Operations and Properties - Linear Algebra

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Question

$A = \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$ .

Which of the following is a consequence of the Cayley-Hamilton Theorem?

(Note: and refer to the two-by-two identity and zero matrices, respectively.)

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Answer

The Cayley-Hamilton Theorem states that a square matrix is a solution of its own characteristic equation. To find this, first find the polynomial formed from the determinant of $\lambda I-A$ . This matrix is

$\lambda I-A$

$= \lambda \begin{bmatrix} 1& 0\ 0& 1 \end{bmatrix}- \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$

Multiply $\lambda$ by each element in the identity:

$= \begin{bmatrix} \lambda& 0\ 0& \lambda \end{bmatrix}- \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$

Subtract elementwise:

$= \begin{bmatrix} \lambda -1 & -4\ 1& \lambda -3 \end{bmatrix}$

Take the determinant by taking the product of the upper-left and lower-right entries, and subtracting the product of the other two:

$\begin{vmatrix} \lambda -1 & -4\ 1& \lambda -3 \end{vmatrix}$

$= \left ( \lambda -1\right )\left ( \lambda -3\right ) - (-4)(1)$

$= \lambda ^{2} -4 \lambda+3 +4$

$= \lambda ^{2} -4 \lambda+7$

The characteristic equation is

$\lambda ^{2} -4 \lambda+7 = 0$

By the Cayley-Hamilton Theorem, the matrix is a solution to this equation, so the correct choice is the above equation, modified for matrix arithmetic:

$A^{2} -4 A+7I =O$

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Question

$A = \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$ .

Which of the following is a consequence of the Cayley-Hamilton Theorem?

(Note: and refer to the two-by-two identity and zero matrices, respectively.)

Tap to reveal answer

Answer

The Cayley-Hamilton Theorem states that a square matrix is a solution of its own characteristic equation. To find this, first find the polynomial formed from the determinant of $\lambda I-A$ . This matrix is

$\lambda I-A$

$= \lambda \begin{bmatrix} 1& 0\ 0& 1 \end{bmatrix}- \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$

Multiply $\lambda$ by each element in the identity:

$= \begin{bmatrix} \lambda& 0\ 0& \lambda \end{bmatrix}- \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$

Subtract elementwise:

$= \begin{bmatrix} \lambda -1 & -4\ 1& \lambda -3 \end{bmatrix}$

Take the determinant by taking the product of the upper-left and lower-right entries, and subtracting the product of the other two:

$\begin{vmatrix} \lambda -1 & -4\ 1& \lambda -3 \end{vmatrix}$

$= \left ( \lambda -1\right )\left ( \lambda -3\right ) - (-4)(1)$

$= \lambda ^{2} -4 \lambda+3 +4$

$= \lambda ^{2} -4 \lambda+7$

The characteristic equation is

$\lambda ^{2} -4 \lambda+7 = 0$

By the Cayley-Hamilton Theorem, the matrix is a solution to this equation, so the correct choice is the above equation, modified for matrix arithmetic:

$A^{2} -4 A+7I =O$

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Question

$A = \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$ .

Which of the following is a consequence of the Cayley-Hamilton Theorem?

(Note: and refer to the two-by-two identity and zero matrices, respectively.)

Tap to reveal answer

Answer

The Cayley-Hamilton Theorem states that a square matrix is a solution of its own characteristic equation. To find this, first find the polynomial formed from the determinant of $\lambda I-A$ . This matrix is

$\lambda I-A$

$= \lambda \begin{bmatrix} 1& 0\ 0& 1 \end{bmatrix}- \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$

Multiply $\lambda$ by each element in the identity:

$= \begin{bmatrix} \lambda& 0\ 0& \lambda \end{bmatrix}- \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$

Subtract elementwise:

$= \begin{bmatrix} \lambda -1 & -4\ 1& \lambda -3 \end{bmatrix}$

Take the determinant by taking the product of the upper-left and lower-right entries, and subtracting the product of the other two:

$\begin{vmatrix} \lambda -1 & -4\ 1& \lambda -3 \end{vmatrix}$

$= \left ( \lambda -1\right )\left ( \lambda -3\right ) - (-4)(1)$

$= \lambda ^{2} -4 \lambda+3 +4$

$= \lambda ^{2} -4 \lambda+7$

The characteristic equation is

$\lambda ^{2} -4 \lambda+7 = 0$

By the Cayley-Hamilton Theorem, the matrix is a solution to this equation, so the correct choice is the above equation, modified for matrix arithmetic:

$A^{2} -4 A+7I =O$

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Question

$A = \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$ .

