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I'm trying to prove that statement:
Prove that $\lambda = 0$ is an eigenvalue if and only if $A$ is singular.

I'm not sure if my proof is totally correct:

Suppose that $\lambda = 0$
if det(A) = $\lambda_1 \cdot \lambda_2 . . .\lambda_n = 0$
then A is singular.

If anyone could show me a better proof that would help.

Mark
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  • What do you think is wrong with your proof? – Nick Jan 10 '14 at 22:50
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    What's correct depends on what facts you are allowed to use. You are using the fact that the determinant in the product of the eigenvalues, and the fact that singular is the same as determinant zero (or maybe that's your definition of singular). Another approach is eigenvalue zero implies non-trivial nullspace implies non-invertible. – Gerry Myerson Jan 10 '14 at 22:51
  • @Nick just my definition of the determinant of A mostly. I just thought I was missing something here. – Mark Jan 10 '14 at 22:57
  • @Gerry I can see how that approach may work, but I'm trying to see if I can find a simpler proof. If not, I may use that. – Mark Jan 10 '14 at 22:58

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$A$ is singular $\iff x\mapsto Ax$ is not injective $\iff$ we can find $x\neq 0$ with $Ax=0\iff 0 $ is an eigenvalue of $A$.

Pedro
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Your proof is right, albeit a little unclearly written. However, you don't need the machinery of the determinant to prove this. If $\lambda=0$ is an eigenvalue of $A$, then this means there's some non-zero vector $v$ with $Av=\lambda v=0v=0$. That is, $\ker A$ is non-trivial, so $A$ is singular.

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Given the fact that the determinant of $A$ is the product of the eigenvalues, then this is sufficient.

Alternatively, suppose that $0$ is an eigenvalue of $A$, with corresponding non-zero eigenvector $v$. If $A$ were non-singular, then we could write

$$v = Iv = A^{-1} Av = A^{-1} 0v = 0$$

Alternatively, if $A$ is singular then $A$ must have a non-trivial null space, since it's not injective (viewed as a linear transformation). Hence $0$ is an eigenvalue.

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I'm not sure if this quite proves it, but this is the "proof" I worked through to answer the same problem I was assigned for HW.

If λ = 0, then det(λI - A)=0 becomes det(-A)=0. the determinant of A (regardless of the fact that it is scaled by -1 in this case) must therefore be zero, which means the matrix A is singular.

Aviv
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  • It seems to be a different approach to a proof, but given how I'm currently learning L.A. myself, I may not see the similarity between the one I wrote and one of the other answers. – Aviv Dec 06 '16 at 22:42
  • That being said, next time I'll be sure to be certain of any future answers before posting them – Aviv Dec 06 '16 at 22:43
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I know that this thread is somewhat old with an accepted answer, but I just took a course in Linear Algebra at my university and I was told that in order to prove iff instead of if you must prove that the implication goes both ways:

(1) Show that $\lambda=0\implies A$ is singular: $$\lambda=0\implies 0=|A-\lambda I|= |A-(0)I|= |A|$$ (2) Show that $A$is singular$\implies0$ is an eigenvalue of $A$: $$|A|=\Pi_{i=1}^{n}\lambda_{i}\ \land|A|=0\implies\lambda_{k} = 0\mid1\leq k\leq n$$ $$\therefore\lambda=0\iff |A|$$

joshuaneigh
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