since \\( \\dim(\\mathcal{P}_3(\\mathbb{R}))=4 \\gt 3=\\dim(\\mathcal{P}_2(\\mathbb{R})) \\), it is impossible for \\( L \\) to be injective.<\/div>\n","notChosenFeedback":"
Answer: <\/strong>The only correct choice is c.<\/div>\n\n
Reasoning for a:<\/strong> \\( \\ker(L) \\) contains the constant polynomials, so \\( \\ker(L)\\neq\\{\\vec{0}\\} \\).<\/div>\n\n
Alternatively:<\/strong> Since \\( \\dim(\\mathcal{P}_3(\\mathbb{R}))=4 \\gt 3=\\dim(\\mathcal{P}_2(\\mathbb{R})) \\), it is impossible for \\( L \\) to be injective.<\/div>\n","tip":""},"text":"
a.<\/strong> \\( L\\colon \\mathcal{P}_3(\\mathbb{R}) \\to \\mathcal{P}_2(\\mathbb{R}) \\), \\( L(p(x))=p'(x) \\).<\/div>\n"},{"correct":false,"tipsAndFeedback":{"chosenFeedback":"
Answer: <\/strong>The only correct choice is c.<\/div>\n\n
Reasoning for b:<\/strong> \\( \\ker(L) \\) contains all matrices \\( A \\) such that \\( A^T=-A \\) (such matrices are called \\textit{skew-symmetric}). For instance, \\( A=\\begin{bmatrix} 0&-1\\\\1&0 \\end{bmatrix} \\) is skew-symmetric hence is in \\( \\ker(L) \\). So \\( \\ker(L)\\neq\\{\\vec{0}\\} \\).<\/div>\n","notChosenFeedback":"
Answer: <\/strong>The only correct choice is c.<\/div>\n\n
Reasoning for b:<\/strong> \\( \\ker(L) \\) contains all matrices \\( A \\) such that \\( A^T=-A \\) (such matrices are called \\textit{skew-symmetric}). For instance, \\( A=\\begin{bmatrix} 0&-1\\\\1&0 \\end{bmatrix} \\) is skew-symmetric hence is in \\( \\ker(L) \\). So \\( \\ker(L)\\neq\\{\\vec{0}\\} \\).<\/div>\n","tip":""},"text":"
b.<\/strong> \\( L\\colon M_{2\\times 2}(\\mathbb{F}) \\to M_{2\\times 2}(\\mathbb{F}) \\), \\( L(A)=A+A^T \\).<\/div>\n"},{"correct":true,"tipsAndFeedback":{"chosenFeedback":"
Answer: <\/strong>The only correct choice is c.<\/div>\n\n
Reasoning for c:<\/strong> For any \\( a+bx+cx^2 \\in \\ker(L) \\), we have \\( L(a+bx+cx^2) = \\begin{bmatrix} a+2b&b\\\\b+c&2a+b \\end{bmatrix} = \\begin{bmatrix} 0&0\\\\0&0 \\end{bmatrix} \\). By equating coefficients, we get \\( a+2b=0,\\ b=0,\\ b+c=0, \\) and \\( 2a+b=0 \\), from which we find that \\( a=b=c=0 \\). Thus \\( \\ker(L) = \\{\\vec{0}\\} \\) and so \\( L \\) is injective.<\/div>\n","notChosenFeedback":"
Answer: <\/strong>The only correct choice is c.<\/div>\n\n
Reasoning for c:<\/strong> For any \\( a+bx+cx^2 \\in \\ker(L) \\), we have \\( L(a+bx+cx^2) = \\begin{bmatrix} a+2b&b\\\\b+c&2a+b \\end{bmatrix} = \\begin{bmatrix} 0&0\\\\0&0 \\end{bmatrix} \\). By equating coefficients, we get \\( a+2b=0,\\ b=0,\\ b+c=0, \\) and \\( 2a+b=0 \\), from which we find that \\( a=b=c=0 \\). Thus \\( \\ker(L) = \\{\\vec{0}\\} \\) and so \\( L \\) is injective.<\/div>\n","tip":""},"text":"
c.<\/strong> \\( L\\colon \\mathcal{P}_2(\\mathbb{F}) \\to M_{2\\times 2}(\\mathbb{F}) \\), \\( L(a+bx+cx^2) = \\begin{bmatrix}a+2b&b\\\\b+c&2a+b\\end{bmatrix} \\).<\/div>\n"}],"overallFeedback":[{"from":0,"to":100}],"behaviour":{"enableRetry":true,"enableSolutionsButton":false,"enableCheckButton":true,"type":"auto","singlePoint":false,"randomAnswers":false,"showSolutionsRequiresInput":true,"confirmCheckDialog":false,"confirmRetryDialog":false,"autoCheck":false,"passPercentage":100,"showScorePoints":true},"UI":{"checkAnswerButton":"Check","submitAnswerButton":"Submit","showSolutionButton":"Show solution","tryAgainButton":"Retry","tipsLabel":"Show tip","scoreBarLabel":"You got :num out of :total points","tipAvailable":"Tip available","feedbackAvailable":"Feedback available","readFeedback":"Read feedback","wrongAnswer":"Wrong answer","correctAnswer":"Correct answer","shouldCheck":"Should have been checked","shouldNotCheck":"Should not have been checked","noInput":"Please answer before viewing the solution","a11yCheck":"Check the answers. The responses will be marked as correct, incorrect, or unanswered.","a11yShowSolution":"Show the solution. The task will be marked with its correct solution.","a11yRetry":"Retry the task. Reset all responses and start the task over again."},"confirmCheck":{"header":"Finish ?","body":"Are you sure you wish to finish ?","cancelLabel":"Cancel","confirmLabel":"Finish"},"confirmRetry":{"header":"Retry ?","body":"Are you sure you wish to retry ?","cancelLabel":"Cancel","confirmLabel":"Confirm"},"question":"
Of the linear maps given below, select all that are injective.<\/p>\n\n
Note:<\/strong> \\( \\mathcal{P}_n(\\mathbb{F}) \\) is the vector space of polynomials in the indeterminate \\( x \\) of degree \\( \\leq n \\) with coefficients in \\( \\mathbb{F} \\).<\/p>\n"}