Which of the following is a consequence of the Cayley-Hamilton Theorem?

(Note: and refer to the two-by-two identity and zero matrices, respectively.)

Tap to reveal answer

Answer

The Cayley-Hamilton Theorem states that a square matrix is a solution of its own characteristic equation. To find this, first find the polynomial formed from the determinant of $\lambda I-A$ . This matrix is

$\lambda I-A$

$= \lambda \begin{bmatrix} 1& 0\ 0& 1 \end{bmatrix}- \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$

Multiply $\lambda$ by each element in the identity:

$= \begin{bmatrix} \lambda& 0\ 0& \lambda \end{bmatrix}- \begin{bmatrix} 1&4 \ -1&3 \end{bmatrix}$

Subtract elementwise:

$= \begin{bmatrix} \lambda -1 & -4\ 1& \lambda -3 \end{bmatrix}$

Take the determinant by taking the product of the upper-left and lower-right entries, and subtracting the product of the other two:

$\begin{vmatrix} \lambda -1 & -4\ 1& \lambda -3 \end{vmatrix}$

$= \left ( \lambda -1\right )\left ( \lambda -3\right ) - (-4)(1)$

$= \lambda ^{2} -4 \lambda+3 +4$

$= \lambda ^{2} -4 \lambda+7$

The characteristic equation is

$\lambda ^{2} -4 \lambda+7 = 0$

By the Cayley-Hamilton Theorem, the matrix is a solution to this equation, so the correct choice is the above equation, modified for matrix arithmetic:

$A^{2} -4 A+7I =O$

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Question

A $5 \times 5$ matrix has as its set of eigenvalues $\left \{ -1000, 1000, 0, -1000i, 1000i \right \}$ .

True, false, or indeterminate: the matrix is singular.

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Answer

A matrix is singular - that is, not having an inverse - if and only if one of its eigenvalues is 0. This is seen to be the case.

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Question

The eigenvalues of a four-by-four matrix are:

$\left \{ 6-\sqrt{5},6+ \sqrt{5}, 6+ 2i , 6-2i \right \}$

Which one is the dominant eigenvalue?

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Answer

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. The absolute values of all four eigenvalues are as follows:

$|6-\sqrt{5 }| = 6-\sqrt{5 }\approx 3.76$

$|6+\sqrt{5 }| = 6+\sqrt{5 }\approx 8.24$

$|6-i\sqrt{5 }| =\sqrt{ 6^{2}+(-\sqrt{5 }) ^{2}}= \sqrt{36+5} = \sqrt{41} \approx 6.40$

$|6-i\sqrt{5 }| =\sqrt{ 6^{2}+(\sqrt{5 }) ^{2}}= \sqrt{36+5} = \sqrt{41} \approx 6.40$

$|6+\sqrt{5 }| > |6-i\sqrt{5 }| = |6+i\sqrt{5 }| >|6-\sqrt{5 }|$ ;

therefore, the dominant eigenvalue is $6+ \sqrt{5}$ .

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Question

A $5 \times 5$ matrix has as its set of eigenvalues $\left \{ -1000, 1000, 0, -1000i, 1000i \right \}$ .

True, false, or indeterminate: the matrix is singular.

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Answer

A matrix is singular - that is, not having an inverse - if and only if one of its eigenvalues is 0. This is seen to be the case.

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Question

The eigenvalues of a four-by-four matrix are:

$\left \{ 6-\sqrt{5},6+ \sqrt{5}, 6+ 2i , 6-2i \right \}$

Which one is the dominant eigenvalue?

