0
$\begingroup$

Suppose we are given a matrix $A$ representing a linear map, where the $A=\begin{pmatrix}1&1&1&3\\1&1&2&4\\1&1&1&3\end{pmatrix}$. I need to find the kernel and image. To begin, we find the kernel:

$\begin{pmatrix}1&1&1&3\\1&1&2&4\\1&1&1&3\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$. This gives the following 2 equations:

$x_1+x_2+x_3+3x_4=0$

$x_1+x_2+2x_3+4x_4=0$

Subtracting, we obtain $x_4=-x_3$. As such the solution set for the kernel is equal to $x_1\begin{pmatrix}1\\0\\0\\0\end{pmatrix}+x_2\begin{pmatrix}0\\1\\0\\0\end{pmatrix}+x_3\begin{pmatrix}0\\0\\1\\-1\end{pmatrix}$, and so the nullity is $3$.

For the image : we need the set of all $A\bf{x}: $ $\bf{x}=\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}$. We do the following:

$\begin{pmatrix}1&1&1&3\\1&1&2&4\\1&1&1&3\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=\begin{pmatrix}a\\b\\c\end{pmatrix}$.

This gives $a=c$ immediately, and $x_4=x_3+a-b$. So the solution set for $\bf x$ is given by $x_1\begin{pmatrix}1\\0\\0\\0\end{pmatrix}+x_2\begin{pmatrix}0\\1\\0\\0\end{pmatrix}+x_3\begin{pmatrix}0\\0\\1\\0\end{pmatrix}+\begin{pmatrix}0\\0\\0\\a-b\end{pmatrix}$.

If we take $a=b+1$ we get the whole of $\mathbb{R}^4$.

Dimension of space $=4$, nullity $=3$, rank $=4$. But this contradicts the rank nullity theorem. Where have I gone wrong?

$\endgroup$

1 Answer 1

1
$\begingroup$

The nullity of the matrix is not $3$. The equation $x_4=-x_3$ is not a sufficient condition for $x$ to be in the kernel. For example, the vector $\begin{pmatrix}1\\1\\0\\0\end{pmatrix}$ satisfies the condition $x_4=-x_3$, but it is not in the kernel of the matrix.


Where you went wrong is when you started with two equations,

$$x_1+x_2+x_3+3x_4=0\\ x_1+x_2+2x_3+4x_4=0,$$

and then reduced the two equations to one equation. That should ring alarm bells immediately. The number of equations you work with should not decrease. If the number of equations decreases, then you can only assume that any solution to the original equations is also a solution to the smaller set of equation, but not the other way around.


For example, take a system of two equations,

$$x_1+x_2=0\\ x_1-x_2=0.$$

From these two equations, if I sum them up, I get $2x_1=0$ and therefore $x_1=0$.

And indeed, if $x_1, x_2$ is a solution to the original system of equations, then $x_1=0$. However, if $x_1=0$, then we cannot say that $x_1,x_2$ is a solution to the original system of equations. For example, $x_1=0, x_2=1$ is not a solution.


To correct your solution, you should work with equations using only reversible steps. You can, for example, transform any set of linear equations to a row echelon form. The steps taken to do so are reversible, so they will guarantee you do not lose any solution.

$\endgroup$
4
  • $\begingroup$ How do I rectify this? $\endgroup$
    – user960711
    Oct 29, 2021 at 13:16
  • $\begingroup$ @alidixon222 You need to extract one more equation from the two original equations. $\endgroup$
    – 5xum
    Oct 29, 2021 at 13:18
  • $\begingroup$ @alidixon222 For example, finding the row echelon form of a matrix guarantees no solutions are lost. I suggest you do that. $\endgroup$
    – 5xum
    Oct 29, 2021 at 13:21
  • $\begingroup$ So in row echelon form we have 2 nonzero rows, so rank = 2. Essentially now I need to show the nullity is 2, how do I do that? $\endgroup$
    – user960711
    Oct 29, 2021 at 14:51

You must log in to answer this question.