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

You are watching: Why is determinant the product of eigenvalues

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.  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.

Share
Cite
Folshort
edited Sep 13 "20 at 10:23 Gaurang Tandon
answered Sep 28 "13 at 7:42 onimonionimoni
\$endgroup\$
11
14
\$egingroup\$
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\$.

See more: Is There A Tak And The Power Of Juju Games For Sale, Video Game / Tak And The Power Of Juju

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.