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Let: $$ A = \left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \\ \end{matrix}\right ] $$

The characteristic equation of this matrix is: $(\lambda +1) (\lambda^2 -4\lambda -5) = 0$

$\implies (A+I)(A^2- 4A -5I) = 0$

But $AB = 0 \require{cancel} \cancel\implies A= 0$ or $B= 0$

But in this case, on solving we can clearly see that $A^2 - 4A - 5A = 0$ (0 denotes null matrix)

So, when does this work? When are we allowed to get cancel factors? In another question I did yesterday, cancelling the factor from the characteristic equation gave me the wrong answer.

Archer
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  • no! in this case $x^2-4x-5$ is a minimal polynomial – Chinnapparaj R Oct 10 '18 at 07:20
  • $AB=0 \Rightarrow A=0$ or $B=0$ does not hold for matrices. Each factor can for example remove separate subspaces. – mathreadler Oct 10 '18 at 07:23
  • The characteristic equation is $(\lambda+1)(\lambda^2-4\lambda-5) = (\lambda+1)^2 (\lambda-5)$. The lowest degree polynomial that has $A$ for a zero is the minimal polynomial - look it up on wikipedia. – Kolja Oct 10 '18 at 07:24
  • @Kolja When does matrix satisfy its minimal polynomial? – Archer Oct 10 '18 at 07:25
  • Every matrix iz a zero of it's minimal polynomial. Do you know how to decompose a matrix into Jordan blocks? – Kolja Oct 10 '18 at 07:27
  • @Kolja No I am not aware of that. I am just in high school atm. – Archer Oct 10 '18 at 07:28
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    Then it might be a little too advanced. To give you an example, consider the matrices $A=\left(\begin{matrix} 1 & 0 \ 0 & 1 \end{matrix}\right)$ and $B=\left( \begin{matrix} 1 & 1 \ 0 & 1 \end{matrix}\right)$. They both have the same characteristic polynomial, but minimal polynomial for $A$ is $x-1$, and for $B$ it's $(x-1)^2$.

    To know more about the difference, i suggest reading on characteristic polynomials and Jordan blocks on wikipedia.

     – Kolja Oct 10 '18 at 07:31
  • @Kolja How are minimal polynomials determined? Like i don't see why a difference is there in the minimal polynomials of A and B. – Archer Oct 10 '18 at 07:34
  • Maybe the easiest example is $\begin{bmatrix}1&0\0&0\end{bmatrix} \begin{bmatrix}0&0\0&1\end{bmatrix} = 0$ – mathreadler Oct 10 '18 at 07:35
  • @mathreadler I am well aware that $AB = 0$ doesnt imply that either of them is 0. I have even mentioned that in my question. – Archer Oct 10 '18 at 07:37
  • @Abcd if the matrix is diagonalizable, then you just take the radical of the characterestic polynomial, i.e. the product of all prime factors - in your case the matrix is diagonalizable, the characteristic polynomial is $(x+1)^2(x-5)$, it's prime factors are $x+1$ and $x-5$, and the minimal polynomial the product of prime factors - $(x+1)(x-5)=x^2-4x-5$. On the other hand, the matrix $B$ from the above comment is NOT diagonalizable, so this doesn't work. In that case you should look up Jordan normal form. Don't expect to understand it in 2 minutes, it takes some time. – Kolja Oct 10 '18 at 08:54

2 Answers2

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Let's review the zero product property: $$xy = 0 \implies x = 0 \text{ or } y = 0.$$ It holds in fields, such as $\mathbb{R}$ or $\mathbb{C}$ because non-zero numbers always have a multiplicative inverse. So, if $x \neq 0$, then $$xy = 0 \implies x^{-1}xy = x^{-1}0 \implies y = 0.$$ (Or $x = 0$, in which case we were done before we even started.)

Matrices don't have this property you can have non-zero matrices that do not have inverses. In fact, if a matrix $A$ does not have an inverse, then there exists a non-zero matrix $B$ such that $AB = 0$. To prove this, consider a non-invertible matrix $A$. Then $\operatorname{ker} A \neq \lbrace 0 \rbrace$, so there must be a non-zero column vector $v$ such that $Av = 0$. If you form $B$ by simply putting $v$ into all of its columns, you get $AB = 0$.

So, $AB = 0 \implies A = 0 \text{ or } B = 0$ if and only if $A$ or $B$ is invertible.

Theo Bendit
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In general $AB=0$ does't imply $A=0$ or $B=0$. In your case, that factor is corresponding to a minimal polynomial and note that the minimal polynomial is the smallest degree polynomial satisfied by $A$ .

For, here the matrix is diagonalizable so the minimal polynomial is a distinct linear factors which is exactly $$(x+1)(x-5)$$ and so $$(A+I)(A-5I)=A^2-4A-5I=0$$

Chinnapparaj R
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