# Show that: If $A \circ B =0$ but $A \neq 0$ and $B\neq 0$ then you have that $\det(A)=0=\det(B)$

Show that: If $A$ and $B$ are real $n \times n$ matrices with $A \circ B=0$ but $A \neq 0$ and $B \neq 0$, then you have that $\det(A)=0=\det(B)$

($0$ is the matrix which only has zeroes)

I'm not sure how to do this correctly but something tells me that a proof by contradiction would be very useful here.

Let $\det(A) \neq 0 \Rightarrow B = Id \circ B = A^{-1} AB = 0 \Rightarrow B=0 \,\,\,\unicode{x21af}$

Let $\det(B) \neq 0 \Rightarrow A = Id \circ A = B^{-1} BA = 0 \Rightarrow A=0 \,\,\,\unicode{x21af}$

And that's why we must have that $\det(A)=0=\det(B)$

I think this should work because we just showed a contradiction for every possibility and this means that the negation of our assumption must be correct.

But will this also cover that both determinants must necessarily be zero? I think it only shows that at least one determinant must be zero... : /

Maybe there is a better way to prove this?

• In second case, you need $A=ABB^{-1}=0$ but otherwise this is fine. – Angina Seng May 8 '18 at 20:20
• Dose $\circ$ means component-wise matrix multiplication? – Mahdi May 9 '18 at 18:52
• @Mahdi Yes! It's also called "Hadamard product". As far as I know, you can only use it on matrices of same size and for this reason the Hadamard product is also always commutative (the normal matrix product isn't). en.wikipedia.org/wiki/Hadamard_product_(matrices) – cnmesr May 9 '18 at 21:49

Your proof using a property of determinant is OK. Here is a second proof:

We have $A\circ B=0$ so for every $X\in\mathcal M_{n,1}(\Bbb R)$:

$$A(BX)=0\implies Im(B)\subset \ker A$$

If $B$ is invertible then $Im(B)=\mathcal M_{n,1}(\Bbb R)=\ker A\implies A=0$ which is a contradiction so $B$ isn't invertible. Moreover, $B\ne0$ so $Im B\ne\{0\}$ and so $\ker A\ne\{0\}$ which means that $A$ isn't invertible. Ofcourse you can translate the non invertibility in term of determinant.

You can use the formule of determinants $$Det(AB)=Det(A).Det(B)$$, thus suppose that $$Det(A)=0$$ and $$Det(B)\neq 0$$ thus $$B$$ is invertible and $$AB=0$$ Multiplying by inverse B concluded that $$A=0$$ absurd.

• \det(A) is a defined MathJax command as well as \cdot – SK19 May 8 '18 at 23:22

Yes, your proof is okay (aside from Lord Shark the Unknown's comment), because you do it like this:

We want to show that $$\det(A)=0$$ and $$\det(B)=0$$. Now for proof by contradiction, assume $$\det(A)\neq 0$$ or $$\det(B)\neq 0$$.

Case 1: $$\det(A)\neq 0$$. Then ... hence contradiction.

Case 2: $$\det(B)\neq 0$$. Then ... hence contradiction.

Since all cases lead to a contradiction, our assumption "$$\det(A)\neq 0$$ or $$\det(B)\neq 0$$" must be wrong, hence $$\det(A)=0$$ and $$\det(B)=0$$.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/2772752/show-that-if-a-circ-b-0-but-a-neq-0-and-b-neq-0-then-you-have-that-d", "fetch_time": 1618295852000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-17/segments/1618038072175.30/warc/CC-MAIN-20210413062409-20210413092409-00442.warc.gz", "warc_record_offset": 482692053, "warc_record_length": 40226, "token_count": 898, "char_count": 2782, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2021-17", "snapshot_type": "latest", "language": "en", "language_score": 0.8760601282119751, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00009-of-00064.parquet"}