Prove that AB=0AB = 0 implies det(A+B)2=det(A−B)2\det\left(A + B\right)^2 = \det\left(A – B\right)^2.

Given matrices A, B \in \mathcal{M}_{n \times n}\left(\mathbb{R}\right)A, B \in \mathcal{M}_{n \times n}\left(\mathbb{R}\right), prove that if AB = 0AB = 0, \det\left(A + B\right)^2 = \det\left(A – B\right)^2\det\left(A + B\right)^2 = \det\left(A – B\right)^2.

This is based on a quiz question from a few weeks ago that I never quite solved, and I forgot my instructors’ solutions as well.

Here are possible ideas I had:

Show that AB = BAAB = BA (except it’s not true).
Show that BA = -BABA = -BA—wait, isn’t this equivalent to (1)?
Show that \forall i,j\, \left(A_{ij} = 0\right) \mathrm{or} \left(B_{ij}=0\right)\forall i,j\, \left(A_{ij} = 0\right) \mathrm{or} \left(B_{ij}=0\right) so that \det\left(A + B\right) = \left(-1\right)^k\det\left(A – B\right)\det\left(A + B\right) = \left(-1\right)^k\det\left(A – B\right) for some kk. This is definitely a sufficient condition, but I’m not sure if it’s necessary.

We have learned diagonalization, but we have not learned the Jordan canonical form yet, so it would be ideal if you avoided that kind of answer.

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1 Answer
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\renewcommand\im{\operatorname{im}}\renewcommand\im{\operatorname{im}}Consider a basis of \mathbb{R}^n\mathbb{R}^n formed by taking a basis of the image \operatorname{im}(B)\operatorname{im}(B) and extending to a basis of \mathbb{R}^n\mathbb{R}^n. Then \im(B) \subseteq \ker(A)\im(B) \subseteq \ker(A), so in this basis, the matrices of AA and BB are block matrices

A = \begin{bmatrix}
0 & A_1 \\ 0 & A_2
\end{bmatrix},\quad
B = \begin{bmatrix}
B_1 & B_2 \\ 0 & 0
\end{bmatrix},

A = \begin{bmatrix}
0 & A_1 \\ 0 & A_2
\end{bmatrix},\quad
B = \begin{bmatrix}
B_1 & B_2 \\ 0 & 0
\end{bmatrix},

where B_1B_1 is r \times rr \times r (where r= \operatorname{rank}(B)r= \operatorname{rank}(B)), A_1A_1 and B_2B_2 are r \times (n-r)r \times (n-r), and A_2A_2 is (n-r) \times (n-r)(n-r) \times (n-r).
Why? If \{v_1,\cdots v_n\}\{v_1,\cdots v_n\} is a basis of \mathbb{R}^n\mathbb{R}^n with \{v_1,\cdots,v_r\}\{v_1,\cdots,v_r\} a basis of \im(B)\im(B), then Bv_iBv_i is in \im(B)\im(B) for all ii, so the last n-rn-r rows of BB are zero. Also, Av_i = 0Av_i = 0 for i \le ri \le r since v_i \in \im(B)v_i \in \im(B), so the first rr columns of AA are zero.

Then

A \pm B = \begin{bmatrix}
\pm B_1 & A_1 \pm B_2 \\ 0 & A_2
\end{bmatrix}

A \pm B = \begin{bmatrix}
\pm B_1 & A_1 \pm B_2 \\ 0 & A_2
\end{bmatrix}

has determinant \pm \det(B_1) \det(A_2)\pm \det(B_1) \det(A_2) (it’s block triangular), so the squares have the same determinant.

  

 

I do not understand the middle step in the first paragraph. Can you explain how this basis produces those block matrices? even if AA or BB has full rank?
– Simon Kuang
2 days ago

  

 

I edited the answer with some explanation. Hopefully it makes it clear. If one of the matrices has full rank then it is invertible, so the other matrix must be zero. (so rr or n-rn-r is zero)
– arkeet
2 days ago