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If a $3 \times 3$ matrix has determinant zero, then is it possible that its rank could be $3$? I think it only could be $2$ or less. I am right or wrong? Please explain.

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Yes you are correct. The rank is the number of non-zero singular values and the determinant is the product of these singular values. – user17762 May 18 '11 at 16:35

Just think about this image: A matrix can be viewed as a set of $n$ vectors of $n$ dimension (imagine the case of $n=3$). Thus the determinant is actually the volumn of the polyhedron spanned by these vectors (example: $(1,0,0)^T,(0,1,0)^T$ and $(0,0,1)^T$ form a unit cube), where the sign (plus or minus) indicates the orientation (right-handed system -- plus, and left-handed system otherwise).

Now you can understand many properties of determinants: Exchanging two columns (or rows) will flip the sign of determinant -- that's because the orientation has changed. Or, if one column (or row) becomes all zeroes, the determinant will also be reduced to zero -- that's because the polyhedron shrinks to a sheet, and thus occupies no space.

As for the rank, the rank of a matrix actually tells you what is the minimum dimension of the space holding all the vectors of the matrix. Take $n=3$ as an example again: If $r(A)=3$, it means, in order to contain all the column vectors of $A$, a space of at least 3 dimension is a must. As a result, they span a nonzero volumn. If $r(A)=2$, it means the vectors are in fact lying in a plane, or $r(A)=1$, lying within a line, either of which holds a zero volumn.

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+1 Great answer! – M Turgeon Jun 1 '12 at 16:49
Just wanted to thank you for this answer. – Gigili Jan 7 '14 at 10:40
that is a really good answer ! – houda May 10 '15 at 22:32

A matrix does have full rank iff it is invertible. The determinant of a matrix is zero iff it is not invertible. Thus if a $3 \times 3$ matrix has a determinant of zero it can not have the rank $3$.

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You are correct; a singular matrix (square matrix with determinant zero) cannot be full rank.

To see this, think about the effect that elementary row operations have on the determinant:

  • If $B$ is obtained from $A$ by exchanging two rows, then $\det(B) = -\det(A)$.

  • If $B$ is obtained from $A$ by multiplying a row by a $\lambda$, then $\det(B) = \lambda\det(A)$. Remember that for this to be an elementary row operation, you must have $\lambda\neq 0$.

  • If $B$ is obtained from $A$ by adding a multiple of one row to another row, then $\det(B)=\det(A)$.

This means that if $B$ is obtained from $A$ by a sequence of elementary row operations, then $\det(B)=0$ if and only if $\det(A)=0$; that is, you cannot go from a matrix with zero determinant to a matrix with nonzero determinant by doing elementary row operations; and neither can you go from a matrix with nonzero determinant to a matrix with zero determinant. (Though the value of the determinant can change from some nonzero number to another, because of the second type of elementary row operation mentioned above, it cannot go from zero to nonzero or from nonzero to zero).

Now suppose that $A$ is $n\times n$ and has rank $n$. Then we know that by doing elementary row operations we can transform $A$ into the $n\times n$ identity matrix. The $n\times n$ matrix has determinant $1$; so if we can transform $A$ into the identity matrix with elementary row operations, then $\det(A)\neq 0$ (because $\det(I_n)\neq 0$). So a matrix of rank $n$ has nonzero determinant. This is logically equivalent to the contrapositive: if $\det(A)=0$, then $A$ does not have rank $n$ (and so has rank $n-1$ or less).

Conversely, if the rank of $A$ is strictly less than $n$, then with elementary row operations we can transform $A$ into a matrix that has at least one row of zeros. The determinant of a matrix with one row of zeros is $0$, so that means that $\det(A)=0$.

In summary, if $A$ is $n\times n$, then $\mathrm{rank}(A)=n$ if and only if $\det(A)\neq 0$.

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What fact are you using to conclude that if an $n \times n$ matrix $A$ has rank $n$ then it can be row reduced to an identity matrix? Intuitively, since $A$ has rank $n$ it must have $n$ linearly independent rows; consequently no row can be "eliminated" by elementary operations. Is this the right track? – ItsNotObvious May 18 '11 at 18:16
@3Sphere: There are many different definitions of rank. One very common one (which is the one I was implicitly using) is that the rank of a matrix is the number of pivots (nonzero rows) in the reduced row-echelon form of $A$; if the matrix is $n\times n$, and the number of pivots in the reduced row-echelon form of $A$ is $n$, then the reduced row-echelon form of $A$ is the identity. (cont) – Arturo Magidin May 18 '11 at 18:20
@3Sphere: (cont) Alternatively, the rank is the dimension of the row space, elementary row operations don't change the row space; zero vectors can be removed from spanning sets without changing the span; and a space spanned by $k$ vectors has dimension at most $k$. So the reduced row-echelon form of $A$ cannot have fewer than $\mathrm{rank}(A)$ nonzero rows (which amounts to what you write as "intuitively"). – Arturo Magidin May 18 '11 at 18:22
Thanks for the clarification. – ItsNotObvious May 18 '11 at 18:28

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