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Given the A vector space $V$ over the field $\Bbb Z_2$ and a linear map $t:V \to V $ .following matrix $T =\begin{bmatrix}3& -1 &1\\-1 & 5 & -1\\ 1 & -1 & 3\end{bmatrix}$ where $T$ is a representation of $t$ Find the Kernel and Image. Finally say whether the following statement is true or false : $\operatorname{Ker} t \oplus \operatorname{Im} t = V $

Idea: rewrite $T$ as such $T==\begin{bmatrix}1& 1 &1\\1 & 1 & 1\\ 1 & 1 & 1\end{bmatrix}$ however I don't know how to precede

With @Jim help I found that kernel $=span(\lbrace\begin{bmatrix} -1\\1\\ 0\end{bmatrix} \begin{bmatrix} -1\\0\\ 1\end{bmatrix} \rbrace )$ However I don't quite know how to find the kernel $\cap$ Image

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Rewriting $t$ as you have is the correct first step. From there recall that as $\mathbb Z_2$ is a field everything you learned in linear algebra still holds.

You can see that the rank of $t$ is $1$ and the image is spanned by the vector $$\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}.$$ You know that the kernel has dimension $2$ so you just need to find two linearly independent vectors that are killed by your rewritten $t$. You can probably just guess them, so I'll let you do that.

As for kernel $\oplus$ image equaling $V$. You now know that the dimensions add to be $3$ so if the sum is direct then it does indeed equal $V$. To see that the sum is direct you need to show that the $0$ vector is the only vector in the intersection. As you know exactly what the image is, just apply $t$ and directly check what's in the kernel. You will find that the sum is indeed direct.

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  • $\begingroup$ I found the basis of kernel is $\lbrace \begin{bmatrix} -1 \\1\\0 \end{bmatrix} , \begin{bmatrix} -1 \\0\\1 \end{bmatrix} \rbrace $. However I don't know how to find the Kernel ∩ Image $\endgroup$ – Mathman Mar 6 '14 at 18:25
  • $\begingroup$ T is $\begin{bmatrix}3& -1 &1\\-1 & 5 & -1\\ 1 & -1 & 3\end{bmatrix}$ not $\begin{bmatrix}1& 1 &1\\1 & 1 & 1\\ 1 & 1 & 1\end{bmatrix}$ $\endgroup$ – ketan Mar 6 '14 at 18:30
  • $\begingroup$ @ketan Yes in $\Bbb R$ not in $\Bbb Z_2$ $\endgroup$ – Mathman Mar 6 '14 at 18:38
  • $\begingroup$ @Jim I found the basis of kernel is I found that kernel $=span(\lbrace\begin{bmatrix} -1\\1\\ 0\end{bmatrix} \begin{bmatrix} -1\\0\\ 1\end{bmatrix} \rbrace )$ However I don't quite know how to find the kernel $\cap$ Image $\endgroup$ – Mathman Mar 6 '14 at 18:40
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    $\begingroup$ @Matthew: The vectors in the image are all of the form $v = x\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$. The vectors in the kernel are those that $t$ sends to zero. So to find the vectors that are in both the image and the kernel figure out for which $x$ does $t(v) = 0$. $\endgroup$ – Jim Mar 6 '14 at 18:58
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kernel ∩ Image is just the intersection of two subspaces , for this
find kernal of $\begin{bmatrix}-1& -1 &1\\1 & 0 & 1\\ 0 & 1 & 1\end{bmatrix}$
this matrix is formed by combining basis vectors of both kernal and img of T

that will be kernel ∩ Image

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  • $\begingroup$ When you pile two bases into a matrix like that, the kernel is a subspace whose dimension is equal to the dimension of the intersection, but the subspace itself need not equal the intersection. It will in this case because the dimension you get is zero, but in general it's misleading to say that this is how you compute the intersection. $\endgroup$ – Jim Mar 6 '14 at 20:09

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