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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.

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

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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.

$A$ 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 $A$.

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.

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.

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