In this problem, allow&nbsp;T1:R2→R2&nbsp;and&nbsp;T2:R2→R2&nbsp;be linear transformations. Find&nbsp;Ker(T1),Ker(T2),Ker(T3)&nbsp;of the respective matrices:A=[1−1−20],B=[15−20]

Step 1T1:R2→R2&nbsp;and&nbsp;T2:R2→R2A=[1−1−20],B=[15−20]&nbsp;are matrices with respect to&nbsp;T1&nbsp;and&nbsp;T2&nbsp;Then if&nbsp;x∈Ker(T1)&nbsp;Then&nbsp;T1(x)=0&nbsp;i.e.&nbsp;Ax=0⇒[1−1−20][x1x2]=[00]x1−x2=0,−2x1=0⇒x1=0x1=x2⇒x2=0So&nbsp;x=(00)&nbsp;only so&nbsp;Ker(T1)={(00)}Let&nbsp;x∈Ker(T2)⇒T2(x)=0&nbsp;and&nbsp;Bx=0⇒[15−20][x1x2]=[00]⇒x1+5x2=0,−2x1=0⇒x1=0&nbsp;and&nbsp;x2=0So&nbsp;x=(00)i.e.&nbsp;Ker(T2)={(00)}Answer is as above

Step by step solution to "In this problem, allowT1:R2→R2andT2:R2→R2be linear transformations. FindKer(T1),Ker(T2),Ker(T3)of the..."

Bergen

Answered question

2020-10-20

In this problem, allow $T_{1} : R^{2} \to R^{2}$ and $T_{2} : R^{2} \to R^{2}$ be linear transformations. Find $K e r (T_{1}), K e r (T_{2}), K e r (T_{3})$ of the respective matrices:
$A = [\begin{matrix} 1 & - 1 \\ - 2 & 0 \end{matrix}], B = [\begin{matrix} 1 & 5 \\ - 2 & 0 \end{matrix}]$

Answer & Explanation

rogreenhoxa8

Skilled2020-10-21Added 109 answers

Step 1
$T_{1} : R^{2} \to R^{2} and T_{2} : R^{2} \to R^{2}$
$A = [\begin{matrix} 1 & - 1 \\ - 2 & 0 \end{matrix}], B = [\begin{matrix} 1 & 5 \\ - 2 & 0 \end{matrix}]$ are matrices with respect to $T_{1} and T_{2}$ Then if $x \in K e r (T_{1})$ Then $T_{1} (x) = 0$ i.e. $A x = 0 \Rightarrow [\begin{matrix} 1 & - 1 \\ - 2 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$
$x_{1} - x_{2} = 0, - 2 x_{1} = 0 \Rightarrow x_{1} = 0$
$x_{1} = x_{2} \Rightarrow x_{2} = 0$
So $x = (\begin{matrix} 0 \\ 0 \end{matrix})$ only so $K e r (T_{1}) = {(\begin{matrix} 0 \\ 0 \end{matrix})}$
Let $x \in K e r (T_{2}) \Rightarrow T_{2} (x) = 0$ and $B x = 0$
$\Rightarrow [\begin{matrix} 1 & 5 \\ - 2 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] \Rightarrow x_{1} + 5 x_{2} = 0, - 2 x_{1} = 0$
$\Rightarrow x_{1} = 0 and x_{2} = 0$
So $x = (\begin{matrix} 0 \\ 0 \end{matrix})$
i.e. $K e r (T_{2}) = {(\begin{matrix} 0 \\ 0 \end{matrix})}$
Answer is as above

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