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Question: Let A be a 3 x 3 matrix with characteristic polynomial $det(A - \lambda I) = -\lambda^3 + 3\lambda - 2$. Find det(A) and tr(A)

I have no idea how to start this question. I know that the properties for det(A) and tr(A) is

If A has eigenvalues $\lambda_1,\lambda_2,\cdots,\lambda_n$, then det(A) = $\lambda_1,\cdots,\lambda_n$ and tr(A) = $\lambda_1 + \lambda_2 + \cdots + \lambda_n$

But I have no idea how to use this definition to answer this question.

$\lambda^3 + 3\lambda - 2 = -(\lambda-1)^2 (\lambda + 2)$

$det(A) = -2 \cdot 1 = -2 ? $

$tr(A) = -2 + 1 = -1 ?$

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    $\begingroup$ You can use Vieta's formula $\endgroup$ Apr 5, 2017 at 3:39
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    $\begingroup$ What does that have to do with Eigenvalues? I mean if you meant to factor them u get $= -(\lambda-1)^2(\lambda + 2)$ $\endgroup$
    – user349557
    Apr 5, 2017 at 4:16
  • $\begingroup$ Expand $(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)$ and compare the coefficients that you end up with the two facts about eigenvalues that you’ve quoted. You’ll see that you can solve the problem without explicitly computing any eigenvalues. $\endgroup$
    – amd
    Apr 5, 2017 at 5:10
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    $\begingroup$ Eigenvalues are 1,1 and -2. So trace(A) should be 0. This can also be directly seen from the characteristic polynomial. Trace is the coefficient of $\lambda^2$ and determinant is nothing but the constant term in the polynomial. $\endgroup$ Apr 5, 2017 at 6:51
  • $\begingroup$ Useful note: A very easy way to get the determinant of any square matrix from the characteristic polynomial is to plug in $\lambda=0$ as $det(A-0I)= det(A)$. Or as the above comments put it, the constant term of the polynomial. $\endgroup$
    – B.A
    Apr 29, 2017 at 2:36

2 Answers 2

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This answer is just to give you a little more insight for this problem.

The general form of characteristic polynomial of $n\times n$ matrix $A$ can be written as follows

Char(A) = $\lambda^n - trace(A) \lambda^{n-1} + $terms of lower degree. Adding to this, the constant term of the characteristic polynomial of the matrix $A$ is $(-1)^n det(A)$.

In your problem the characteristic polynomial of the matrix $A$ can be written as $\lambda^3 - 3\lambda + 2 = 0$

From the above argument it is clear that $Trace(A) = 0$ (Since coefficient of $\lambda^2 = 0$) and $det(A) = (-1)^3 \times (constant~ term) = (-1)^3 \times 2 = -2$.

This result can be useful in the computation of trace and determinant of the matrix without computing its eigenvalues. As in many cases you may find it difficult to factorize the polynomial in $\lambda$ to compute the eigenvalues. Also, there is a result that

For higher degrees poynomial, no general formula exists (or more precisely, no formula in terms of addition, subtraction, multiplication, division, arbitrary constants and $n-th $roots). This result is proved in Galois theory and is known as the Abel-Ruffini theorem. For more details you can click here.

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The characteristics equation of $A$ is given by: \begin{align*} \left|A-\lambda I\right|&=0\\ \implies -\lambda^3 + 3\lambda - 2&=0\hspace{25pt}\cdots(i)\\ \implies(\lambda-1)^2(\lambda+2)&=0 \end{align*} So the eigen values are given by: $\lambda_1=1,\ \lambda_2=1,\ \lambda_3=-2$.

Now $Tr(A)=$ Sum of eigen values $=\lambda_1+\lambda_2+\lambda_3=0$, and

$det(A)=$ Product of eigen values $=\lambda_1\cdot\lambda_2\cdot\lambda_3=-2$.

Or, you can apply directly Vita's formula in (i) which gives:

Sum of roots of polynomial $=Tr(A)=-\dfrac{0}{-1}=0$, and

Product of roots of polynomial $=det(A)=-\dfrac{-2}{-1}=-2$.

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