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Answer

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. The absolute values of all four eigenvalues are as follows:

$|6-\sqrt{5 }| = 6-\sqrt{5 }\approx 3.76$

$|6+\sqrt{5 }| = 6+\sqrt{5 }\approx 8.24$

$|6-i\sqrt{5 }| =\sqrt{ 6^{2}+(-\sqrt{5 }) ^{2}}= \sqrt{36+5} = \sqrt{41} \approx 6.40$

$|6-i\sqrt{5 }| =\sqrt{ 6^{2}+(\sqrt{5 }) ^{2}}= \sqrt{36+5} = \sqrt{41} \approx 6.40$

$|6+\sqrt{5 }| > |6-i\sqrt{5 }| = |6+i\sqrt{5 }| >|6-\sqrt{5 }|$ ;

therefore, the dominant eigenvalue is $6+ \sqrt{5}$ .

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Question

A $5 \times 5$ matrix has as its set of eigenvalues $\left \{ -1000, 1000, 0, -1000i, 1000i \right \}$ .

True, false, or indeterminate: the matrix is singular.

Tap to reveal answer

Answer

A matrix is singular - that is, not having an inverse - if and only if one of its eigenvalues is 0. This is seen to be the case.

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Question

The eigenvalues of a four-by-four matrix are:

$\left \{ 6-\sqrt{5},6+ \sqrt{5}, 6+ 2i , 6-2i \right \}$

Which one is the dominant eigenvalue?

Tap to reveal answer

Answer

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. The absolute values of all four eigenvalues are as follows:

$|6-\sqrt{5 }| = 6-\sqrt{5 }\approx 3.76$

$|6+\sqrt{5 }| = 6+\sqrt{5 }\approx 8.24$

$|6-i\sqrt{5 }| =\sqrt{ 6^{2}+(-\sqrt{5 }) ^{2}}= \sqrt{36+5} = \sqrt{41} \approx 6.40$

$|6-i\sqrt{5 }| =\sqrt{ 6^{2}+(\sqrt{5 }) ^{2}}= \sqrt{36+5} = \sqrt{41} \approx 6.40$

$|6+\sqrt{5 }| > |6-i\sqrt{5 }| = |6+i\sqrt{5 }| >|6-\sqrt{5 }|$ ;

therefore, the dominant eigenvalue is $6+ \sqrt{5}$ .

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Question

A $5 \times 5$ matrix has as its set of eigenvalues $\left \{ -1000, 1000, 0, -1000i, 1000i \right \}$ .

True, false, or indeterminate: the matrix is singular.

Tap to reveal answer

Answer

A matrix is singular - that is, not having an inverse - if and only if one of its eigenvalues is 0. This is seen to be the case.

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Question

The eigenvalues of a four-by-four matrix are:

$\left \{ 6-\sqrt{5},6+ \sqrt{5}, 6+ 2i , 6-2i \right \}$

Which one is the dominant eigenvalue?

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Answer

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. The absolute values of all four eigenvalues are as follows:

$|6-\sqrt{5 }| = 6-\sqrt{5 }\approx 3.76$

$|6+\sqrt{5 }| = 6+\sqrt{5 }\approx 8.24$

$|6-i\sqrt{5 }| =\sqrt{ 6^{2}+(-\sqrt{5 }) ^{2}}= \sqrt{36+5} = \sqrt{41} \approx 6.40$

$|6-i\sqrt{5 }| =\sqrt{ 6^{2}+(\sqrt{5 }) ^{2}}= \sqrt{36+5} = \sqrt{41} \approx 6.40$

$|6+\sqrt{5 }| > |6-i\sqrt{5 }| = |6+i\sqrt{5 }| >|6-\sqrt{5 }|$ ;

therefore, the dominant eigenvalue is $6+ \sqrt{5}$ .

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Question

The characteristic equation of a three-by-three matrix is

$\lambda ^{3}+ 10\lambda - 57= 0$

Which of the following is its dominant eigenvalue?

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Answer

The eigenvalues of a matrix are the zeroes of its characteristic equation.

We can try to extract one or more zeroes using the Rational Zeroes Theorem, which states that any rational zero must be the positive or negative quotient of a factor of the constant and a factor of the leading coefficient. Since the constant is 57, and the leading coefficient of the polynomial is 1, any rational zeroes must be one of the set $\left \{ 1,3,19 , 57 \right \}$ divided by 1, with positive or native numbers taken into account. Thus, any rational zeroes must be in the set

$\left \{\pm 1, \pm 3, \pm19 ,\pm 57 \right \}$

By trial and error, it can be determined that 3 is a solution of the equation:

$\lambda ^{3}+ 10\lambda - 57= 0$

$3^{3}+ 10 (3) - 57= 0$

True.

