Is there any simple way of telling if a matrix $A$ is singular or not?

I saw somewhere the following:

If $rank(A)<n \Rightarrow$ $A$ is singular.

If $rank(A)\geqslant n \Rightarrow$ A is non-singular.

But I don't know how to verify if is this is always true. Also, how does the determinant relate to the singularity of a matrix?

  • 1
    $\begingroup$ $\text{det}(A)=0$ if and only if $A$ is square and singular. $\endgroup$ – Adrian Keister Feb 27 '18 at 19:08
  • $\begingroup$ For small size of $A$ the determinant is easy to compute, see for example the rule of Sarrus. Then you know quickly whether $A$ is singular or not. $\endgroup$ – Dietrich Burde Feb 27 '18 at 19:10
  • $\begingroup$ @AdrianKeister Isn't that only if $A$ does not have an inverse? $\endgroup$ – Amateur Mathematician Feb 27 '18 at 19:11
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    $\begingroup$ There are a bunch of equivalent conditions: $\det(A)=0$ is equivalent to $A$ singular, equivalent to $A$ does not have an inverse, equivalent to $\text{rank}(A)<n$ (assuming $A$ is $n\times n$), equivalent to non-trivial kernel, and the list goes on and on. $\endgroup$ – Adrian Keister Feb 27 '18 at 19:12
  • $\begingroup$ In exact arithmetic, the determinant or the rank are useful. In inexact arithmetic, things are slightly murkier, and one has to be contented with determining the condition number. (See e.g. this.) $\endgroup$ – J. M. is a poor mathematician Feb 28 '18 at 3:22

An n by n square matrix A is per definition singular if it is not invertible. There are several ways of determining this. As Adrian Keister pointed out, A is singular if and only if it's determinant is equal to zero, which can relatively easy be computed for n=1,2,3 in the general case(by hand). Using this method for lager n generally becomes rather tedious.

Now, if you are familiar with the "standard" way of calculating the inverse of a matrix A (row reducing [A| i] to [i|A^-1] ), this process requires that A is row equivalent to the n by n identity matrix i. So A is singular if and only if it is not row equivalent to the n by n identity matrix i. This test is easy by hand for n=1,2,3,4, 5 in the general case, and might be tedios for larger n.

Furthermore, if you have learned about how to determine if a set of column vectors are linearly independent, you know that they are linearly independent if and only if the matrix consisting of each vector as a column is row equivalent to the identity matrix, that is; their span is equal to n if and only if this mentioned matrix is row equivalent to i. If we think about this matrix as a complete unit, the dimension of it's colunmn space (it's rank) is equal to n if and only if the matrix is row equivalent to i. Combining these we get that a n by n square matrix A is singular (non-invertible ) if and only if its rank is not equal to n, the amount of columns of A. And since it's rank can't be greater than n it thus has to be less than n.

If the given matrix A has a row or column consisting entirely of zeros it is automatically singular, in correspondence with the previous points mentioned.

There is a huge bunch of things that are equivalent to a n by n matrix A being invertible, see for example http://mathworld.wolfram.com/InvertibleMatrixTheorem.html If either of these conditions fails to hold for a square matrix A, then it is not invertible. :)


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