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Could anyone help explain what is the geometric meaning of singular matrix ? What's the difference between singular and non-singular matrix ? I know the definition, but couldn't understand it very well.

Thanks.

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The singular matrix has no inverse matrix. This means that the operator described by the matrix is not invertible. –  superM Jul 3 '12 at 6:52
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Look at $\mathbb{R^3}$, for example. A singular $3\times 3$ real matrix maps $\mathbb{R^3}$ into a proper subspace of $\mathbb{R^3}$ (a plane through the origin, a line through the origin, or, most degenerately, to just the origin). A non-singular matrix maps $\mathbb{R^3}$ onto itself. –  André Nicolas Jul 3 '12 at 6:58
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4 Answers

up vote 9 down vote accepted

A matrix can be thought of as a linear function from a vector space $V$ to a vector space $W$. Typically, one is concerned with $n\times n$ real matrices, which are linear functions from $\mathbb R^n$ to $\mathbb R^n$. An $n\times n$ real matrix is non-singular if its image as a function is all of $\mathbb R^n$ and singular otherwise. More intuitively, it is singular if it misses some point in $n$-dimensional space and non-singular if it doesn't.

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It's right of course, but "if it misses some point" may give a bit of a bad picture: as soon as it misses any point at all, it necessarily has to miss almost all of them. Which IMO gives a better impression of a singular linear mapping. –  leftaroundabout Jul 3 '12 at 8:46
    
@leftaroundabout I don't get the miss almost all comment. For example if $Im(f)$ is $n/2\leq k<n$ dimensional subspace of $\mathbb{R}^n$, $f$ hits more than half of $\mathbb{R}^n$ in some sense, and not "misses almost all". –  rschwieb Jul 3 '12 at 12:16
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@rschwieb in some sense it hits more than half of the points, but not in a canonical sense. Of course, these spaces have all uncountable-infinitely many points in them; "how many points" in $\mathbb{R}^n$ generally refers to the Lebesgue measure, which is zero for any proper subspace. Almost all points lie outside of it may be a somewhat colloquial way of saying this, but it's widely accepted. –  leftaroundabout Jul 3 '12 at 12:43
    
@leftaroundabout Ok thanks for clarifying. Measure theory was very far from my mind in this algebraic problem. –  rschwieb Jul 3 '12 at 13:46
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If you are in $\Bbb{R}^3$, say you have a matrix like

$$\left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33} \end{array}\right].$$

Now you can think of the columns of this matrix to be the "vectors" corresponding to the sides of a parallelepiped. If this matrix is singular i.e. has determinant zero, then this corresponds to the parallelepiped being completely squashed, a line or just a point.

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It could also be an 1D object, or even a 0D object (zero matrix). –  celtschk Jul 3 '12 at 7:13
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11 upvotes for such an answer? I don't get Math.SE..... –  user38268 Jul 15 '12 at 8:52
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You can think of a $n \times n$ normal matrix as a linear transformation. It 'stretches' or 'scale' a vector in each of its eigenvector directions by factor of eigenvalues, which could be a complex number.

If $m$ eigenvalues are zeros, you can interpret it as the matrix wipes out $m$ dimensions in the $n$ dimensional vector space, or $m$ dimensions are squashed or collapse, as described in BenjaLim's answer. The annihilated $m$ dimensions is called the null space or kernel of the matrix, and the remained $n-m$ dimensions is the co-image of the matrix, whose dimension ($n-m$ in this case) is the rank of the matrix.

If the matrix has null space (zero eigenvalues), you cannot invert it because information in the null space are lost, just as you cannot invert $0 \times a = b$ by $a = b/0$. The best you can do is to recover the information in the co-image space by pseudo-inverse.

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A nonsingular $n \times n$ matrix with real (or complex) entries corresponds to a linear transformation of ${\mathbb R}^n$ (or ${\mathbb C}^n$) to itself that is one-to-one and onto. A singular matrix corresponds to a linear transformation that is neither one-to-one nor onto.

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