Coefficients of the characteristic equation

[spaghetti3451]

let's say you are given the following matrix T:

T₁₁ T₁₂ . . . T_1n
T₂₁ T₂₂ . . . T_2n
. . .
. . .
. . .
T_n1 T_n2 . . . T_nn

You want to find the characteristic equation for the matrix, which is
$C_{n} \lambda^{n} + C_{n-1} \lambda^{n-1} + ... + C_{1} \lambda + C_{0} = 0.$

Now, $C_{n} = (-1)^{n}$, $C_{n-1} = (-1)^{n-1} Tr(T)$, and $C_{0} = det(T)$.

I have been having a pretty hard proving these three formulae. The first one looks intuitively obvious and I think I have a fairly good understanding of why the second one has the structure that it has. But the complexity of the matrix notation and the fact that there are so many elements in the matrix is halting my attempts to precisely determine each of the formulae from the matrix. I would appreciate any help. Thanks.

 

[DonAntonio]

let's say you are given the following matrix T:

T₁₁ T₁₂ . . . T_1n
T₂₁ T₂₂ . . . T_2n
. . .
. . .
. . .
T_n1 T_n2 . . . T_nn

You want to find the characteristic equation for the matrix, which is
$C_{n} \lambda^{n} + C_{n-1} \lambda^{n-1} + ... + C_{1} \lambda + C_{0} = 0.$

Now, $C_{n} = (-1)^{n}$, $C_{n-1} = (-1)^{n-1} Tr(T)$, and $C_{0} = det(T)$.

I have been having a pretty hard proving these three formulae. The first one looks intuitively obvious and I think I have a fairly good understanding of why the second one has the structure that it has. But the complexity of the matrix notation and the fact that there are so many elements in the matrix is halting my attempts to precisely determine each of the formulae from the matrix. I would appreciate any help. Thanks.

The characteristic polynomial, not equation (unless equalled to zero or some other number) is the determinant of $T-xI$ .

If we also know that the determinant is the sum of products of n numbers, each one of which is formed with exactly one element from each row and each column of the matrix, we get at once that one of such products in the above determinant is $(-x)(-x)\cdot...\cdot (-x)=(-1)^nx^n$ , and there is no other product as above with n x's, so this last is the greatest degree term in the pol., and you get $C_n$ .

Now you try to deduce the other two.

DonAntonio

 

[ajkoer]

Co = Det (T) is easy. The last term Co is always the product of the roots. The roots of the characteristic equation for the matrix are the eigenvalues. As the product of the eigenvalues is equal to the Determinant, we are done. Proof of the last statement:

Det (T) = Det (P'EP) = Det(P'PE) = Det (I*E) = Det (E) = e1*e2*...

where E is a diagonal matrix with the eigenvalues on the diagonal of matrix T and P'P = I.

Similarly, the second term is always the minus sum of the roots. Note, rescaling the term becomes C(n-1)/C(n) = -1 as was required.

 

[morphism]

Co = Det (T) is easy. The last term Co is always the product of the roots. The roots of the characteristic equation for the matrix are the eigenvalues. As the product of the eigenvalues is equal to the Determinant, we are done. Proof of the last statement:

Det (T) = Det (P'EP) = Det(P'PE) = Det (I*E) = Det (E) = e1*e2*...

where E is a diagonal matrix with the eigenvalues on the diagonal of matrix T and P'P = I.

Similarly, the second term is always the minus sum of the roots. Note, rescaling the term becomes C(n-1)/C(n) = -1 as was required.

One minor difficulty with this approach is that the OP didn't specify the field over which they're working, so one might have to enlarge it so that it contains the eigenvalues of T, and then one has to prove that det and Tr once taken over this enlarged field will give the same values as over the original field.

A more serious problem, however, is that your proof seems to assume that T is diagonalizable. This issue really needs to be addressed. (Suggestion: If we're working over a field that contains the eigenvalues of T then, although T may not be diagonalizable over this field, it's still possible to find a matrix P such that $PTP^{-1}=E$ is upper triangular. This would salvage your argument.)

 

[blue_raver22]

I understand your frustration with trying to prove these formulae. However, there are some steps you can take to make the process easier.

First, let's break down the characteristic equation and understand what each term represents. The characteristic equation is a polynomial equation that allows us to find the eigenvalues of a matrix, which are the values of lambda that satisfy the equation. The degree of the polynomial is equal to the size of the matrix (n), and the coefficients represent certain properties of the matrix.

Now, let's look at each of the coefficients:

1. C_{n}: This coefficient is the leading coefficient of the polynomial and is always equal to (-1)^n. This is because the characteristic equation is derived from the determinant of the matrix, which is always multiplied by (-1)^n based on the size of the matrix.

2. C_{n-1}: This coefficient is equal to (-1)^{n-1} times the trace of the matrix (Tr(T)). The trace of a matrix is the sum of its diagonal elements, and it represents the sum of the eigenvalues. This is why the coefficient is multiplied by (-1)^{n-1}, as it takes into account the sign of the eigenvalues.

3. C_{0}: This coefficient represents the determinant of the matrix (det(T)). The determinant of a matrix is a scalar value that represents the scaling factor of the matrix. It is also equal to the product of all the eigenvalues of the matrix.

Now, in order to prove these formulae, you can use mathematical induction. Start with a 2x2 matrix and prove the formulae for n=2, and then continue to larger matrices. You can also use properties of determinants and traces to prove these formulae.

I hope this helps and good luck with your proof! If you need further assistance, don't hesitate to seek out a math tutor or consult with a colleague who may have more expertise in this area.