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Let $T: \Bbb R^3 \rightarrow \Bbb R^3$ be a linear transformation satisfying \begin{align*} T(0,1,1) =& (-1,1,1) \\ T(1,0,1) =& (1,-1,1) \\ T(1,1,0) =& (1,-1,0) . \end{align*}

Is it necessary true that $\ker(T) = \operatorname{Sp}\{(1,-1,1)\}$ ? Well, I tried to say that we know that $\operatorname{Im}(T) = \operatorname{Sp}\{T(0,1,1),\,T(1,0,1),\,T(1,1,1)\}$

So, $\operatorname{Im}(T) = \operatorname{Sp}\{(-1,1,1),\,(1,-1,1),\,(1,-1,0)\}$ which means $\operatorname{Sp}\{(1,-1,1)\} \in \operatorname{Im}(T)$ and also $(1,1,1)$ is linearly independent by $2$ other vectors which are in $\operatorname{Im}(T)$.

Now, how can I prove that $\operatorname{Sp}\{(1,1,1)\}$ not inside $\ker(T)$? or maybe $\operatorname{Sp}\{(1,1,1)\} \in \ker (T)$ which makes it $\operatorname{Sp}\{(1,1,1)\} = \ker (T)$?

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    $\begingroup$ Hint: use the dimensions' theorem to show $\;\dim \ker T=1\;$ ... $\endgroup$ – DonAntonio May 31 '14 at 18:10
  • $\begingroup$ I did show that $dimImT = 2$, and because $dimR^3 = 3$. Thus, $dimKerT=1$. and because of that I can say that $(1,-1,1)$ CAN be a basis to $KerT$. Thus, it can be in its span because it's a sub-space of $R^3$ and also a basis to $KerT$. Thus, $KerT = Sp(\{1,-1,1)\})$ $\endgroup$ – Ilan Aizelman WS May 31 '14 at 18:32
  • $\begingroup$ Thank you very much @DonAntonio $\endgroup$ – Ilan Aizelman WS May 31 '14 at 18:33
  • $\begingroup$ @DonAntonio Well, are you sure I can say that? I think I need to show somehow that T(Sp(\{(1,-1,1)\}) = 0 and then I can say that it equals to $KerT$. but how can I show that? Or am I all wrong and what I wrote is enough? $\endgroup$ – Ilan Aizelman WS May 31 '14 at 19:20
  • $\begingroup$ Of course you must show that vector is in the kernel, Ilan!! I thought you already did that...I didn´t check that. $\endgroup$ – DonAntonio May 31 '14 at 19:22
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We can simply put our vectors in the matrix and do row operations in the following way - we are trying to get a basis in the right part: $$\left( \begin{array}{ccc|ccc} 0 & 1 & 1 &-1 & 1 & 1\\ 1 & 0 & 1 & 1 &-1 & 1\\ 1 & 1 & 0 & 1 &-1 & 0 \end{array} \right)\sim \left( \begin{array}{ccc|ccc} 0 & 1 & 1 &-1 & 1 & 1\\ 0 &-1 & 1 & 0 & 0 & 1\\ 1 & 1 & 0 & 1 &-1 & 0 \end{array} \right)\sim \left( \begin{array}{ccc|ccc} 1 & 2 & 1 & 0 & 0 & 1\\ 0 &-1 & 1 & 0 & 0 & 1\\ 1 & 1 & 0 & 1 &-1 & 0 \end{array} \right)\sim \left( \begin{array}{ccc|ccc} 1 & 1 & 0 & 1 &-1 & 0\\ 0 &-1 & 1 & 0 & 0 & 1\\ 1 & 2 & 1 & 0 & 0 & 1 \end{array} \right)\sim \left( \begin{array}{ccc|ccc} 1 & 1 & 0 & 1 &-1 & 0\\ 0 &-1 & 1 & 0 & 0 & 1\\ 1 & 3 & 0 & 0 & 0 & 0 \end{array} \right)$$

What have we found out by doing this?

Notice that at the beginning we have $a|b$, where $T(a)=b$ in each row. This property is not changed using row operations.

So we see that $T(1,3,0)=(0,0,0)$. This means that $(1,3,0)\in\operatorname{Ker} T$.

We have also found out that the image is generated by the vectors $(1,-1,0)$ and $(0,0,1)$. (If we look only on the right part, then we have tried to row reduce the matrix consisting of images of basis vectors.) Since these vectors are linearly independent, we get that $\dim\operatorname{Im} T =2$. By rank-nullity theorem we know that $\dim\operatorname{Ker} T=1$.

