14

I need to determine whether this matrix is injective \begin{pmatrix} 2 & 0 & 4\\ 0 & 3 & 0\\ 1 & 7 & 2 \end{pmatrix}

Using gaussian elimination, this is what I have done: \begin{pmatrix} 2 & 0 & 4 &|& 0\\ 0 & 3 & 0 &|& 0\\ 1 & 7 & 2 &|& 0 \end{pmatrix}

Divide row1 by 2, and then minus row3 by values of row1: \begin{pmatrix} 1 & 0 & 2 &|& 0\\ 0 & 3 & 0 &|& 0\\ 0 & 7 & 0 &|& 0 \end{pmatrix}

Divide row 2 by 3, divide row 3 by 7 and minus row 3 by row2: \begin{pmatrix} 1 & 0 & 2 &|& 0\\ 0 & 1 & 0 &|& 0\\ 0 & 0 & 0 &|& 0 \end{pmatrix}

Am I doing this correctly? How do I show that the matrix is (not) injective? I was thinking along the lines of "$x + z \ne 0$."

Peter LeFanu Lumsdaine
  • 11,200
eplosivemilton
  • 141
  • Notice the last column is twice the first. So figure out some $v$ such that $A v = 0$. – copper.hat Mar 03 '13 at 18:26
  • 2
    You have just shown that the rank of your matrix is $2$, so the nullity is $1$ by the rank-nullity theorem, hence it is not injective. Or directly, you are one step away to prove that the nullspace is ${(-2t,0,t);;;t\in\mathbb{K}}$. Or even more directly, see copper.hat's comment. – Julien Mar 03 '13 at 18:26

3 Answers3

22

The formal definition of injective is, that a function is injective, if $f(x)=f(y)\implies x=y$. Maybe it is at first not very intuitive that for linear functions it is the same as the triviality of the null space. But in fact it is, as $f(0 \cdot x)=0 \cdot f(x)=0$ we know that a linear function which is injective always have trivial null space.
To proof that a linear function is injetive if it's nullspace is trivial we take any $x,y$ with $f(x)=f(y)$. This is equivalent to $f(x)-f(y)=0$ as $f$ is injective it is the same as $f(x-y)=0$. But the null space is trivial so $x=y$ which implies the function to be injective.

In general a linear function is injective if it's columns are linear indepedent. As the first and the third are not linear independent your matrix is not injective. This is because the columns of your matrix are the images of the unit vectors. If they are not linear independent, the null space can't be trivial.

To formulate it with ranks, a matrix is injective if the rank is the same as the number of columns you have.

For a square matrix you can take the determinant, as here injective, surjective and bijective are equivalent.

Dominic Michaelis
  • 20,177
  • 2
    you have a small typo, the part where you go $f(x) - f(y)=0$ -> $f(x-y)=0$ this is due to linearity of $f$, not having anything to do with injectivity – user3180 Mar 02 '21 at 23:25
6

Compute the rank of your matrix; since the last row is all zeros, it has rank 2.

rank + nullity $ = 3 \implies$ nullity $= 1.$

Hence, you matrix cannot be injective.

amWhy
  • 210,739
4

Look at vectors of the form $\left(\begin{array}{c}2x\\ 0\\ -x\end{array}\right)$. What does your matrix do to these vectors? What does injective mean?

Alexander Gruber
  • 28,037