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Let V be a vector space with basis $u_1, u_2, u_3$ and let $T : V → V$ be the linear transformation such that

$T(u_1)=u_1 −u_2 −u_3$
$T(u_2)=2u_2 −6u_3$
$T(u_3)=−u_1 +3u_2 −5u_3$

Find bases for the null space and the image of T and determine the rank and nullity of T .

I think I solved for the null space and nullity first.
After solving the system of equations with the solutions equal to $0$, I came up with $u_3=k$ which gave $u_1=4k$ and $u_2=3k$ which gives $(4k,3k,k)$ nullity 3. Was this the correct method (gaussian elimination)?

And for solving for the basis of the image would I use the same method but instead of having $0$ in the right-most column for the solutions I would have 1,2,-5 (in descending order) corresponding to the coefficient in the given equation.
*After a second look I'm almost sure this is incorrect since the coefficients aren't the same as the T($u_n$)

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The matrix of $T$ with respect to the given basis is $$ A=\begin{bmatrix} 1 & 0 & -1 \\ -1 & 2 & 3 \\ -1 & -6 & -5 \end{bmatrix} $$ Let's perform Gaussian elimination: $$ \begin{bmatrix} 1 & 0 & -1 \\ -1 & 2 & 3 \\ -1 & -6 & -5 \end{bmatrix} \xrightarrow{\begin{aligned}R_2&\gets R_2+R_1\\R_3&\gets R_3+R_1\end{aligned}} \begin{bmatrix} 1 & 0 & -1 \\ 0 & 2 & 2 \\ 0 & -6 & -6 \end{bmatrix} \xrightarrow{R_2\gets\tfrac{1}{2}R_2} \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & -6 & -6 \end{bmatrix} \xrightarrow{R_3\gets R_3+6R_2} \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} $$ The rank of this matrix is $2$ and a basis for the column space is given by the first two column of the original matrix.

This implies that $\{T(u_1),T(u_2)\}$ is a basis for the image of $T$.

A basis for the null space of the matrix is given by the single vector $$ \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} $$ so a basis for the null space for $T$ is given by the single vector $u_1-u_2+u_3$.

Just recall that the image/null space of $T$ correspond to the column space/null space of the matrix $A$ via the coordinate map.

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  • $\begingroup$ Thank you for the answer, I'm confused as to what I solved for in my case. Does the ($4k,3k,k$) have any significance? $\endgroup$ – Justtryingtodomath Feb 16 '18 at 14:50
  • $\begingroup$ @Justtryingtodomath Sorry, it means nothing. $\endgroup$ – egreg Feb 16 '18 at 14:54
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$imT=\{(T(u_1),T(u_2),T(u_3)):u_1,u_2,u_3 \in V \}=\{(u_1-u_2-u_3,2u_2-6u_3,-u_1+3u_2-5u_3):u_1,u_2,u_3 \in V\}=\{u_1(1,0,-1)+u_2(-1,2,3)+u_3(-1,-6,-5):u_1,u_2,u_3 \in V \}$

$=<(1,0,-1),(-1,2,3),(-1,-6,-5)>$,but $-4(1,0,-1)=3(-1,2,3)+(-1,-6,-5) $ so

$imT=<(1,0,-1),(-1,2,3)>\Rightarrow dim(imT)=2$

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