Possible Ranks of BA Given That AB Has Rank 2

Suppose $A$ and $B$ are real $4 \times 4$ matrices with $\operatorname{rank}(AB)=2$. Find every value that $\operatorname{rank}(BA)$ can take. Prove the upper and lower bounds, and show that each integer in the stated range can be achieved.

Constraints & Assumptions

• No invertibility, symmetry, or commutativity assumption is made.
• A complete solution needs constructions or a general existence argument, not just rank inequalities.

Clarifying Questions to Ask

• Are explicit matrices required for the attainable ranks?
• May standard rank-nullity and image/kernel formulas be used?

Hint: compare images with kernels. Use $\operatorname{rank}(AB)=\operatorname{rank}(B)-\dim(\operatorname{image}(B)\cap\ker(A))$, and apply the corresponding formula to $BA$.

What a Strong Answer Covers

• Why rank four for $BA$ is impossible.
• Why rank zero is not excluded by $\operatorname{rank}(AB)=2$.
• Constructions for all attainable intermediate ranks.
• Correct use of rank, image, and kernel dimensions.

Follow-up Questions

• How does the range change for $n \times n$ matrices with $\operatorname{rank}(AB)=r$?
• What additional conclusion follows if either factor is invertible?

Overview: For real $4 \times 4$ matrices with $\operatorname{rank}(AB)=2$, determine the possible range of $\operatorname{rank}(BA)$.

Community answers

Answer by shpg

What the clarifying questions mean

For “Are explicit matrices required for the attainable ranks?” There are two possible types of proof, constructive and non-constructive, so this is really asking: “Do I need to produce the matrices, or is it enough to prove that they exist?”

For “May standard rank-nullity and image/kernel formulas be used?” This asks whether the full set of linear algebra tools is available, or whether the proof must stay within Sylvester's inequality and basic rank bounds.

Answer

Upper bound: this follows from Sylvester's inequality.

$\operatorname{rank}(AB)=2 \ge \operatorname{rank}(A)+\operatorname{rank}(B)-n$

where $n=4$ because both matrices are $4 \times 4$. Therefore

$\operatorname{rank}(A)+\operatorname{rank}(B)\le 6.$

Also,

$\operatorname{rank}(BA)\le\min(\operatorname{rank}(A),\operatorname{rank}(B))\le\frac{\operatorname{rank}(A)+\operatorname{rank}(B)}{2}\le 3.$

Lower bound: attainable ranks can be shown by explicit matrices.

Let

$A=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}.$

Case $\operatorname{rank}(BA)=2$: take $B=I_4$, the $4 \times 4$ identity matrix. Then $AB=A$, so $\operatorname{rank}(AB)=2$, and $BA=A$, so $\operatorname{rank}(BA)=2$.

Case $\operatorname{rank}(BA)=1$: take

$B=\begin{pmatrix}1&1&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0\end{pmatrix}.$

Then $BAe_1=Be_1=e_1$, $BAe_2=Be_2=e_1$, and $BAe_3=BAe_4=0$, so $\operatorname{rank}(BA)=1$. Also $ABe_1=Ae_1=e_1$, $ABe_2=Ae_1=e_1$, $ABe_3=Ae_2=e_2$, and $ABe_4=0$, so $\operatorname{rank}(AB)=2$.

Case $\operatorname{rank}(BA)=0$: take

$B=\begin{pmatrix}0&0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0\end{pmatrix}.$

Then $BAe_1=Be_1=0$, $BAe_2=Be_2=0$, and $BAe_3=BAe_4=0$, giving $\operatorname{rank}(BA)=0$. Moreover, $ABe_3=Ae_1=e_1$, $ABe_4=Ae_2=e_2$, and $ABe_1=ABe_2=0$, so $\operatorname{rank}(AB)=2$.

Follow-up questions

General $n \times n$ case: if $\operatorname{rank}(AB)=r$, then $0\le\operatorname{rank}(BA)\le r$. If either $A$ or $B$ is invertible, then $\operatorname{rank}(BA)=\operatorname{rank}(AB)=r$.