What is the geometric meaning of singular matrix

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.

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.

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 there are $m$ zero eigenvalues and the $m$ independent corresponding eigenvectors, 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.