1
$\begingroup$

I can understand the dimension of column space of matrix is the no. of independent column vectors But why is the dimension of nullspace = no. of free variables?

$\endgroup$
5
$\begingroup$

For an $m \times n$ matrix, $A$, the Rank-Nullity theorem says that: $$ \text{column rank}(A) + \text{nullity}(A) = n.$$ where $\text{nullity}(A)$ is the dimension of the null space of $A$.

When you find the reduced row echelon form of a matrix, the max number of independent columns (i.e. the column rank) is the number of pivot columns (columns containing a leading one for some row). Notice now that free variables correspond to the columns without pivots. So the number of free variables is $n - \text{column rank}(A)$.

So \begin{align*} \text{nullity}(A) &= n - \text{column rank}(A) \\ &= \text{number of free variables}. \end{align*}

$\endgroup$
1
  • $\begingroup$ It seems weird the free variables are the dependent/non-pivot columns. $\endgroup$ Nov 28 '16 at 23:53
0
$\begingroup$

As you had mentioned, $\text{Col } A$ is the set of all linear combinations of $A$'s columns, and so the basis for the vector space $\text{Col } A$ is the set of linearly independent columns of $A$. And so, $\text{rank } A$ is the number of linearly independent columns of $A$.

Now, how about $\text{Nul } A$? $\text{Nul } A$ is the set of all solutions to the homogenous equation $A\mathbf{x} = \mathbf{0}$, and it is said to be defined implicitly, i.e. you'll have to check if a vector $\mathbf{u}$ is in $\text{Nul } A$ by checking if $A\mathbf{u}$ is indeed $\mathbf{0}$. Getting an explicit description of $\text{Nul } A$ amounts to solving for $A\mathbf{x} = \mathbf{0}$, and doing so will give you the spanning set of $\text{Nul } A$, which is automatically linearly independent. The number of vectors in this linearly independent spanning set is equal to the number of free variables in $A$. Give it a try: row reduce, say, a 3 x 4 matrix.

Also, by the rank theorem,

$$\text{rank } A + \text{nullity } A = \text{number of columns in } A$$

That is, because $\text{rank } A$ is the number of $A$'s pivot columns (i.e. the number of linearly independent columns in $A$), then $\text{nullity } A$ is the number of non-pivot columns of $A$ (i.e. the number of free variables in $A$). Hope this helps!

$\endgroup$
0
$\begingroup$

It is definition of dimension that dimension of every space is the number of free variable in it .e.g in a plane you need 2 free variable to define it so the dimension of plane is 2. As Ken Duna proved we know that dimension of the image of A (the space that A can span it ) is the number of independent columns . actually is it obvious because the in $ Ax=b $ (which x and b are vectors) you can see b is made from linear combinations of columns of A and weights are elements of x . In the other word (b = the first columns of A times first element of x + second column of A times second element of x + ....+ last column of A times last element of x. Note that here multiplication is multiplication of vector and scalar and sum is sum of vectors) so $$ b = \sum_{i=0}^n \alpha_i v_i $$ which $v_i$'s are columns of b and $\alpha_i$'s are elements of vector x. knowing that b is span of columns of $A$ now we can say if we want to solve $b = 0$ then we should look at how many of $\alpha_i$'s are free variable e.g if all of them should be zero then dimension of nullspace is zero or if one of them is free and others determined based on that then dimension of nullspace is one . finally consider columns of A. if all of them are linearly independent then all of the $\alpha_i$'s should be zero (dimension of null space would be zero). if some of them are dependent(they can be created by linear combination of other columns) (e.g 3 of them are dependent) then exactly you have 3 $\alpha$ which are free then dimension of nullpace would be 3

$\endgroup$

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.