If rank$(A) = 2$, then $A^2 \neq 0_3$

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 $$A$$.

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.

• What is $O_3$? The 3 by 3 zero matrix? Oct 21 '18 at 22:10
• @JavaMan yes, it is.
– user606835
Oct 21 '18 at 22:10
• Can you use Jordan decompositon?
– ALG
Oct 21 '18 at 22:11
• @ALG No, I want a solution that does not involve things from advanced linear algebra, just basic matrix manipulation.
– user606835
Oct 21 '18 at 22:12

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.

• +1 nice answer! Oct 21 '18 at 22:22

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, $$A^2 \neq 0_3$$.