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Show that there is no matrix whose row space and nullspace both contain the same nonzero vector.

I've defined the row space and nullspace of a matrix $A$ as such:

$A=\pmatrix{a_1\\...\\a_n}$ where $a_i$ is the $i$th row of $A$ and $u^T$ is a vector of the nullspace of $A$.

Then $Au^T=0 \implies \pmatrix{a_1u^T\\...\\a_nu^T}=0$ where $a_iu^T=0$ for all $i$.

That's as far as I've gotten up to... which honestly isn't very far, haha. Can someone please show me how I should go on from here?

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  • $\begingroup$ You might want to take a look at this, as well. $\endgroup$ Apr 19, 2017 at 17:57
  • $\begingroup$ @thanasissdr thanks for the link, I will check it out! $\endgroup$
    – iamaweed
    Apr 20, 2017 at 3:11

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You're almost there. The only thing you need to observe now is that the rowspace is the orthogonal complement of the nullspace and no vector can be orthogonal to itself :)

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  • $\begingroup$ I'm sorry, I'm quite slow on the uptake! Do you mind elaborating how I can go about proving that observation? Is the orthogonal complement proof the only method I can use for this question? $\endgroup$
    – iamaweed
    Apr 19, 2017 at 17:15
  • $\begingroup$ @amakelov With only exception the zero vector. $\endgroup$ Apr 19, 2017 at 17:45
  • $\begingroup$ @thanasissdr: yep, the zero vector is in its own orthogonal complement $\endgroup$
    – amakelov
    Apr 19, 2017 at 17:52
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    $\begingroup$ @iamaweed so, you know what the orthogonal complement is right? Then by definition, the kernel is the space of vectors orthogonal to the rows, so it is the space of vectors orthogonal to all linear combinations of the rows too, i.e. it is the orthogonal complement of the row space. The orthogonal complement proof is pretty much the only one i can think of. $\endgroup$
    – amakelov
    Apr 19, 2017 at 17:54
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    $\begingroup$ @amakelov Oh, I get what you mean now! Thanks so much for the explanation!! $\endgroup$
    – iamaweed
    Apr 20, 2017 at 3:10
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The row space is given by ${\cal R} A^T$.

Suppose $x \in {\cal R} A^T \cap \ker A$.

Then there is some $y$ such that $x=A^T y$ and also, $A x = 0$.

Substitute the above into $\|x\|^2 = x^T x$ (that is, replace the $x$ by $A^T y$ and use the other equality). What does this say about $x$?

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  • $\begingroup$ This makes it easier to understand, too! Thanks for your input, upvoted! (: $\endgroup$
    – iamaweed
    Apr 20, 2017 at 3:13

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