R-2-R-2-x-y-2x-3y-3y-find-L-R-2-find-ker-and-the-im-f-o-f-

Question Number 215525 by zetamaths last updated on 09/Jan/25

Φ : R^{2} \to R^{2}

(x .; y) ∣\to (2 x + 3 y : 3 y)

f i n d

Φ \in L (R^{2})

f i n d k e r (Φ) a n d t h e i m (Φ)

f o f = ?

Answered by MrGaster last updated on 09/Jan/25

\ker (Φ) = {x, y} \in R^{2} ∣ Φ (x, y) = (0, 0)}

(\begin{matrix} 2 & 3 \\ 0 & 3 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})

{\begin{cases} 2 x + 3 y = 0 \\ 3 y = 0 \end{cases}

y = 0 \Rightarrow 2 x = 0 \Rightarrow x = 0

kel (Φ) = {(0, 0)} ✓

im (Φ) = {Φ (x, y) ∣ (x, y) \in R^{2}}

im (Φ) = sqan ((\begin{matrix} 2 \\ 0 \end{matrix}), (\begin{matrix} 3 \\ 3 \end{matrix}))

Reduse to echelon form :

(\begin{matrix} 2 & 3 \\ 0 & 3 \end{matrix}) \sim (\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix})

im (Φ) = sqan (((\begin{matrix} 2 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 3 \end{matrix}))

= sqan ((\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix}))

= R^{2} ✓

Conclusion :

\ker (Φ) = {(0, 0)}

im (Φ) = R^{2}

f = 2

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