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I know you could just say because the $\det = 0$.

But during the introduction of determinants the professor said, obviously if two columns of the matrix are linearly dependent the matrix can't be inverted, therefore it is zero. He made it sound like it is an intuitive thing, a simple observation, but I always have to resort to the properties of determinants to show it.

How does one trivially see that you can not invert a matrix without a full rank?

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    You may find helpful resources [here](https://www.google.com/#q=invertible+matrix+theorem). – 2017-02-06
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    That is just repeating what I told you. I know about determinant properties and about characteristics of invertible matrices, I want to know about an intuitive way to see that to help think about invertible matrices – 2017-02-06
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    Ok. But note that 2 columns of a matrix being linearly dependent is not equivalent to the matrix having less than full rank. You may want to clarify which one you mean (and use "intuitive" since that's not the same as "trivially see[ing]"). The first implies the second but it's possible to have full rank without having only two columns being linearly dependent. For example, $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix}$ has less than full rank because the third column is the sum of the first two, but no two columns of $A$ are linearly dependent on each other. – 2017-02-06

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Suppose that the columns of $M$ are $v_1, \ldots, v_n$, and that they're linearly dependent. Then there are constants $c_1, \ldots, c_n$, not all $0$, with $$ c_1 v_1 + \ldots + c_n v_n = 0. $$ If you form a vector $w$ with entries $c_1, \ldots, c_n$, then (1) $w$ is nonzero, and (2) it'll turn out that $$ Mw = c_1 v_1 + \ldots + c_n v_n = 0. (*) $$ (You should write out an example to see why this first equality is true). Now we also know that $$ M0 = 0. (**) $$ So if $M^{-1}$ existed, we could say two things: $$ 0 = M^{-1}0 \ (**)\\ w = M^{-1} 0\ (*) $$ But since $w \ne 0$, these two are clearly incompatible. So $M^{-1}$ cannot exist.

Intuitively: a nontrivial linear combination of the columns is a nonzero vector that's sent to $0$, making the map noninvertible.

But when you really get right down to it: proving this, and things like it, help you develop your understanding, so that statements like this become intuitive. Think about something like "the set of integers that have integer square roots". I say that it's intuitively obvious that $19283173$ is not one of these.

Why is that "obvious"? Because I've squared a lot of numbers, and all the squares have a last digit that's either $0, 1, 4, 5, 6,$ or $9$ (because those are the last digits of squares of single-digit numbers). Now that I've told you that, my statement about "intuitively obvious" is obvious to you, too. But until you'd at least learned a little about integer squares by investigating them, your intuition wasn't as good as mine. Sometimes "intuition" is just another name for "applied experience."

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You can think of a matrix as a linear mapping and the rank of the matrix corresponds to the dimension of the image of the mapping. So if you cut off some dimension, you can hardly lift it up back. (Imagine for example the following: a line in a plane; now you linearly transform the plane into a 1-dimensional space (a line) by projecting the plane onto, say, $x$-axis; show this to someone else - this other person cannot tell you, what the original line was, which corresponds to non-existence of the inverse map.)