Linear Algebra: characteristic polynomials and trace

[MrsM]

The question is : Is it true that two matrices with the same characteristic polynomials have the same trace?

I know that similar matrices have the same trace because they share the same eigenvalues, and I know that if two matrices have the same eigenvalues, they have the same trace. But I am struggling on linking the information in my head.

Your help is appreciated. Thanks :)

 

[PeroK]

The question is : Is it true that two matrices with the same characteristic polynomials have the same trace?

I know that similar matrices have the same trace because they share the same eigenvalues, and I know that if two matrices have the same eigenvalues, they have the same trace. But I am struggling on linking the information in my head.

Your help is appreciated. Thanks :)

The trace is related to a particular coefficient in the characteristic polynomial. Can you see which one?

 

[fresh_42]

Hint: Develop the determinant in the characteristic polynomial along the first row and see which term involves the trace.

 

[MrsM]

Dear Fresh_42 and Perok,

Thank you for responding, the question included no specific matrix. It is a question of concept. I suppose I could try to make one. But I was hoping I could answer using concepts.

Perok, I think it is for X^2, Y^2, Z^2 in a 3x3 matrix.

Fresh_42, I don't understand what you mean by developing the determinant along the first row to get the trace. I thought I added the eigenvalues to get the trace.

 

[PeroK]

Dear Fresh_42 and Perok,

Thank you for responding, the question included no specific matrix. It is a question of concept. I suppose I could try to make one. But I was hoping I could answer using concepts. To answer your question Perok, I think it is for X^2, Y^2, Z^2 in a 3x3 matrix.

I'm not sure what you mean by that. Try to calculate the characteristic polynomial for a general $2 \times 2$ matrix. Using that can you do it for a general $3 \times 3$ matrix. Which coefficients of the polynomial seem to be following a pattern? Is one of them the trace? Can you generalise to $n \times n$ matrices using induction?

 

[fresh_42]

The characteristic polynomial is defined by $\det (A-\lambda \cdot I)$. And the determinant is the sum of the products along all even diagonals (from top left to bottom right) minus the sum of the products along all odd diagonals (from top right to bottom left). It can be computed with the Laplace formula which makes it accessible to proofs by induction.

 

[MrsM]

Fresh,
I don't understand, but I'm definitely trying. I've been up since 3am and have only completed 5 out of 30 questions on this assignment. I am looking at dozens of determinant calculations I have done over the past few weeks using Laplace. They are done correctly, but I am still struggling with how it relates to trace and the polynomial.

Perok, I did the following calculations per your suggestion, but fail to see the connection you are nudging me towards. please see below:
A is now
1, -3;
-3, 9;
the_Display_Matrix is now
1, -3;
-3, 9;
The determinant of the_Display_Matrix is
0;
the_Display_Matrix is now
0;
The trace of the_Display_Matrix is 0.
The trace of A is 10.
The determinant of A is
0;
B is now
1, 0, 1;
-2, 1, -3;
1, 2, 0;
The determinant of B is
1;
The trace of B is 2.
The eigenvalues of B are
0.17609282675202476 + 0 i
0.9119535866239881 + 2.201627508452126 i
0.9119535866239881 + -2.201627508452126 i
The characteristic polynomial of B is given by
det( B - z I ) = 1 + (-6) z + (2) z^2 + (-1) z^3

 

[PeroK]

Fresh,
I don't understand, but I'm definitely trying. I've been up since 3am and have only completed 5 out of 30 questions on this assignment. I am looking at dozens of determinant calculations I have done over the past few weeks using Laplace. They are done correctly, but I am still struggling with how it relates to trace and the polynomial.

Perok, I did the following calculations per your suggestion, but fail to see the connection you are nudging me towards. please see below:
A is now
1, -3;
-3, 9;
the_Display_Matrix is now
1, -3;
-3, 9;
The determinant of the_Display_Matrix is
0;
the_Display_Matrix is now
0;
The trace of the_Display_Matrix is 0.
The trace of A is 10.
The determinant of A is
0;
B is now
1, 0, 1;
-2, 1, -3;
1, 2, 0;
The determinant of B is
1;
The trace of B is 2.
The eigenvalues of B are
0.17609282675202476 + 0 i
0.9119535866239881 + 2.201627508452126 i
0.9119535866239881 + -2.201627508452126 i
The characteristic polynomial of B is given by
det( B - z I ) = 1 + (-6) z + (2) z^2 + (-1) z^3

I have no idea what all that means. The characteristic equation of the general $2 \times 2$ matrix:
$$\left( \begin{array} \
a & b \\
c & d \end{array} \right)$$
Is:
$$\left| \begin{array} \
a-\lambda & b \\
c & d-\lambda \end{array} \right| = \lambda^2 - (a+d)\lambda + ad-bc$$
Do any of these coefficients look familiar?

 

[MrsM]

I found this, because you said to look at a 2x2 and a 3x3, and see if they have something in common, and I am trying to see the correlation, but aside from that they both have something involving Trace in the middle, I don't. I appreciate your help, but at this point I have spent about 2 hours on a problem, and I should just move on to the next one. Thank you anyhow.

(3)
where

is the matrix trace of

and

is its determinant.

Similarly, the characteristic polynomial of a

matrix is

which can also be written explicitly in terms of traces as

 

[fresh_42]

Yes, and it generalizes to all matrices: the trace is the second highest coefficient, the determinant the lowest.

 

[MrsM]

So does that mean since they have the same coefficients, the equation is the same? What if the values of x and y are different? Does that make the equation still the same?

And since the coefficients are the same, and the trace is represented by the second highest one, does that mean if two matrices have the same characteristic polynomial, the trace is the same? I would say yes, but my brain is so fried right now from going in circles... I just don't really know

 

[fresh_42]

Equality of two polynomials means equality of each coefficient. So if $\det(A-\lambda \cdot I)=\det(B-\lambda \cdot I)$ then $tr(A) = tr(B)$, as the traces are the second highest coefficient each.