2
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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 $\operatorname{Row}(A)\cap\operatorname{Col}(A) = \{0\}$.

$\endgroup$

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.