7
$\begingroup$

Why does $\det A=0$ imply that there exists a non-zero solution to the homogeneous linear equations determined by $A$?

Also is it and if and only if case? That is if there is a non zero solution to the set of linear equations does the matrix determined by the coefficients automatically have a determinant of zero?

$\endgroup$
4
$\begingroup$

Having $\det A=0$ means the linear equations are not independent, hence there are more unknowns than equations, and you can choose arbitrarily a number of these unknowns – in particular, you can choose them non-zero.

Conversely, a set of non-zero solutions can be interpreted as the coefficients of a non-trivial relation between the columns of the matrix, which implies its determinant is $0$.

$\endgroup$
4
$\begingroup$

If $\det A\ne 0$, we can quite explicitly write down the invere matrix $A^{-1}$. Then $Av=0$ implies $v=A^{-1}Av=A^{-1}0=0$.

And if there is no non-zero solution to $Av=0$ then the $n$ (assuming $A$ is $n\times n$) vectors $v_i:=Ae_i$, $1\le i\le n$, with $(e_1,\ldots,e_n)$ the standard basis are linearly independent, hence a basis of $n$-dimensional space, hence $v_i\mapsto e_i$ defines a linear map and the corresponding matrix $B$ is obviously inverse to $A$ (i.e., $AB$ and $BA$ both correspond to the identity map). Then $1=\det(AB)=\det(A)\det(B)$, hence $\det(A)\ne0$.

$\endgroup$
2
$\begingroup$

For each of the three elementary row operations, applying it either does nothing to the determinant (in the case of a $R_1 \mapsto R_1 + 6 R_3$ sort of operation) or it multiplies by a non-zero scalar ($-1$ for a flip-flop, and whatever scalar you're multiplying the row by for a scalar multiplication).

So: elementary row operations can change the determinant of a matrix, but they can't change whether that determinant is zero or not.

Therefore, given a square matrix with $\det A = 0$, when we echellonize it, we must get a matrix with some $0$s on the diagonal (because a diagonal matrix with no $0$s on the diagonal doesn't have determinant $0$). It's then obvious that the corresponding homogeneous equation has a non-zero solution (corresponding to the row of $0$s in the matrix).

Conversely, if there exists a non-zero solution, then our matrix echellonizes to a matrix with a row full of zeroes, so the determinant of that matrix is zero; hence the determinant of our original matrix was zero.

$\endgroup$
1
$\begingroup$

$0=\det(A)=\det(A-0\cdot I)$, hence $0$ is an eigenvalue of $A$.

$\endgroup$
0
$\begingroup$

If the determinant of $A$ is non-zero then $A$ is bijective and the inverse of $A$ exists. This means: $$ A x = 0 \iff x = A^{-1} 0 = 0 $$ Due to the bijection there is only the zero solution.

If there is a non-zero solution $x$ to the homogenous system, we have $$ 0 = Ax = \sum_i x_i \, a_i \Rightarrow a_j = \frac{1}{x_j} \sum_{i\ne j} x_i a_i $$ for some non-zero component $x_j$. To ease the following, let us assume $j=1$.

The determinant is an alternating multi linear form, the linearity gives $$ \text{det}(A) = \text{det}(a_1,\ldots,a_n) = \text{det}\left(\frac{1}{x_1} \sum_{i\ne 1} x_i a_i, a_2,\ldots,a_n\right) = \frac{1}{x_1}\sum_{i\ne 1} x_i \text{det}(a_i, a_2,\ldots,a_n) = 0 $$ because the first and $i$-th argument contain the same vector $a_i$. The determinant vanishes in such a case, because one might switch the equal arguments which would keep the value and because of the alternating property it would need to attain the negative value as well and this leaves only zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.