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  • $\begingroup$ So now because of $dimKerT = 1$ I can assume that $Sp(\{(1,-1,1)\}) \in KerT$ which means $Sp(\{(1,-1,1)\}) = KerT$ $\endgroup$ – Ilan Aizelman WS May 31 '14 at 20:02
  • $\begingroup$ Why do you keep saying that $(1,-1,1)$ is in kernel? If I did not make a mistake, I have computed above that $\operatorname{Ker} T=\operatorname{Sp}\{(1,3,0)\}$, so clearly $(1,-1,1)$ does not belong there. I have also tried to compute $T(1,-1,1)$ here. Again, I might have made mistake, but the result I got is not zero vector. $\endgroup$ – Martin Sleziak May 31 '14 at 20:06
  • $\begingroup$ So $KerT \ne Sp(\{1,-1,1)\})$ ? $\endgroup$ – Ilan Aizelman WS May 31 '14 at 20:09
  • $\begingroup$ In another words: The vectors $(1,3,0)$ and $(1,-1,1)$ are linearly independent. So if we know that the dimension of kernel is one, they cannot be both in the kernel. $\endgroup$ – Martin Sleziak May 31 '14 at 20:10
  • $\begingroup$ Yes, I am saying that $\operatorname{Ker}T= \operatorname{Sp}\{(1,3,0)\} \ne\operatorname{Sp}\{(1,-1,1)\}$. $\endgroup$ – Martin Sleziak May 31 '14 at 20:11
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The matrix associated to $T$ with respect to the basis $\mathscr{B}=\{(0,1,1),(1,0,1),(1,1,0)\}$ on the domain and the canonical basis on the codomain is $$ A=\begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & 1 & 0 \end{bmatrix} $$ The matrix associated to $T$ with respect to the canonical basis on both the domain and the codomain is $$ B=AS^{-1} $$ where $$ S=\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} $$ and $$ S^{-1}=\begin{bmatrix} -1/2 & 1/2 & 1/2 \\ 1/2 & -1/2 & 1/2 \\ 1/2 & 1/2 & -1/2 \end{bmatrix} $$ so that $$ B=\begin{bmatrix} 3/2 & -1/2 & -1/2 \\ -3/2 & 1/2 & 1/2 \\ 0 & 0 & 1 \end{bmatrix} $$

Can you compute the null space of $B$? That is, the set of vectors $v$ such that $Bv=0$, which is the kernel of $T$.

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  • $\begingroup$ Too complicated to understand, I'll stick to Don's answer. Thx! $\endgroup$ – Ilan Aizelman WS May 31 '14 at 18:51
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Let the basis for the domain be $B=\{v_1,v_2,v_3\}=\{(0,1,1),\;(1,0,1),\;(1,1,0)\}$. Let $w_1,w_2,w_3$ be the respective images of $v_i's$ under $T$.

A simple observation shows that: the set $\{w_1, w_2\}$ is linearly independent (as they are not multiples of each other) whereas $\{w_1,w_2,w_3\}$ is a dependent set because $w_1-w_2+2w_3=0$. This means the dimension of the range space is exactly $2$, hence the kernel will be of dimension $1$ (by the rank-nullity theorem).

In fact we can now get the basis vector for the kernel as well:

Since $w_1-w_2+2w_3=0$, this means $T(v_1-v_2+2v_3)=0$. Thus the vector $v_1-v_2+2v_3=(1, 3, 0)$ forms the basis of the kernel.

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  • $\begingroup$ if the base of Kernel is $(1,3,0)$ , how can I show that $KerT = Sp(1,-1,1)$ ? $\endgroup$ – Ilan Aizelman WS May 31 '14 at 19:10
  • $\begingroup$ The kernel of $T$ does not contain the vector $(1,-1,1)$. In fact you can verify that $$T(1,-1,1)=(3/2, -3/2, 1) \neq (0,0,0)$$. $\endgroup$ – Anurag A Jun 1 '14 at 1:25
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You know that ${\rm im}\; f=\langle (-1,1,1),(1,-1,1),(1,-1,0)\rangle$. This is the same as $$\langle (-1,1,1)\color{red}{+(1,-1,0)},(1,-1,1),(1,-1,0)\rangle=\langle (0,0,1),(1,-1,1),(1,-1,0)\rangle$$ which in turn is the same as $$\langle (0,0,1),(1,-1,1)\color{blue}{-(0,0,1)},(1,-1,0)\rangle=\langle (0,0,1),(1,-1,0),(1,-1,0)\rangle$$

Thus the dimension of the image is $2$. By the rank-nullity theorem, the dimension of the kernel is $1$. It follows that if you find $v\in\ker T$ nonzero, $\langle v\rangle =\ker T$. Can you check whether $(1,-1,1)\in \ker T$?

