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I would like to know if there is any proof without using the fact that:

$$\lambda_1\cdot\lambda_2\cdot\ldots\cdot\lambda_{n-1}\cdot\lambda_n = det(A)$$

I managed to prove that if $\lambda = 0$ then, A is singular, by:

$$ det(A-\lambda\cdot I)=0 \Rightarrow det(A-0\cdot I)=0 \Leftrightarrow det(A)=0 $$

However I can't prove that if A is singular then, $\lambda = 0 $.

I talked to my teacher and he told me that maybe I could prove by induction that:

$$ p(\lambda)=a_n\lambda^n+\ldots+a_1\lambda+a_0 $$

where,

$$a_0=det(A)$$

Then:

$$ p(\lambda)=a_n\lambda^n+\ldots+a_1\lambda\Rightarrow p(\lambda)=\lambda(a_n\lambda^{n-1}+\ldots+a_1) $$

which guarantees that $\lambda=0$ is an eigenvalue.

If anyone could help, I would greatly appreciate it.

Victor Pinto
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2 Answers2

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A is singular iff there exists a nonzero vector X such that AX=0 iff 0 is an eigenvalue.

No determinants required here, just the definition ;)

Béranger Seguin
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The following answer is not as short and elegant as the one by @Béranger Seguin, but it I think it is interesting\important to understand it since it provides a good sense of what happens to eigenvalues of singular matrix.

Let $A$ be singular matrix, then $A$ is equivalent to a matrix $A'$ with a row of zeros. Let $\phi(I)=\phi(I)_1\phi_{n-1}(I)...\phi_1(I)$ be a sequence of elementary operation which one do to obtain $A'$ from $A$, i.e. $A'=\phi(I)A=\phi(I)_1\phi_{n-1}(I)...\phi_1(I)A$

the following polynomial has the same roots as $\det(A-\lambda I)$ $$\det\phi(I) \det(A-\lambda I)=\det(\phi(I)(A-\lambda I)) =det(A'-\lambda \phi(I))$$ since $A'$ has a row of zeros $A'-\lambda \phi(I)$ has a row $i$ with $-\lambda$ on a main diagonal. Calculate determinant by this row to get $$det(A'-\lambda \phi(I)) = -\lambda M_{ii} $$ where $M_{ii}$ denote a minor. Note that $-\lambda M_{ii}=0$ has a solution $\lambda=0$.

Michael Medvinsky
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