From the method given to me in my textbook, to find the spanning set of a null space. we row reduce our matrix to find the basic solutions of the parameters. However, it seems like the process is exactly the same when trying to find our basis for the null space. Since the definition of a basis is that it spans the set and it is linearly independent, is the basis of a null space the same as its spanning set?
Is the basis of a null space the same as its spanning set
linear algebralinear-transformationsmatrices
Related Solutions
The null space of $A$ is the set of solutions to $A{\bf x}={\bf 0}$. To find this, you may take the augmented matrix $[A|0]$ and row reduce to an echelon form. Note that every entry in the rightmost column of this matrix will always be 0 in the row reduction steps. So, we may as well just row reduce $A$, and when finding solutions to $A{\bf x}={\bf 0}$, just keep in mind that the missing column is all 0's.
Suppose after doing this, you obtain $$ \left[\matrix{1&0&0&0&-1 \cr 0&0&1&1&0 \cr 0&0&0&0&0 \cr 0&0&0&0&0 \cr }\right] $$
Now, look at the columns that do not contain any of the leading row entries. These columns correspond to the free variables of the solution set to $A{\bf x}={\bf 0}$ Note that at this point, we know the dimension of the null space is 3, since there are three free variables. That the null space has dimension 3 (and thus the solution set to $A{\bf x}={\bf 0}$ has three free variables) could have also been obtained by knowing that the dimension of the column space is 2 from the rank-nullity theorem.
The "free columns" in question are 2,4, and 5. We may assign any value to their corresponding variable.
So, we set $x_2=a$, $x_4=b$, and $x_5=c$, where $a$, $b$, and $c$ are arbitrary.
Now solve for $x_1$ and $x_3$:
The second row tells us $x_3=-x_4=-b$ and the first row tells us $x_1=x_5=c$.
So, the general solution to $A{\bf x}={\bf 0}$ is $$ {\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right] $$
Let's pause for a second. We know:
1) The null space of $A$ consists of all vectors of the form $\bf x $ above.
2) The dimension of the null space is 3.
3) We need three independent vectors for our basis for the null space.
So what we can do is take $\bf x$ and split it up as follows:
$$\eqalign{ {\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right] &=\left[ \matrix{0\cr a\cr 0\cr 0\cr 0}\right]+ \left[\matrix{c\cr 0\cr 0\cr 0\cr c}\right]+ \left[\matrix{0\cr 0\cr -b\cr b\cr 0}\right]\cr &= a\left[ \matrix{0\cr1\cr0\cr 0\cr 0}\right]+ c\left[ \matrix{1\cr 0\cr 0\cr 0\cr 1}\right]+ b\left[ \matrix{0\cr 0\cr -1\cr 1\cr 0}\right]\cr } $$ Each of the column vectors above are in the null space of $A$. Moreover, they are independent. Thus, they form a basis.
I'm not sure that this answers your question. I did a bit of "hand waving" here. What I glossed over were the facts:
1)The columns of the echelon form of $A$ that do not contain leading row entries correspond to the "free variables" to $A{\bf x}={\bf 0}$. If the number of these columns is $r$, then the dimension of the null space is $r$ (again, if you know the dimension of the column space, you can see that the dimension of the null space must be the number of these columns from the rank-nullity theorem).
2) If you split up the general solution to $A{\bf x}={\bf 0}$ as done above, then these vectors will be independent (and span of course since you'll have $r$ of them).
Yes. The following three terms are equivalent (for a vector space!):
- A linearly independent spanning set.
- A minimal spanning set.
- A maximal linearly independent set.
The first obviously implies the second and third. To see that 2. implies 1., suppose that if $\{x_1,\ldots,x_m\}$ is a minimal spanning set, but not a basis. Then, for some constants $\alpha_1,\ldots,\alpha_m$, not all zero, we have that
$$\displaystyle \alpha_1 x_1+\cdots+\alpha_m x_m=0$$
So, assume that $\alpha_1\ne 0$. Then,
$$x_1=\frac{-\alpha_2}{\alpha_1}x_2+\cdots+\frac{-\alpha_m}{\alpha_1}x_m$$
Thus, $\{x_2,\ldots,x_m\}$ is a spanning set (why?) and thus this contradicts minimality.
You and try to prove that 3 implies 1.
Best Answer
Every basis is a spanning set, but not every spanning set is a basis. The difference is that a spanning set need not be linearly independent.
More exactly, a basis is a minimal spanning set, that is, if you remove any vector from it, the resulting set no longer is a spanning set.