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How do I find the null space of this matrix and its dimension (nullity)? It's a $3\times5$ matrix \begin{bmatrix} 1 & 0 & 2 & 1 & 1 \\ 0 & 0 & 1 & 5 &-2 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}

I know the RREF will be \begin{bmatrix} 1 & 0 & 0 & -9 & 5\\ 0 & 0 & 1 & 5 & -2\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} But the number of free variables should be 3 but the 2nd column has all zeros so what should I do about that? When I get the null space, my dimension is 2 but it's supposed to be 3..

egreg
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    You're one step away from the RREF. Then solve the easy equations. – egreg Mar 01 '19 at 18:08
  • I know the RREF will be row 1: [1 0 0 -9 5] row 2: [0 0 1 5 -2] row 3: [0 0 0 0 0] But the number of free variables should be 3 but the 2nd column has all zeros so what should I do about that? When I get the null space, my dimension is 2 but it's supposed to be 3.. –  Mar 01 '19 at 18:13
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    The second variable is free. So you have three of them. Please, add the details to your questions, so it can be accepted as a valid one. – egreg Mar 01 '19 at 18:14
  • See https://math.stackexchange.com/q/1521264/265466. – amd Mar 01 '19 at 19:41

2 Answers2

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You find correctly the RREF, which is equivalent to the linear system \begin{cases} x_1=9x_4-5x_5\\ x_3=-5x_4+2x_5 \end{cases} The second variable is free as well. The bound variables (non free) are those corresponding to the pivots (leading $1$). In this case number one and three. The others are free. You can write the system as \begin{cases} x_1=0x_2+9x_4-5x_5\\ x_3=0x_2-5x_4+2x_5 \end{cases} if you prefer.

A basis is given by the three vectors $$ \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 9 \\ 0 \\ -5 \\ 1 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} -5 \\ 0 \\ 2 \\ 0 \\ 1 \end{bmatrix} $$ The idea is to find the solutions where, in turn $$ \begin{cases} x_2=1 \\ x_4=0 \\ x_5=0 \end{cases} \qquad \begin{cases} x_2=0 \\ x_4=1 \\ x_5=0 \end{cases} \qquad \begin{cases} x_2=0 \\ x_4=0 \\ x_5=1 \end{cases} $$ The three vectors so obtained (computing the corresponding values of $x_1$ and $x_3$) will certainly be linearly independent, so a basis of the null space, whose dimension is “number of columns (that is, of variables) minus the rank”.

egreg
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Recall the definition of the null space of a matrix. The null space of an $m\times{}n$ matrix is the set of all solutions to the homogeneous equation \begin{equation} A\vec{x}=\vec{0} \end{equation}

And what this really means is that we're looking for the vectors $\vec{x}$ such that when we multiply the matrix $A$ with this vector, we get the zero vector!

So before we even row-reduce the matrix, we can atleast figure out where $\vec{x}$ lives, now in our case, $A$ is an $3\times{}5$ matrix, and so my vector $\vec{x}$ must have $5$ rows in order for the multiplication to be defined. We can easily find out the dimension of the null-space of $A$ (nullity of $A$) after row-reducing, and we don't even need the reduced form to figure that out,

The dimension of the null-space is equal to the number of columns of $A$ that $\textbf{do not}$ get a pivot after row-reducing to echelon form. Now notice that in our case, even though the second column has only $0$'s in it, it still doesn't contain a pivot. And so the columns that do not get a pivot (have a free variable) are the second, fourth and fifth column and hence the dimension of $\text{Nul}(A)=3$. Egreg in an earlier post mentioned how you can compute such a basis, but I would also like to add on that for those vectors that form a basis for the null-space, you can easily check that they do belong in the null-space by simply multiplying the matrix $A$ with said vector. And that should give you the zero vector. Used many words in my explanation, hope this helps.

kareem bokai
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