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Let $A$ be a real $3 \times 3 $ matrix such that rank$(A) = 2$.

Prove that $A^2 \neq 0_3$. where $0_3$ represents the null matrix of order $3$.

I am looking for a solution involving only basic manipulation using matrices. I already have a better solution using the range and the nullity of $A$.

Thank you in advance!

Edit. No Sylvester's inequality, Jordan form or range+nullity / linear transformations. At most use the definition of the rank as the dimension of the column/row space.

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  • $\begingroup$ What is $O_3$? The 3 by 3 zero matrix? $\endgroup$
    – JavaMan
    Oct 21 '18 at 22:10
  • $\begingroup$ @JavaMan yes, it is. $\endgroup$
    – user606835
    Oct 21 '18 at 22:10
  • $\begingroup$ Can you use Jordan decompositon? $\endgroup$
    – ALG
    Oct 21 '18 at 22:11
  • $\begingroup$ @ALG No, I want a solution that does not involve things from advanced linear algebra, just basic matrix manipulation. $\endgroup$
    – user606835
    Oct 21 '18 at 22:12
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Let us write $A=(C_1,C_2,C_3)$ where $C_i$ is the column $i$ of $A.$

Assume $A^2=0.$ That is, we have that $C_1,C_2,C_3$ are two linearly independent solutions of the system $Ax=0.$ But since $A$ is of order $3$ and has rank $2$ this is not possible.

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  • $\begingroup$ +1 nice answer! $\endgroup$ Oct 21 '18 at 22:22
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If the rank of $A$ is $2$, then there exists a subspace spanned by the non-null vector $v_1$ such that:

$$Av_1 = 0,$$

i.e. $v_1 \in \ker(A).$ On the other hand, there must be at least another non-null vector $v_2$ which is an eigenvector of $A$ for an eigenvalue $\lambda_2 \neq 0$, i.e.

$$Av_2 = \lambda_2 v_2.$$

Of course, $v_1 \in \ker(A^2).$ Moreover:

$$A^2v_2 = A(Av_2) = A(\lambda_2 v_2) = \lambda_2 Av_2 = \lambda_2^2 v_2.$$

This means that $\ker(A^2) \neq \mathbb{R}^3.$ Therefore, $A^2 \neq 0_3$.

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