5
$\begingroup$

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 $\mathrm{Rref}(A)$...

$\endgroup$
  • 2
    $\begingroup$ it really doesn't matter whether you divide by 2 or not. I suppose there is some good to removing common divisors. Since the basis for the range is one dimensional the basis for the kernel must be two dimensional. It comes from the 0 vector in the middle column of the matrix you first wrote. it is necessary because multiples of this new vector are independent of multiples of the basis element you found. $\endgroup$ – user71352 Jul 17 '13 at 0:25
6
$\begingroup$

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 $(1,0,0)^T+2(0,0,1)^T=(1,0,2)^T\in\ker(A)$

$\endgroup$
  • $\begingroup$ Good and Enough hint. $\endgroup$ – mrs Nov 21 '13 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.