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Our exercise is to find all solutions to the equation $Ax = 0$, among others for the following matrix

$$A =\begin{pmatrix} 6 & 3 & -9 \\ 2 & 1 & -3 \\ -4 & -2 & 6 \end{pmatrix}.$$

This amounts to finding the kernel, and obviously, the rows of the matrix are multiples of each other, so we can reduce the equations to:

$$6x_1 + 3x_2 - 9x_3 = 0.$$

Choosing $x_3 = 0$, $x_1 = 1$ and $x_2 = -2$ would fulfill the equation in my opinion.

However, the official solution is described as the following set:

$$\lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}$$

with $\mu, \lambda \in$ R.

My questions now:

Where does the vector (and its multiples) $\lambda \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$ come from? Also, since after the reduction we are left with only one row, the dimension of the kernel should accordingly be $1$ - however $\left[\begin{smallmatrix} 0 \\ 3 \\ 1 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 1 \\ -2 \\ 0 \end{smallmatrix}\right]$ (from $x_3 = 0$, $x_1 = 1$ and $x_2 = -2$) seem to be two different basis vectors - what have I done or understood wrongly?

Many thanks

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  • $\begingroup$ Actually, the dimension of the kernel should be 3-1=2. It is the range that is of dimension 1. Specifically, your reasoning of "the matrix is only left with only one row after the reduction" (more technically: there is only 1 linearly independent row) actually leads to the conclusion that the dimension of the range is 1, not the kernel. $\endgroup$
    – suncup224
    Aug 3, 2015 at 11:12

3 Answers 3

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The dimension of the subspace given by $6x_1+3x_2-9x_3=0$ is $2$. This is the equation of a plane in $\mathbb{R}^3$.

The number of independant equations defining a subspace is not the dimension of this subspace. It is $n-$ dimension of the subspace. If you had $2$ independant equations left, it would be a line, because $3-2=1$ and the dimension of a line is $1$.

Then, once you know that the dimension of the subspace is $2$, all you have ti do to find a basis is finding two linearly independant vectors satisfying its equation. Just looking at it, it is obvious that $(1, 1, 1)$ is a solution. In order to find the other, you can just fix two coordinates (e.g. $x=1$ and $z=1$ and deduce the third one using the equation.

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You correctly found that $$\ker A= \{ (x_1,x_2,x_3) \in \mathbb R^3 :6x_1 +3x_2 - 9x_3 = 0\},$$ which is equivalent to $$\ker A = \left\{\left( x_1, x_2, \frac {2x_1}3 +\frac {x_2}3\right): x_1, x_2 \in \mathbb R\right\}.$$

If we move forward one step we have that $$\ker A = \left\{x_1 \cdot \left(1,0,\frac 23\right) +x_2 \left(0,1,\frac 13\right) : x_1,x_2 \in \mathbb R\right\}.$$

That means that the kernel is spanned by the $2$ linearly independent vectors $(1,0,\frac 23),\,(0,1,\frac 13)$ (thus, $\dim\ker A = 2)$, or equivalently the kernel of $A$ is spanned by any $2$ linearly independent vectors which can be written as a linear combination of the 2 above vectors.

  • First case: Consider $x_1 = 0, x_2 = 3\implies e_1 = (0,3,1).$

  • Second case: Consider $x_1 =1 , x_2 =1 \implies e_2 =(1,1,1). $

We can easily check that $e_1,e_2$ are linearly independent vectors.

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First, note that $$\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1\\1\\1\end{pmatrix} - \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}, $$ So your missing solution $\left[\begin{smallmatrix} 1 \\ -2 \\ 0 \end{smallmatrix}\right] $ is, in fact, a special case of the general solution $ \lambda \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right] + \mu \left[\begin{smallmatrix} 0 \\ 3 \\ 1 \end{smallmatrix}\right] $ with $\lambda =1$ and $\mu = -1$.

Second, the general solution can be obtained from reduced equation $6x_1 + 3x_2 - 9x_3 = 0$ by imposing certain assumptions:

  • For example, if we assume $x_1 = 0$, we will get the following $$ \begin{cases} x_1 = 0 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases} \implies \begin{cases}x_1 = 0 \\ 3x_2 - 9x_3 = 0 \end{cases} \implies \begin{cases}x_1 = 0 \\ x_2 = 3 x_3\end{cases} $$ Assuming parametrization $x_3 = \mu$, we get $$ \begin{cases} x_1 = 0 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases} \implies \begin{cases}x_1=0 \\ x_2 = 3 x_3 \\ x_3 = \mu & - \operatorname{parameter}\end{cases} \iff \begin{cases}x_1 = 0 \\ x_2 = 3\mu \\ x_3 = \mu \end{cases} $$ We can rewrite this solution in the vector form: $$ \begin{cases}x_1 = 0 \\ x_2 = 3\mu \\ x_3 = \mu \end{cases} \iff x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 3\mu \\ \mu \end{bmatrix} \iff x = \begin{bmatrix} 0 \\ 3 \\ 1\end{bmatrix} \mu $$

  • Similarly, imposing condition $x_1 = x_2$, we write $$ \begin{cases} x_1 = x_2 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases} \implies \begin{cases}x_1 = x_2 \\ 6x_1 + 3x_2 - 9x_3 = 0 \end{cases} \implies \begin{cases}x_1 = x_2 \\ x_1 = x_3\end{cases} $$ Assuming parametrization $x_1 = \lambda$, we get $$ \begin{cases} x_1 = \lambda & -\operatorname{parameter}\\ x_2 = x_1 \\ x_3 = x_1 \end{cases} \implies \begin{cases} x_1 = \lambda \\ x_2 = \lambda \\ x_3 = \lambda \end{cases} $$ The vector form of the solution then looks like $$ \begin{cases} x_1 = \lambda \\ x_2 = \lambda \\ x_3 = \lambda \end{cases} \implies x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} \lambda \\ \lambda \\ \lambda \end{bmatrix} \iff x= \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix} \lambda $$

Third, Since parametrized solutions $x = x(\lambda)$ and $x = x(\mu)$ are linearly independent, and since the equation $ 6x_1 + 3x_2 - 9x_3 = 0$ is linear, the general solution can be written as the sum of two independent parametrized particular solutions, i.e. $$ x = x_\lambda + x_\mu = \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix} \lambda + \begin{bmatrix} 0 \\ 3 \\ 1\end{bmatrix} \mu $$


Finally, Note that we could have imposed different assumptions on $x$, and that would result in different parametrized solutions. However, as long as these solutions will be linearly independent, the will still span the same two-dimentional space of solutions which we have now.

Thus, the kernel of $A$, which is also a space of all solutions of the system $Ax = 0$, is the linear span of vectors $\left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 0 \\ 3 \\ 1 \end{smallmatrix}\right]$: $$ \ker A = \operatorname{span} \left(\; \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3 \\ 1\end{bmatrix} \;\right) = \left\{\; \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3 \\ 1\end{bmatrix} \; \right\} = \left\{\ \vec v = \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix} \lambda + \begin{bmatrix} 0 \\ 3 \\ 1\end{bmatrix} \mu \mathrel{\bigg|} \lambda, \mu \in \mathbb R\ \right\} $$

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  • $\begingroup$ Thank you very much - very helpful! $\endgroup$
    – Pugl
    Aug 3, 2015 at 12:23

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