# Linear Algebra:Rank

1. Feb 21, 2013

1. The problem statement, all variables and given/known data
Let A be the following 3 × 3 matrix:
A =([4 2 6],[2 1 3],[2 1 3])
i) Find the rank of A
ii) Show that there exists an 3 × 2 matrix W, of rank 2, such that AW = 0.
iii) Construct one such matrix W.

2. Relevant equations

3. The attempt at a solution

I think the answer to part 1 is rank(A)=1 since row 2 and row 3 are the same and row 1 is just twice row 2 or twice row 3. So there is only one independent row.

I'm not sure how to prove the existence of W however. Do I use the rank-nullity theorem in some way? i.e. rank(a)+nullity(a)=n where A is a mxn matrix. In this case the nullity(A)=1. Any help would be much appreciated.

Last edited: Feb 21, 2013
2. Feb 21, 2013

### HallsofIvy

Staff Emeritus
Yes, the rank of A is 1. Frankly, I would "prove" the existence of W by finding W! (Especially since the third part of the problem asks you to find it.)

You want a "3 x 2" matrix, W, such that AW= 0. Okay:
$$\begin{bmatrix}4 & 2 & 6 \\ 2 & 1 & 3 \\ 2 & 1 & 3\end{bmatrix}\begin{bmatrix}a & b \\ c & d \\ e & f\end{bmatrix}= \begin{bmatrix}4a+ 2c+ 6e & 4b+ 2d+ 6f \\ 2a+ c+ 3e & 2b+ d+ 3f \\ 2a+ c+ 3e & 2b+ d+ 3f \end{bmatrix}= \begin{bmatrix}0 & 0 \\ 0 & 0 \\ 0 & 0\end{bmatrix}$$

So we have the equations 4a+ 2c+ 6e= 0, 4b+ 2d+ 6f= 0, 2a+ c+ 3e= 0, 2b+ d+ 3f= 0, 2a+ c+ 3e= 0, and 2b+d+ 3f= 0. Of course, the third and fifth equations are the same as are the fourth and sixth equation. In fact, those reduce to only two independent equations. Solve those so that the W matrix has rank 2.

3. Feb 21, 2013

### jbunniii

Yes, that's right.
You have $\textrm{rank}(A) = 1$, and $n = 3$, so $\textrm{nullity}(A) = 2$, not $1$. That means that you can find two linearly independent vectors, say $w_1$ and $w_2$, in the kernel (null space) of $A$. That is to say, they will satisfy $Aw_1 = 0$ and $Aw_2 = 0$. If you construct your $W$ matrix so that its two columns consist of $w_1$ and $w_2$, then you can easily check that this will give you $AW = 0$.

4. Feb 21, 2013

### jbunniii

Halls beat me to it! And his method will be more useful for the next part as it shows you how to find $w_1$ and $w_2$. Mine is just an existence proof, but hopefully it makes it clear why it should be possible to find such a $W$.

5. Feb 23, 2013