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I'm afraid there could be a task like that in my exam. What way would you recommend me if there was a quadratic matrix given, such as $A= \begin{pmatrix} 2 & 0 & 4\\ 0 & 3 & 0\\ 1 & 7 & 2 \end{pmatrix}$?

There are several (for me confusing) ways doing it I think.

The one we had in our readings is to check if the column vectors are linearly independent (or something like that :S). But I think this would only tell us whether the linear mapping is injective.

But I think there is another, faster way with rank? I hope you can explain with this example?

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But I think there is another, faster way with rank? I hope you can explain with this example?

For square matrices, you have both properties at once (or neither). If it has full rank, the matrix is injective and surjective (and thus bijective). You could check this by calculating the determinant: $$\begin{vmatrix} 2 & 0 & 4\\ 0 & 3 & 0\\ 1 & 7 & 2 \end{vmatrix} = 0 \implies \mbox{rank}\,A < 3$$ Hence the matrix is not injective/surjective.


In general for an $m \times n$-matrix $A$:

  • If the matrix has full rank ($\mbox{rank}\,A = \min\left\{ m,n \right\}$), $A$ is:
    • injective if $m \ge n = \mbox{rank}\,A$, in that case $\dim \ker A = 0$;
    • surjective if $n \ge m = \mbox{rank}\,A$;
    • bijective if $m=n=\mbox{rank}\,A$.
  • If the matrix does not have full rank ($\mbox{rank}\,A < \min\left\{ m,n \right\}$), $A$ is not injective/surjective.

See also:

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  • $\begingroup$ Thank you! For non-square matrix, could I also do this: If the dimension of the kernel $= 0 \Rightarrow$ injective. If rank = dimension of matrix $\Rightarrow$ surjective ? $\endgroup$
    – tenepolis
    Commented Mar 23, 2017 at 12:04
  • $\begingroup$ @tenepolis Yes, I extended the answer a bit. $\endgroup$
    – StackTD
    Commented Mar 23, 2017 at 13:20

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