Show that the determinant of $A$ is equal to the product of its eigenvalues

Show that the determinant of a matrix $A$ is equal to the product of its eigenvalues $\lambda_i$.

So I’m having a tough time figuring this one out. I know that I have to work with the characteristic polynomial of the matrix $\det(A-\lambda I)$. But, when considering an $n \times n$ matrix, I do not know how to work out the proof. Should I just use the determinant formula for any $n \times n$ matrix? I’m guessing not, because that is quite complicated. Any insights would be great.

Answer

Suppose that $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A$. Then the $\lambda$s are also the roots of the characteristic polynomial, i.e.

$$\begin{array}{rcl} \det (A-\lambda I)=p(\lambda)&=&(-1)^n (\lambda – \lambda_1 )(\lambda – \lambda_2)\cdots (\lambda – \lambda_n) \\ &=&(-1) (\lambda – \lambda_1 )(-1)(\lambda – \lambda_2)\cdots (-1)(\lambda – \lambda_n) \\ &=&(\lambda_1 – \lambda )(\lambda_2 – \lambda)\cdots (\lambda_n – \lambda)
\end{array}$$

The first equality follows from the factorization of a polynomial given its roots; the leading (highest degree) coefficient $(-1)^n$ can be obtained by expanding the determinant along the diagonal.

Now, by setting $\lambda$ to zero (simply because it is a variable) we get on the left side $\det(A)$, and on the right side $\lambda_1 \lambda_2\cdots\lambda_n$, that is, we indeed obtain the desired result

$$ \det(A) = \lambda_1 \lambda_2\cdots\lambda_n$$

So the determinant of the matrix is equal to the product of its eigenvalues.

Attribution
Source : Link , Question Author : onimoni , Answer Author : Gaurang Tandon

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