# Show that $A \cdot \textbf{0}_{n \times p}=\textbf{0}_{m \times p}$

Suppose $$A \in \mathbb{F}^{m \times n}$$. Show that $$A \cdot \textbf{0}_{n \times p}=\textbf{0}_{m \times p}$$ without computing individual entries. You can use that $$\mathbb{F}^{r \times s}$$ forms a vector space for all integers $$r, s \geq 1$$.

This is a no brainer question if we could compute the entries individually. But I can't get my head around as to how we can do this without component by component calculation. Any help would be nice.

• Maybe $AA = AA$ so $A (A-A) = A0 = 0$? Aug 15, 2022 at 4:51
• Maybe you need to say something about how the image of the $\mathbf{0}$ matrix has to be just $\{0\}$. And then of course $A$ sends $0$ to $0$? If this is homework, it probably depends on what you just learned...
– SBK
Aug 15, 2022 at 4:52
• @copper.hat ... if $m=n=p$ Aug 15, 2022 at 4:59
• @BrianMoehring Thanks, I missed that. Aug 15, 2022 at 5:15

If you're allowed to use the algebraic properties of matrices, then observe that: $$A0 = A(0+0) = A0 + A0$$

Since $$\mathbb{F}^{r \times s}$$ is a vector space, an additive inverse $$-(A0)$$ exists. Hence: $$A0 = A0 + 0 = A0 + (A0+(-A0)) = (A0+A0)+(-A0) = A0+(-A0) = 0$$ as was desired.

• @Prem OP wrote specifically that "You can use that $\mathbb{F}^{r\times s}$ forms a vector space for all integers $r,s\geq 1$." Aug 15, 2022 at 9:19
• It's not circular at all; there's nothing about the axioms of a vector space which talk about multiplication of two vectors. Aug 15, 2022 at 9:28

$$A \cdot \textbf{0}_{n \times p}=\textbf{0}_{m \times p}$$

For all $$v\neq 0$$ $$A\cdot \textbf{0}_{n \times p} v =0$$ and then use the uniqueness of the zero element.

• No it's using associativity. $(A0)v=A(0v)=0$. Aug 15, 2022 at 9:39
• The terms $A0$ only look equal, but are in fact not the same. In the question the 0 stood for the n by p matrix with all zeroes. In Suzu's answer, when $A0$ occurs the second time (as you correctly point out), it stands for the zero vector of p-dimensional space. It is a short deduction to show that a matrix maps the zero vector always to the zero vector. Aug 15, 2022 at 16:56
• Sounds good, @8bc3457f , but a Doubt : How do we conclude $(0v)=0$ without calculating the elements component by component ( which OP forbids ) ? We have converted Matrix Issue into Vector Issue. Maybe , we should go further , to convert this Vector Issue into Scalar Issue by $|0v|=|0||v|=0$ !!
– Prem
Aug 15, 2022 at 18:57
• I expect that using $0v=0$ is permitted in the OP's situation. Nevertheless here an algebraic proof: $0v = (0+0)v = 0v + 0v$ so by eliminating $0v$ once, we obtain $0 = 0v$ as we wanted. (Here I used "0" for both the zero vector and for the zero matrix.) Aug 15, 2022 at 19:48
• Applying a norm to the vectors should be avoided in this situation, because this is imposes additional structure on the vector spaces which isn't always available. The first equality of your argument using the norm is only a $\le$-inequality in general. But that's no problem here. On the other hand a pedantic reader might interpret the "0 !!" at the end as a double factorial (0!!=1) which would make your proof wrong ;) Aug 15, 2022 at 19:52

Let us use $$0$$ to mean either the LHS Zero Matrix or the RHS Zero Matrix, with Correct Size.

$$A(0) = A(0+0) = A(0)+A(0)$$
[[ This is because 0+0=0 & Multiplication is Distributive over Addition ]]

$$A(0)-A(0) = A(0)$$
[[ This is by adding $$-A(0)$$ to LHS & RHS ]]

$$0=A(0)$$
[[ This is by using $$X-X = 0$$ on LHS ]]

$$A0=0$$
[[ This is by exchanging LHS & RHS ]]

We use Minimal Properties in this Derivation.

• This proof uses the same argument as @MordeusMorgenstern, arranged differently. Cyclic arguments are no problem in any of the answers, because $\mathbb{F}^{r\times s}$ being a vector space is not concerned with matrix multiplication. Aug 15, 2022 at 9:34