# Find a $4\times 4$ matrix whose reduced row echelon form has two leading ones and whose row space intersects its column space only at the origin.

Find a $4\times 4$ matrix whose reduced row echelon form has two leading ones and whose row space intersects its column space only at the origin.

Would $$\begin{bmatrix}1& 0& 1&0\\0& 1& 0&1\\0& 0& 0&0\\0&0&0&0\end{bmatrix}$$

be an answer?

## 1 Answer

Yes, it is an example of such a matrix.

First of all, it is $4\times 4$.

It is already in reduced row echelon form so it is easy to see that it has two leading ones.

If we call the matrix $A$, then we have $\operatorname{Row}(A) = \operatorname{span}\{(1, 0, 1, 0), (0, 1, 0, 1)\}$ and $\operatorname{Col}(A) = \operatorname{span}\{(1, 0, 0, 0), (0, 1, 0, 0)\}$. Now suppose $v \in \operatorname{Row}(A)\cap\operatorname{Col}(A)$; as $v \in \operatorname{Row}(A)$, $v = (a, b, a, b)$ for some $a, b \in \mathbb{R}$, and as $v \in \operatorname{Col}(A)$, $v = (c, d, 0, 0)$ for some $c, d \in \mathbb{R}$. Setting $(a, b, a, b) = (c, d, 0, 0)$, we find that $a = b = c = d = 0$, so $v = 0$. That is $\operatorname{Row}(A)\cap\operatorname{Col}(A) = \{0\}$.