2

So we know that $A$ is not invertible $\iff$ det$(A)=0\iff \lambda=0$ is an eigenvalue of $A$. But the negation doesn't equal to the title statement. How would you prove the title question?

CoolKid
  • 2,738
  • 1
    Hint: How can you express "$A$ is not invertible" in terms of the nullspace of $A$? – Fryie Mar 07 '16 at 01:11
  • 1
    Hint: The determinant of A is the product of its eigenvalues. – SplitInfinity Mar 07 '16 at 01:12
  • 2
    your fact "$A$ is invertible iff $\det A=0$" is wrong. It's the opposite, actually – Alex Mathers Mar 07 '16 at 01:13
  • 1
    @SplitInfinity that's not true - it's generally only true for matrices that have a Jordan Canonical Form. There are matrices (over the reals) that do not have eigenvalues, yet the determinant still has to be defined. – Fryie Mar 07 '16 at 01:16
  • 1
    @SplitInfinity The determinant over any algebraically complete field is the product of its eigenvalues. Consider $\begin{pmatrix}0 & -1\ 1 & 0 \end{pmatrix}$ over $\mathbf{R}$. – Nobody Mar 07 '16 at 01:16
  • Strictly speaking, the determinant condition for invertibility is $A$ is invertible $\Leftrightarrow$ $\det(A)$ is invertible. If $A$ is a matrix over a field, this indeed means $\det(A) \neq 0$, but $A$ could also be a matrix over a ring instead, where there are other non-invertible elements. – Fryie Mar 07 '16 at 01:22
  • I'm somewhat at a loss to know how to review this Question. The valid problem is covered by the proposed duplicate, but the OP here seems not to recognize the contrapositive form. Voting to close as unclear. – hardmath Mar 07 '16 at 22:19

2 Answers2

4

Straight from the definition... $\lambda = 0$ is an eigenvalue if and only if there is a $v \neq 0$ such that $Av = 0$. But this happens if and only if $\ker A \neq \{ 0 \}$, which happens if and only if $A$ is not invertible.

Geometrically, an eigenvalue of zero means that the transformation $A$ compresses the line spanned by $v$ to a point, i.e., applying $A$ reduces dimensions. Thus it cannot have full rank.

Alex Provost
  • 20,991
0

  • If $det(A)=0$ then $0$ is an eigenvalue.

  • If $0$ is an eigenvalue then $det(A)=0$. Recall that for a given conditional its contrapositive is always true.

Nitin Uniyal
  • 7,946