Find basis for $ImT$ and basis for $kerT$.
$v_1=(1,1,1)$
$v_2=(0,0,1)$
$v_3=(0,1,1)$
$B=(v_1,v_2,v_3) \in R^3$
My solution
$[T]_B=\left[\begin{array}{cccc}
1 & 2 & 1 \\
-2 & 0 & 1 \\
1 & 6 & 4 \\
\end{array}\right]$
Lets find $[T]_E$.
$T(1,1,1)=(1,1,1)+(0,0,-2)+(0,1,1)=(1,2,0)$
$T(0,0,1)=(2,2,2)+(0,0,0)+(0,6,6)=(2,8,8)$
$T(0,1,1)=(1,1,1)+(0,0,1)+(0,4,4)=(1,5,6)$
Therefore :
$T(1,0,0)=T(1,1,1)-T(0,1,1)=(1,2,0)-(1,5,6)=(0,-3,-6)$
$T(0,1,0)=T(0,1,1)-T(0,0,1)=(1,5,6)-(2,8,8)=(-1,-3,-2)$
$T(0,0,1)=(2,8,8)$
Hence,
$[T]_E=\left[\begin{array}{cccc}
0 & -1 & 2 \\
-3 & -3 & 8 \\
-6 & -2 & 8 \\
\end{array}\right]$
Let's find $ImT$ :
$\left[\begin{array}{cccc}
0 & -3 & -6 \\
-1 & -3 & -2 \\
2 & 8 & 8 \\
\end{array}\right] \rightarrow … \rightarrow \left[\begin{array}{cccc}
1 & 0 & -4 \\
0 & 1 & 2 \\
0 & 0 & 0 \\
\end{array}\right]$
Therefore $B_{ImT}=\{(4,-2,1)\}$
For $KerT$ calculate $[T]_E$
$[T]_E=\left[\begin{array}{cccc}
0 & -1 & 2 \\
-3 & -3 & 8 \\
-6 & -2 & 8 \\
\end{array}\right] \rightarrow … \rightarrow \left[\begin{array}{cccc}
1 & 0 & -\frac{2}{3} \\
0 & 1 & -2 \\
0 & 0 & 0 \\
\end{array}\right]$
Therefore $B_{KerT} = \{(\frac{2}{3},2,1)\}$
But $KerT+ImT \neq 3$
Any ideas, what went wrong? Thanks.
Best Answer
If $\big\{\vec v_1= (1,1,1), \, \vec v_2= (0,0,1),\, \vec v_3=(0,1,1)\big\}$ form a basis of $\mathbb R^3$, then you can show that any $(x,y,z)\in \mathbb R^3$ can be written as $(x,y,z)=x\cdot \vec v_1 +(z-y) \cdot\vec v_2 +(y-x) \cdot \vec v_3$.
So, $T(x,y,z)\begin{array}[t]{l}= x\cdot T(\vec v_1) + (z-y)\cdot T(\vec v_2) + (y-x)\cdot T(\vec v_3)\\= x\cdot(1,2,0)+(z-y)\cdot (2,8,8) +(y-x)\cdot (1,5,6) \\ =(-y+2z,\,-3x-3y+8z,\, -6x-2y+8z)\end{array}$
$\ker T=\big\{(x,y,z)\in \mathbb R^3:T(x,y,z)=(0,0,0)\big\}=\ldots= \big\{\left(x,3x,\frac {3x}{2}\right):x \in \mathbb R\big\}\\\implies \dim \ker T=1$
$ImT=\big\{(X,Y,Z)\in \mathbb R^3:X=-y+2z,\, Y=-3x-3y+8z,\, Z= -6x-2y+8z,\hspace {5pt} x,y,z \in \mathbb R\big\}$
You can easily show that $Z=2Y-4X$.
So, $ImT = \big\{ (X,Y,2Y-4X): X,Y \in \mathbb R\big\}=\big\{X\cdot (1,0,-4)+Y\cdot (0,1,2):X,\, Y \in \mathbb R\big\}$.
That means $ImT=\big\langle (1,0,-4),\, (0,1,2) \big \rangle$, plus $(1,0,-4),\, (0,1,2)$ are linearly independent $\implies \dim ImT=2$.
Thus, we have $\dim \ker T+\dim ImT=\dim\mathbb R^3$.