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The only thing I can think of I that suppose A is \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{pmatrix} After subtracted from $0$, we get \begin{pmatrix} -a & -b & -c \\ -d & -e & -f \\ -g & -h & -i \\ \end{pmatrix}

Using Cofacter expansion we get Characteristic Polynomial = -det(A). And if the matrix becomes 2$\times$2, Characteristic Polynomial=det(A). Since I am not too familiar with the theorems, how is it related to invertibility?

Could someone give me hint?

CoolKid
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4 Answers4

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By definition of eigenvalue, a matrix $A$ has $\lambda$ as eigenvalue if and only if $A-\lambda I$ is not invertible.

Indeed, $\lambda$ is an eigenvalue if and only if there exists $v\ne0$ such that $Av=\lambda v$, which is equivalent to saying that $(A-\lambda I)v=0$. So another equivalent way to say it is “the null space of $A-\lambda I$ is not the trivial subspace” and, by the rank-nullity theorem, “the matrix $A-\lambda I$ is not invertible”.

In the particular case of $\lambda=0$ you have your statement.

(Note: $I$ denotes the identity matrix of the suitable size.)

egreg
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There is a theorem that states the following: the determinant of a matrix is equal to the product of its eigenvalues (taking algebraic multiplicity into account). Look at this Math.SE question.

This means that if a matrix has an eigenvalue zero, then its determinant is zero.

Now, it is a known result that matrices with determinant zero are not invertible.

Community
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Pedro A
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Let $f$ be the endomorphism associated to $A$. $A$ having $0$ as an eigenvalue is equivalent to $\ker f$ is non-trivial, which is equivalent to $f$ being non-injective, and a fortiori being non bijective. This translates into $A$ being non-invertible.

Bernard
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For this proof, use the definition of an eigenvalue:

Let $A$ be an $n × n$ matrix. A number $\lambda$ is called an eigenvalue of $A$ provided that there exists a nonzero vector v in $R^n$ such that $$Av = \lambda v$$

Proof

Suppose that $\lambda = 0$ is an eigenvalue of $A$.

Then there exists a nonzero vector $v$ such that $Av = 0$

Thus A is not invertible (else the homogeneous equation would have a unique solution, the trivial one)

Conversely,

Suppose $A$ is not invertible. Then the homogeneous equation $Ax = 0$ has a nontrivial solution $v$. Since $Av = 0 = 0v$, then $0$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda = 0$.

QED

Kevin Zakka
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