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Math Help - singular vs non singular matrix

  1. #1
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    singular vs non singular matrix

    Good day guys, I just need some advice on how to determine the difference between a singular and a non singular matrix. Also would it be correct if I classified the matrix below as a reduced row echelon form because each leading 1 is to the right of the previous and there is a zero above and below every leading one?

    A =
    1 2 0 0
    0 0 1 0
    0 0 0 1


    Kind regards
    Hyperion
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  2. #2
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    Re: singular vs non singular matrix

    Yes, the matrix is in rref.

    Singular matrices have some non-zero vector they map to the zero-vector (via matrix multiplication). Non-singular matrices only map the zero-vector to the zero-vector (they have a trivial null space).

    The matrix you have is singular because:

    A(2,-1,0,0) = (0,0,0)

    (other definitions of singular exist).

    In this particular case, it is easy to tell that A will be singular, because rank(A) = 3, whereas dim(dom(A)) = 4. You can't fit all of a 4-dimensional space inside a 3-dimensional one without zeroing out at least one line.
    Thanks from Hyperion and topsquark
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  3. #3
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    Re: singular vs non singular matrix

    Ah ok I see, I think I'm gonna need to spend a bit more time with this to get the hang of it, but thanks so much for the help I appreciate it
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  4. #4
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    Re: singular vs non singular matrix

    Let's say all the entries in a our matrix come from a field, $F$. If this seems too general for you, just imagine I mean the rational numbers, or the reals.

    An $m \times n$ matrix $A$ gives us a function: $F^n \to F^m$ via "left-multiplication":

    $v \mapsto Av$.

    A matrix is singular/non-singular if the function we get is non-injective/injective.

    To see this, suppose the function is non-injective. This means we have two different vectors, $u,v$ with $Au = Av$. Thus: $Au - Av = A(u-v) = 0$.

    Since $u \neq v$, $u - v \neq 0$. Thus $u-v$ is a non-zero vector $A$ maps to the 0-vector, so $A$ is singular.

    On the other hand, if $Au = Av$ always means $u = v$, Then if $Au = 0 = A0$, we must have $u = 0$, so the ONLY vector $A$ maps to the zero vector, IS the zero vector.

    Put another way: singular matrices "lose information", they take a space where you need to specify $n$ things, and turn it into a space where you need to specify $m < n$ things. Obviously, some "collapsing" takes place.

    This "collapsing" is formally called the "nullity" of the matrix. An extreme example is the 0-matrix, which loses ALL information, by sending everything to a single vector, the zero vector.

    The "formal" statement of this is the so-called "rank-nullity theorem" which says:

    rank(A) + nullity(A) = dim(dom(A)).

    We can interpret this as:

    Information kept + Information lost = total information we started with.

    "Dimension" then, tracks information. In the plane, we need to know two pieces of information to locate a point:

    "Upness" and "overness" (for rectangular coordinates) -or-
    "Outness" and "angle" (for polar coordinates)

    This number, 2, characterizes how much information a vector carries. Thus real numbers (on a line) carry just ONE piece of information, their location on said line.

    This is why to UNIQUELY solve a system of equations in $n$ unknowns, we need at least $n$ equations. If we have fewer (so the rank is less than $n$), there's "not enough information" to recover a unique solution.

    For example, given the matrix in this example, if we have the system of equations:

    x + 2y = 0
    z = 0
    w = 0

    We know from the last two equations, that any solution is of the form: (x,y,z,w) = (a,b,0,0). The first equation tells us any solution is of the form: (-2y,y,z,w).

    So a "general solution" is of the form: (-2y,y,0,0) = y(-2,1,0,0) (the example I gave has y = -1). This isn't enough information to get a UNIQUE solution, the missing information in this case is: "what is y?"
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