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  • $\begingroup$ That's exactly what there is left to find, how can I find that $v \in KerT$ and see that it's $(1,-1,1) \in KerT$? I did manage to find KerT base which is (1,3,0) but it doesn't give me anything. Can you edit your answer and show me how to find that v? Much appreciated! $\endgroup$ – Ilan Aizelman WS May 31 '14 at 19:42
  • $\begingroup$ @IlanAizelmanWS Yes, it gives you all you need. You know $\ker T$ is one dimensional thus any nonzero vector in it spans $\ker T$. $\endgroup$ – Pedro Tamaroff May 31 '14 at 19:55
  • $\begingroup$ Oh, so because $dimKerT = 1$ I can assume that if $v \ne 0 $then $v \in KerT$. Thus, if $v = (1,-1,1)$ then I can say that $Sp(1,-1,1) \in KerT$ . Thus, $Sp(1,-1,1) = KerT$ . But other people here told me I still need to prove it and I can't assume that even if $dimKerT = 1$. $\endgroup$ – Ilan Aizelman WS May 31 '14 at 19:58
  • $\begingroup$ No. That is wrong. What you can say is that if $v\neq 0$ and $\color{red}{v\in\ker T}$ then $\ker T=\langle v\rangle$. $\endgroup$ – Pedro Tamaroff May 31 '14 at 20:03
  • $\begingroup$ and $v \ne 0$ is true because its basis isn't $(0,0,0)$? $\endgroup$ – Ilan Aizelman WS May 31 '14 at 20:05
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Hint.

Choose the basis $B = \{(0,1,1),\;(1,0,1),\;(1,1,0)\}$.

What does the matrix representation $[T]^B_E$ of the linear map $T$ with respect to the bases $B$ and $E$ (the standard basis) look like?

Recall that the kernel of a matrix is invariant under basis transformation.

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  • $\begingroup$ If I find that matrix, how will that matrix help me? $\endgroup$ – Ilan Aizelman WS May 31 '14 at 19:18
  • $\begingroup$ @IlanAizelmanWS Then we can determine its kernel and answer those questions concerning the kernel. $\endgroup$ – Mussé Redi May 31 '14 at 20:54
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Assuming you mean the vector $\;(111)\;$ wrt the usual, canonical basis, first write it as a linear combination of the given basis:

$$\begin{pmatrix}\;\;1\\-1\\\;1\end{pmatrix}=a\begin{pmatrix}0\\1\\1\end{pmatrix}+b\begin{pmatrix}1\\0\\1\end{pmatrix}+c\begin{pmatrix}1\\1\\0\end{pmatrix}\iff$$

$$\begin{align*}b+c=1\\{}\\a+c=-1\\{}\\a+b=1\end{align*}\;\implies\;\;b-c=2\implies b=\frac32\;,\;\;a=c=-\frac12$$

Thus we get

$$T\begin{pmatrix}1\\-1\\1\end{pmatrix}=\frac12\left(-T\begin{pmatrix}0\\1\\1\end{pmatrix}+3T\begin{pmatrix}1\\0\\1\end{pmatrix}-T\begin{pmatrix}1\\1\\0\end{pmatrix}\right)=\ldots$$

If the above is zero you've found one non-zero element in the kernel and thus a basis for it, otherwise you need to find another vector that actually is in the kernel.

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  • $\begingroup$ I'm still confused, why finding $T(1,1,1)$ when we need to find $T(1,-1,1)$? and also why do I need to multiply all of them by 0.5 and lastly, the equation you built equals to$ T(1,1,1) = (1.5,-1.5,1)$, but I still don't understand how it helps to find $(1,-1,1) \in KerT$ $\endgroup$ – Ilan Aizelman WS May 31 '14 at 19:51
  • $\begingroup$ Also, I did manage to find that (1,3,0) is a basis to KerT according to Anurag's answer. but it doesn't help me. $\endgroup$ – Ilan Aizelman WS May 31 '14 at 19:52
  • $\begingroup$ Tyop, Ilan: typo. $\endgroup$ – DonAntonio May 31 '14 at 21:51

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