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I am reading a proof of the result: a square matrix $A$ is invertible if and only if $\lambda = 0$ is not an eigenvalue of $A$.

Proof goes as follows: Assume that $A$ is an $n\times n$ matrix and observe that $\lambda = 0$ is the solution of the characteristic equation of $A$ given by

$det (\lambda I - A) = \lambda^n + c_1 \lambda^{n-1} + \ldots c_n = 0 $.

On setting $\lambda = 0$ in above equation, we get $det(-A) = c_n$ or $(-1)^n det(A) = c_n$ Now it follows that $det(A) = 0$ if and ony if $a_n \neq 0$.

I am curious to know that if I want to write the characteristic equation of $A$ as $det( A - \lambda I)$ then what would be the general form of characteristic equation in terms of polynomial in $\lambda$.

Can I still write the characteristic equation of $A$ as

$det( A - \lambda I)$= $\lambda^n + c_1 \lambda^{n-1} + \ldots c_n = 0 $.

Please clarify my doubt. As writing in this form changes the fact that the constant term of characteristic polynomial is $(-1)^n det(A)$, where $n$ is the order of the matrix.

Thank you

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The characteristic polynomial of a matrix $n$ is $\det(A-\lambda I_n)$, this polynomial is always a degree $n$ monic polynomial in $\lambda$, and the roots of this polynomial are exactly the eigenvalues of $A$.

So $0$ is an eigenvalue of $A$ if and only if $\det(A-0I_n)=\det(A)=0$ if and only if $A$ is not invertible.

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  • $\begingroup$ If $n$ is odd, then $\det(A - \lambda I_n)$ is not a monic polynomial in $\lambda$. $\endgroup$ – Sheldon Axler Apr 28 '17 at 5:54
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For any order-$n$ square matrix $M$, $\det(-M)=(-1)^n\det(M)$. You can prove this for yourself by noting that multiplying a row of column of a matrix by a scalar $c$ multiplies its determinant by the same number. Thus, the two polynomials that you get differ only in sign. However, a polynomial $p[x]$ has the same roots as $-p[x]$ since you can multiply both sides of the equation $-p[x]=0$ by $-1$ without changing its solutions.

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