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How can I show injectivity of a function taking and giving out matrices Like $g_{1}: M_{2 \times 5}(\Bbb{R}) \rightarrow M_{5 \times 5}(\Bbb{R})$ defined by $g(Z) = BZ$

for a $2 \times 5$ matrix $Z$ and $B_{5 \times 2}$matrix

I did suppose $g(A) = g(D)$ that is $BA = BD$ but then I cannot apply inverse of $B$ as it is not a square matrix? How can I show $A=D$ for injectivity?

How can I proceed?

Martin Sleziak
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BAYMAX
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  • If you suppose $g(B)=g(D)$ then you'd have that $BC=DC$ given that you've defined $g(B)=BC$.

    Anyway, do you know anything specific about the matrix $C$?

     – Theo C. Oct 25 '18 at 03:53
  • @TheoC. this is actually a followup question to - https://math.stackexchange.com/questions/2968888/injectivity-of-g-1m-2-times-5-bbbr-rightarrow-m-5-times-5-bbbr, u can see here there is a specific matrix $C$. – BAYMAX Oct 25 '18 at 09:37

1 Answers1

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We have $g(D)=DC$, hence

$$g(B)=g(D) \iff BC=DC.$$

If $C$ is invertible, then we get

$$g(B)=g(D) \iff B=D$$

and $g$ is injective in this case.

If $C$ is not invertible, the $g$ is not injective.

Fred
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