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I need to demonstrate that Ker(AB)=${\vec{0}}$, just like Ker(A) and Ker(B). I have a not-so-elegant way of showing that (I think): A and B are n * p and p * m, and it has to be the case that n $\geq$ p $\geq$ m, and rref's of A and B are either equal to the identity matrix, or are of form I and row(s) of 0's on the bottom. So, product of rref's of A and B will equal a matrix that is either I or I with row(s) of 0's on the bottom, and is n * m where n $\geq$ m, so it has more equations than variables. Multiplication of matrices of the aforementioned forms can only yield a matrix of the aforementioned form (I or I with row(s) of zeros), so it must also have Ker=${\vec{0}}$. Is there a more simple (though not too advanced, as we are barely mid-way through the course) way of showing that this is true? Also, while I know that Ker(A)=Ker(rref(A)) for any matrix A, I am not sure if I can say that Ker(rref(A) * rref(B))=Ker(AB). Is this statement true?

Thanks!

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    Both A and B are injective, so is AB. Then Ker(AB)=02011-01-28

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Composition of one-to-one functions is one-to-one.

Alternatively and directly: let $\mathbf{v}\in\mathrm{Ker}(AB)$. Then $A(B(\mathbf{v}))=\mathbf{0}$, so $B(\mathbf{v})\in\mathrm{ker}(A)=\{\mathbf{0}\}$. Therefore, $B(\mathbf{v})=\mathbf{0}$, so $\mathbf{v}\in\mathrm{ker}(B) = \{\mathbf{0}\}$. Therefore, $\mathbf{v}=\mathbf{0}$.

This proves that $\mathrm{ker}(AB)\subseteq\{\mathbf{0}\}$, which suffices.

No need to muck about with sizes of matrices.

Added: (as per request in the comment) I assume that "rref" means "row reduced echelon form". It is true that $\mathrm{Ker}(A) = \mathrm{Ker}(\mathrm{rref}(A))$, because we obtain $\mathrm{rref}(A)$ be left-multiplying $A$ by invertible elementary matrices, and this does not change the nullspace (equivalently, elementary row operations do not change the nullspace). However, it is false that $\mathrm{Ker}(AB) = \mathrm{Ker}(\mathrm{rref}(A)\mathrm{rref}(B))$; if you think about it, you'll see that the problem is that those invertible matrices will now also be between $A$ and $B$. Here is an explicit example: take $$A = \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right),\qquad B = \left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right).$$ Then the reduced row echelon form of $B$ is $A$, so $\mathrm{rred}(A)\mathrm{rred}(B) = AA = A$. However, $AB$ is the zero matrix. So $\mathrm{Ker}(AB) = \mathbb{R}^2$, but $\mathrm{ker}(\mathrm{rred}(A)\mathrm{rred}(B)) = \mathrm{ker}(A) = \{(0,y)\mid y\in\mathbb{R}\}\neq \mathrm{ker}(AB)$.

The reason it does not work is that if $C$ is invertible, then $\mathrm{ker}(CA) = \mathrm{ker}(A)$ (use the same argument as above). But $\mathrm{ker}(AC) = C^{-1}(\mathrm{ker}(A))$, so this may change the kernel. Since $\mathrm{rred}(A) = CA$ for some invertible matrix $C$, and $\mathrm{rred}(B) = DB$ for some invertible matrix $D$, you get $\mathrm{rred}(A)\mathrm{rred}(B) = CADB$, so that $D$ in the middle can mess you up.

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    Thank you Arturo (and everyone else). I managed to work out this solution after completing the assigned readings actually, it makes sense and was pretty obvious. Could you please comment on "Also, while I know that Ker(A)=Ker(rref(A)) for any matrix A, I am not sure if I can say that Ker(rref(A) * rref(B))=Ker(AB). Is this statement true?" just out of my curiosity?2011-01-28
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    @InterestedQuest: Added to the body, since I will write some examples.2011-01-28
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Hint: Assume that $\text{Kernel}(AB)$ is non-trivial. Prove by contradiction by making use of the fact that the nullspace of $A$ and $B$ are trivial.

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    Why bother with an argument by contradiction, when a direct one will do?2011-01-28
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    @Arturo: Right. I had the same argument in mind as yours.2011-01-28
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    Yes, and the "contradiction" arises only at the end. It's a classic example of an "unnecessary" argument by contradiction: assume the conclusion you want is false. Ignore that assumption and prove the conclusion you want is true. Then say "Contradiction!", and conclude the result is true. You can just delete the first and last lines, and get a direct proof. I keep telling my students to keep an eye out for that kind of thing.2011-01-28
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    @Arturo: Good one. Will make note of it in future.2011-01-28
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You can use the rank nullity theorem.