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

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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 $n imes n$ matrix, I do not know how to work out the proof. Should I just use the determinant formula for any $n imes n$ matrix? I"m guessing not, because that is quite complicated. Any insights would be great.


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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.

$$eginarrayrcl det (A-lambda I)=p(lambda)&=&(-1)^n (lambda - lambda_1 )(lambda - lambda_2)cdots (lambda - lambda_n) \ &=&(-1) (lambda - lambda_1 )(-1)(lambda - lambda_2)cdots (-1)(lambda - lambda_n) \ &=&(lambda_1 - lambda )(lambda_2 - lambda)cdots (lambda_n - lambda)endarray$$

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 $lambda_1 lambda_2cdotslambda_n$, that is, we indeed obtain the desired result

$$ det(A) = lambda_1 lambda_2cdotslambda_n$$

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


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Folshort
edited Sep 13 "20 at 10:23
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Gaurang Tandon
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answered Sep 28 "13 at 7:42
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onimonionimoni
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I am a beginning Linear Algebra learner and this is just my humble opinion.

One idea presented above is that

Suppose that $lambda_1,ldots lambda_2$ are eigenvalues of $A$.

Then the $lambda$s are also the roots of the characteristic polynomial, i.e.

$$det(A−lambda I)=(lambda_1-lambda)(lambda_2−lambda)cdots(lambda_n−lambda)$$.

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 $lambda_1lambda_2ldots lambda_n$, that is, we indeed obtain the desired result

$$det(A)=lambda_1lambda_2ldots lambda_n$$.

I dont think that this works generally but only for the case when $det(A) = 0$.

Because, when we write down the characteristic equation, we use the relation $det(A - lambda I) = 0$ Following the same logic, the only case where $det(A - lambda I) = det(A) = 0$ is that $lambda = 0$. The relationship $det(A - lambda I) = 0$ must be obeyed even for the special case $lambda = 0$, which implies, $det(A) = 0$

UPDATED POST

Here i propose a way to prove the theorem for a 2 by 2 case. Let $A$ be a 2 by 2 matrix.

$$ A = eginpmatrix a_11 & a_12\ a_21 & a_22\endpmatrix$$

The idea is to use a certain property of determinants,

$$ eginvmatrix a_11 + b_11 & a_12 \ a_21 + b_21 & a_22\endvmatrix = eginvmatrix a_11 & a_12\ a_21 & a_22\endvmatrix + eginvmatrix b_11 & a_12\b_21 & a_22\endvmatrix$$

Let $ lambda_1$ and $lambda_2$ be the 2 eigenvalues of the matrix $A$. (The eigenvalues can be distinct, or repeated, real or complex it doesn"t matter.)

The two eigenvalues $lambda_1$ and $lambda_2$ must satisfy the following condition :

$$det (A -Ilambda) = 0 $$Where $lambda$ is the eigenvalue of $A$.

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Therefore, $$eginvmatrix a_11 - lambda & a_12 \ a_21 & a_22 - lambda\endvmatrix = 0 $$

Therefore, using the property of determinants provided above, I will try to decompose the determinant into parts.

$$eginvmatrix a_11 - lambda & a_12 \ a_21 & a_22 - lambda\endvmatrix = eginvmatrix a_11 & a_12 \ a_21 & a_22 - lambda\endvmatrix - eginvmatrix lambda & 0 \ a_21 & a_22 - lambda\endvmatrix= eginvmatrix a_11 & a_12 \ a_21 & a_22\endvmatrix - eginvmatrix a_11 & a_12 \ 0 & lambda \endvmatrix-eginvmatrix lambda & 0 \ a_21 & a_22 - lambda\endvmatrix$$

The final determinant can be further reduced.

$$eginvmatrix lambda & 0 \ a_21 & a_22 - lambda\endvmatrix = eginvmatrix lambda & 0 \ a_21 & a_22 \endvmatrix - eginvmatrix lambda & 0\ 0 & lambda\endvmatrix$$

Substituting the final determinant, we will have

$$eginvmatrix a_11 - lambda & a_12 \ a_21 & a_22 - lambda\endvmatrix = eginvmatrix a_11 & a_12 \ a_21 & a_22\endvmatrix - eginvmatrix a_11 & a_12 \ 0 & lambda \endvmatrix - eginvmatrix lambda & 0 \ a_21 & a_22 \endvmatrix + eginvmatrix lambda & 0\ 0 & lambda\endvmatrix = 0 $$

In a polynomial$$ a_nlambda^n + a_n-1lambda^n-1 ........a_1lambda + a_0lambda^0 = 0$$We have the product of root being the coefficient of the term with the 0th power, $a_0$.

From the decomposed determinant, the only term which doesn"t involve $lambda$ would be the first term

$$eginvmatrix a_11 & a_12 \ a_21 & a_22 \endvmatrix = det (A)$$

Therefore, the product of roots aka product of eigenvalues of $A$ is equivalent to the determinant of $A$.

I am having difficulties to generalize this idea of proof to the $n$ by $$ case though, as it is complex and time consuming for me.