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Find a basis for the intersection of two kernels of linear maps $\\mathbb Q^3\\to \\mathbb Q^3$ and $\\mathbb Q^3\\to \\mathbb Q$.

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Wir betrachten die Abbildungen $f = \\mathbf f_A\\colon \\mathbb Q^3 \\to \\mathbb Q^3$ und $g = \\mathbf f_B\\colon \\mathbb Q^3\\to \\mathbb Q$ mit

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\\[ A = \\var{F},\\qquad B = \\var{G} \\]

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a)

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Wegen der Dimensionsformel addieren sich in beiden Fällen die Dimensionen von Kern und Bild zu $3$.

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Die Dimension des Kerns ist die Anzahl der frei wählbaren Variablen in der Lösungsmenge des zugehörigen homogenen linearen Gleichungssystems. Die Dimension des Bildes ist die Dimension des von den Spaltenvektoren erzeugten Untervektorraums. Beide Dimensionen kann man in diesem Beispiel \"durch genaues Hinschauen\" erkennen (bei $g$ ist die einzige Frage, ob alle Einträge der Matrix $0$ sind, oder nicht; die Aufgabe ist so eingerichtet, dass $f$ Rang $1$ oder $2$ hat).

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b)

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Der Kern von $f$ ist die Lösungsmenge des Gleichungssystems $Ax=0$, analog für $g$ und $B$. Der Durchschnitt der Kerne ist also die Lösungsmenge des homogenen Gleichungssystems mit Koeffizientenmatrix $\\begin{pmatrix} A\\\\ B\\end{pmatrix}$ (als Blockmatrix geschrieben), wir nehmen also die durch $A$ und die durch $B$ gegebenen Gleichungen zusammen. Eine Basis der Lösungsmenge kann man dann nach dem üblichen Verfahren bestimmen.

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Die reduzierte Zeilenstufenform von $\\begin{pmatrix} A\\\\ B\\end{pmatrix}$ ist $\\var{reduced_row_echelon_form(matrix(list(F)+list(G)))}$.

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Geben Sie die folgenden Dimensionen an:

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$\\dim {\\rm Ker}(f) = $ [[0]]

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$\\dim {\\rm Im}(f) = $ [[1]]

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$\\dim {\\rm Ker}(g) = $ [[2]]

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$\\dim {\\rm Im}(g) = $ [[3]]

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Bestimmen Sie eine Basis von ${\\rm Ker}(f)\\cap {\\rm Ker}(g)$. Geben Sie Ihre Antwort als eine Matrix an, deren Spalten eine solche Basis bilden.

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[[0]]

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