Consider polynomials A(z) and B(z) defined by A(z)=z^2+a1 z+a2=(z-λ1)(z-λ2)=z^2-(λ1+λ2) z+λ

Consider polynomials A(z) and B(z) defined by ...

Consider polynomials $A(z)$ and $B(z)$ defined by $$ \begin{aligned} & A(z)=z^2+a_1 z+a_2=\left(z-\lambda_1\right)\left(z-\lambda_2\right)=z^2-\left(\lambda_1+\lambda_2\right) z+\lambda_1 \lambda_2 \\ & B(z)=b_1 z+b_2=b\left(z-\lambda_3\right)=b z-b \lambda_3 \end{aligned} $$ where $b=b_1$. Define the Sylvester matrix $\mathbf{E}$ by $$ \mathbf{E}=\left[\begin{array}{llll} a_2 & 0 & b_2 & 0 \\ a_1 & a_2 & b_1 & b_2 \\ 1 & a_1 & 0 & b_1 \\ 0 & 1 & 0 & 0 \end{array}\right] $$ Show that the determinant of $\mathbf{E}$ can be given by $$ |\mathbf{E}|=b^2\left(\lambda_1-\lambda_3\right)\left(\lambda_2-\lambda_3\right) $$

Consider polynomials $A(z)$ and $B(z)$ defined by
$$
\begin{aligned}
& A(z)=z^2+a_1 z+a_2=\left(z-\lambda_1\right)\left(z-\lambda_2\right)=z^2-\left(\lambda_1+\lambda_2\right) z+\lambda_1 \lambda_2 \\
& B(z)=b_1 z+b_2=b\left(z-\lambda_3\right)=b z-b \lambda_3
\end{aligned}
$$
where $b=b_1$. Define the Sylvester matrix $\mathbf{E}$ by
$$
\mathbf{E}=\left[\begin{array}{llll}
a_2 & 0 & b_2 & 0 \\
a_1 & a_2 & b_1 & b_2 \\
1 & a_1 & 0 & b_1 \\
0 & 1 & 0 & 0
\end{array}\right]
$$

Show that the determinant of $\mathbf{E}$ can be given by
$$
|\mathbf{E}|=b^2\left(\lambda_1-\lambda_3\right)\left(\lambda_2-\lambda_3\right)
$$

Key Concepts

Determinant Properties

Understanding the properties of determinants is crucial in simplifying and evaluating structured matrices like the Sylvester matrix. Properties such as multilinearity, the effect of row and column operations, and the behavior under block matrix partitions are instrumental in reducing the complexity of the determinant computation. These properties allow one to express the determinant in a factored form involving differences of roots, as seen in the final result b²(?? - ??)(?? - ??).

Polynomial Factorization

This concept involves expressing a polynomial in terms of its roots. It establishes a direct connection between a polynomial's coefficients and its zeros. For instance, the quadratic polynomial A(z) is represented both in its expanded form and as a product of linear factors, revealing that the sum and product of its roots determine its coefficients. This dual representation is key when relating the roots to other polynomial expressions or matrices built from these coefficients.

Sylvester Matrix

The Sylvester matrix is a structured matrix constructed from the coefficients of two polynomials. It is designed such that its determinant, known as the resultant, vanishes if and only if the two polynomials share a common root. The structure of the matrix reflects the degrees of the involved polynomials and aligns the coefficients in a specific pattern to facilitate this property, making it an essential tool in elimination theory and algebraic geometry.

Resultant of Polynomials

The resultant is a scalar quantity derived from the coefficients of two polynomials, which indicates whether they have a common root. It is computed as the determinant of the Sylvester matrix associated with the polynomials. If the resultant is zero, the polynomials share a common root; otherwise, they do not. In the context of the problem, expressing the determinant in terms of the polynomial roots confirms the absence or presence of common roots based on parameter values.

Transcript

00:01 Hi there, so for this problem, the polynomial f of seed is defined as the following.

00:06 F of seed is equal to seed to the phi minus six times six to the four plus 15, sit to the three, and this minus 34, sit to the square, and this plus 36, six.

00:39 And the last term in here is minus 48.

00:45 Now the question in here for part a is to show that the equation f of seed is equal to zero and has roots of the form seed is equal to landa times i, where i is real and hence factorize this function.

01:13 This function in here.

01:15 So what we need to do is to substitute that seed is equal to lambda times i in each of these terms in here.

01:26 So when we do that, we obtain the following.

01:28 We obtain that f of seed.

01:31 Then in this case, it's lambda times i is equal to i times lambda to the 5 minus 15, lambda to the 3 and this plus 36 lambda.

01:49 Now remember that you need to take into account that lambda to the square is minus 1 and then lambda to the 3 is equal to minus i and so on.

02:03 So with that set, we will have this is the first term.

02:07 This is the imaginary part and the real part is going to be minus 6 times lambda to the 4 plus 3 times 34 lambda to the square minus 48.

02:20 Now for lambda to be a root, both parts of this equation that is the real part and the imaginary part must be zero.

02:30 So that can happen.

02:33 One option, one solution for this, is when lambda is equal to zero.

02:39 But the other solution comes from the quadratic formula that is that the square of lambda should be equal to 15 plus or minus the square root of 225 minus 104 and 144 and this divided by 2.

03:00 So from this we obtain two values that are going to be 12 or 3.

03:10 And of course this is only for the for the imaginary part.

03:19 And also the other solutions that must be satisfied is when three times lambda to the 4 minus 17, lambda to the square plus 24 is equal to 0.

03:36 So from this, we will obtain that this is just simply lambda to the square is equal to 17.

03:44 This is using the, the quadratic formula.

03:53 Square root of this and this divided by six.

03:56 So from this we obtain two values, three or 16 divided by six.

04:03 Now, only when lambda square is equal to three satisfies both, because we can recognize that that is a root that is common in here.

04:20 And in this case, so in both cases, it satisfies this.

04:25 So it's going to be a root, only when lambda square is equal to 3...