True or False? If the square matrix $A=[A_{*1},A_{*2},...A_{*n}]$ has the property that $AA_{*j}=0$ for every $j$, then $A=0$. Note that $A_{*j}$ denotes the $j$th column in the matrix $A$.
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False. Consider $A = \left[\begin{array}{cc} 1 & -1\\ 1 & -1\end{array}\right]$.
Michael Albanese
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Or the nilpotent Jordan block $A=\pmatrix{0&1\ 0&0}$, which I think is a more classical example. – user1551 Sep 18 '13 at 07:45
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Hint: If $b_1,..,b_n$ column vectors and $B=[b_1|\dots|b_n]$ is the matrix out of them, then by the matrix multiplication rule, we get $$AB=[Ab_1\,|\,Ab_2\,|\dots|\,Ab_n]$$ (of course, if the dimension of $b_i$'s equals to the number of columns of $A$).
So, we are talking about the question $A^2=0 \overset{?}{\implies} A=0$.
Berci
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still confused? Why would I add another matrix B? Aren't I only dealing with the one matrix? – mike Sep 18 '13 at 03:03
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1@mike Berci is trying to explain that your problem is equivalent to asking whether $A^2=0$ implies $A=0$. – user1551 Sep 18 '13 at 07:43
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It's a general useful rule. You're right, in this case you need only the $B=A$ case. The other answer gives you examples for $2\times 2$ matrices $A\ne 0$ with $A^2=0$. – Berci Sep 18 '13 at 15:49