Find a basis for the range and kernel of $T$.
$$A =\begin{bmatrix}
2 & 0 & -1\\
4 & 0 & -2\\
0 & 0 & 0
\end{bmatrix} $$
Attempt at Solving for Basis of Range:
On finding the basis for the range, I know that the range is the same thing as the column space. So, finding the basis of the column space should be equivalent to finding the basis of the range. I got the following after reducing:
$$\mathrm{Rref}(A) =\begin{bmatrix}
1 & 0 & -\frac 12\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix} $$
Because I only have a pivot in column $1$, my corresponding column in the original $A$ (thus my basis for the range), would be:
$$\mathrm{Rref}(A) =\begin{bmatrix}
2\\
4\\
0
\end{bmatrix} $$
The solution given in the text says that this should actually be:
$$\mathrm{Rref}(A) =\begin{bmatrix}
1\\
2\\
0
\end{bmatrix}$$
I can see that the book solution only differs from mine in that their solution seems to have been divided by $2$. However, my question is – why are they dividing by 2? I don't understand why the original column is being altered.
Attempt at Solving for Basis of Kernel:
In solving for the kernel, I know that the basis of the kernel should be the same as the basis for the nullspace. From the $\mathrm{Rref}(A)$ above, I got the following equation:
$(X_1) = (\frac12X_3)$. Letting $X_3$ equal one, I got the following matrix for the basis of the kernel:
$$\text{Basis of Kernel} =\begin{bmatrix}
\frac12\\
0\\
1
\end{bmatrix}$$
This answer checks out with my solution in the text, but the text also provides the following solution:
=\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
I'm guessing this comes from the zero row in my $\mathrm{Rref}(A)$. But, why is this done? I thought the number of bases came from the number of independent variables in the $\mathrm{Rref}(A)$…
Best Answer
Notice that if a vector $v$ span a subspace then for all $\lambda\ne 0$, $\lambda v$ span also this subspace so in your example we have $$\mathrm{Im}(A)=\mathrm{span}((2,4,0)^T)=\mathrm{span}((1,2,0)^T)=\mathrm{span}((-100,-200,0)^T)$$
For the kernel: By the rank nullity theorem we have $\dim \ker(A)=2$ and we can see easily from the matrix $A$ (since the second column is zero) that $(0,1,0)^T\in \ker(A)$ so we should find another vector in the kernel: since the first column of the matrix is $(-2)$ times the third column so $(1,0,0)^T+2(0,0,1)^T=(1,0,2)^T\in\ker(A)$