# Zero determinant - Can we make a zero column?

1. ### Himanshu

67
I just wanted to know that the following statement is always true or not.

After I expand the determinant I get the value of the determinant as zero, ie. I know that the value of the determinant zero.

Then with the help of row or column transformations can we transform the determinant into another one that contains at least one row or one column, all whose elements are zero.

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3. ### ozymandias

82
Row and column transformations apply to matrices, not determinants.
However, if you substitute "matrix" for "determinant" in the above quote, it is true. It basically means that your (square) matrix represents a degenerate set of linear equations, i.e., one that has an infinite number of solutions.

Assaf.
Physically Incorrect

4. ### Himanshu

67
The Row and column transformations that I was talking about is of determinant. I think I should rephrase "Row and column transformations" as "Row and column operations".

5. ### ozymandias

82
My apologies, slight terminology misunderstanding on my part :).
We are talking about the same thing. I was thinking more in the direction of elementary operations used in Gaussian elimination, which are more or less the same thing you're talking about.
The answer remains: yes, you are correct. In fact, you can even make a stronger statement (if-and-only-if).

Assaf.
Physically Incorrect

6. ### Himanshu

67
Yes. I thought the same. I have tried it on few examples. But how do I prove it in general. I mean if it is a theorem there must be a proof for it.

By the way I was going through the An Energy Conservation Puzzle on your website. It's really whacking my brain out. Is the solution very simple(ie. does it require only brainwork).

7. ### ozymandias

82
Hey Himanshu,

There are proofs, of course, but as with anything in mathematics, how easy they are depends on what you assume you know.
Proving zero-column-->det(A)=0 is easy, since we can make that column the first, and hence all of the terms in the determinant will have some element in the row (a zero) multiplying them.
The converse, det(A)=0 --> zero-column, is a bit trickier. A heuristic argument (but not a proof) relies on a theorem stating that a square matrix A is invertible iff det(A) is non-zero. This means that, if det(A)=0, A is non-invertible, so we don't have a unique solution to A*v=b (had A been invertible, the solution would've been v=A^(-1)*b). This corresponds, by Gaussian elimination, to a row or column having all-zeros.
I'm afraid that for a full, rigorous proof you'll have to consult a linear algebra textbook.

Regarding the puzzle - good luck :). It can be quite conceptually challenging. There are no cheap shots involved, I can assure you of that. There is no friction you can blame, either ;). The first question is easier, and I'll give you a hint - look up the full statement of the theorem of conservation of (mechanical) energy.
The second part is quite more difficult, conceptually speaking.
Happy riddling :).

Assaf.
Physically Incorrect