How to prove that if the determinant of the matrix is zero then at least one eigenvalue must be zero?

Question

For a matrix $A$, if $\det(A)=0,$ prove and provide an example that at least one eigenvalue must be zero.

At first, I tried using the identity that the product of eigenvalues is the determinant of the matrix, so it follows that at least one must be zero for the determinant to be zero. Is this correct? Could I also prove it by using $(A-\lambda I)X=0$, for some $X\neq 0?$

If $\lambda=0,$ then we have $AX=0$, but I can't say $\det(A)\cdot \det(X)=0$ because $X$ is not a square matrix and doesn't have a determinant. How would I continue?

Answer 1

Let $p(x)=\det(A-xI)$ the char. polynomial of $A$. Then $p(0)=\det(A)=0$, hence $0$ is a root of $p$ and therefore an eigenvalue of $A$.

Answer 2

The determinant of the matrix $A$ also is the determinant of the endomorphism $\mathbb{R}^n \rightarrow \mathbb{R}^n$ (or more generally $k^n$) defined by multiplication by $A$. To say that $A$ has determinant $0$ is to say that this endomorphism is not injective.

Answer 3

Here an elementary way:

$\det(A) = 0 \Rightarrow$ the columns of $A =(c_1 \ldots c_n)$ are linearly dependent $\Rightarrow$ there is a non-zero vector $v = (v_1 \ldots v_n)^T$ such that $v_1c_1 + \cdots v_n c_n = \vec{0} \Rightarrow Av = \vec{0} = 0\cdot v \Rightarrow 0$ is an eigenvalue of $A$.

Answer 4

You are correct that the product of eigenvalues is the determinant, with appropriate muliplicities.
This readily follows from the Jordan normal form.

Once it's known, the problem is solved, as a product in the base field becomes zero iff one of the factors is $0$.

Answer 5

Since matrix $A$ over a field and det$A$ is equal to the product of eigenvalues, by using a property of the field that if $ab=0 \Rightarrow$ either $a=0$ or $b=0.$

Answer 6

I do not know what you know about the determinant and how you think of it, but the determinant of a square matrix $A$ is zero iff the matrix is not invertible, and that is equivalent to the kernel being non-trivial, which means that $Ax=0$ for some $x\ne0$.