# if Det(Mnxn) = 0, can Det(M(n-1)x(n-1))>0?

Say you have a square matrix in RnR^n that is linearly dependent. The resulting determinant of this matrix would, of course, be zero.

Let us assume that only one of the vectors is actually linearly dependent, and as such, eliminating it would result in an independent matrix. The consequence of this is that we would no longer have a square matrix, as the number of elements in each vector would be equal to n, but the resulting number of vectors would be equal to n-1, thus we would not be able to find the determinant of this resulting matrix as it is not square.

My question is, could there exist a situation where all of the vectors of this resulting matrix could have one element â€œeliminatedâ€‌ so to speak, allowing the resulting matrix to be transformed into a square matrix that could then have its determinant calculated?

My idea is that even if a N-dimensional shape â€œcollapsedâ€‌ in on itself as a consequence of linear dependence, maybe it could be the case that the resulting collapsed shape could, in a lesser dimension, still have a determinate value, or volume, if the remaining vectors are linearly independent.

So for example, take the unit cube in R3R^3 which is parametrized by the vectors (1,0,0) (0,1,0) and (0,0,1). This shape is made up of linearly independent vectors and thus has a determinant value that does not equal zero ( it equals one). But lets say that for the third vector, instead of (0,0,1), we had (1,1,0). The resulting matrix would then contain linearly dependent vectors and thus the determinant for this shape in R3R^3 would equal zero.

However, if we use row reduction to eliminate the third row, we are left with the vectors (1,0,0) (0,1,0) and (0,0,0), which can be expressed as a 2×3 matrix. But since a whole column of this 2×3 matrix is just zeros as well, can it be further reduced to a 2×2 matrix, of which we can then find the determinant for the shape in R2R^2? Or do the row operations affect the determinant in a way that Iâ€™m not properly accounting for?

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If you start with a singular matrix (determinant zero), then remove one column so that the remaining columns are independant, and then remove one row, that can lead to a matrix with non-zero determinant. You are right in your example. But the determinant you get in the end of course depends on the row you removed. I’m not sure what your question actually is.
– Simon
2 days ago

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