It follows that $\lambda - 3$ is a factor. Divide $\lambda ^{3}+ 10\lambda - 57$ by $\lambda - 3$ ; the quotient can be found to be $\lambda ^{2} +3 \lambda+19$ , which is prime.

The characteristic equation is factorable as

$(\lambda ^{2} +3 \lambda+19)(\lambda - 3) = 0$

We already know that $\lambda_{1}= 3$ is an eigenvalue; the other two eigenvalues are the zeroes of $\lambda ^{2} +3 \lambda+19$ , which can be found by way of the quadratic formula:

$\lambda _{2}= \frac{-3+ \sqrt{3^{2}-4(1)(19)}}{2} = \frac{-3 + \sqrt{-67}}{2} =- \frac{3} {2}+ \frac{ \sqrt{67}}{2} i$

$\lambda _{3}= \frac{-3- \sqrt{3^{2}-4(1)(19)}}{2} = \frac{-3 - \sqrt{-67}}{2} =- \frac{3} {2}-\frac{ \sqrt{67}}{2} i$

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. Calculate the absolute values of all three eigenvalues:

$\left |\lambda_{1} \right |= \left |3 \right | = 3$

$\left |\lambda _{2} \right| =\left | - \frac{3} {2}+ \frac{ \sqrt{67}}{2} i \right |$

$= \sqrt{\left ( - \frac{3} {2} \right )^{2}+\left ( \frac{ \sqrt{67}}{2} \right ) ^{2}}$

$= \sqrt{\frac{9}{4}+\frac{67}{4}}$

$= \sqrt{\frac{76}{4}}$

$= \sqrt{19}$

$\approx 4.4$

$\left |\lambda _{3} \right| =\left | - \frac{3} {2}- \frac{ \sqrt{67}}{2} i \right |$

$= \sqrt{\left ( - \frac{3} {2} \right )^{2}+\left (- \frac{ \sqrt{67}}{2} \right ) ^{2}}$

$= \sqrt{\frac{9}{4}+\frac{67}{4}}$

$= \sqrt{\frac{76}{4}}$

$= \sqrt{19}$

$\approx 4.4$

Therefore,

$\left |\lambda _{2} \right| = \left |\lambda _{3} \right| > \left |\lambda _{1} \right|$ .

Since no single eigenvalue has absolute value strictly greater than that of the other two, the matrix has no dominant eigenvalue.

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Question

$C(-\infty , \infty)$ , the set of all continuous functions defined on $(-\infty, \infty)$ , is a vector space under the usual rules of addition and scalar multiplication.

True or false: The set of all functions of the form

where is a real number, is a subspace of $C(-\infty , \infty)$ .

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Answer

Let $S = \left \{ f(x)| f(x) = |nx | , n \in \mathbb{R}\right \}$ .

We can show through counterexample that this is not a subspace of $C(-\infty , \infty)$ .

Let . This is an element of .

One condition for to be a subspace of a vector space is closure under scalar multiplication. Multiply by scalar . The product is the function

.

$g(x) \ne S$ .

This violates a criterion for a subspace, so is not a subspace of $C(-\infty , \infty)$ .

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Question

A matrix has $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ as its set of eigenvalues.

Give the set of eigenvalues of $A^{-1}$ .

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Answer

If is nonsingular, the set of eigenvalues of $A^{-1}$ is exactly the set of reciprocals of eigenvalues of . The eigenvalues of are $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ , so the eigenvalues of these numbers are the reciprocals of these. The reciprocal of is

$\frac{1}{1 + i} = \frac{1(1-i)}{(1 + i)(1-i)} = \frac{1 - i}{2} = \frac{1}{2}- \frac{1}{2}i$ .

Similarly,

$\frac{1}{1 - i} = \frac{1}{2}+ \frac{1}{2}i$

$\frac{1}{-1 + i} =- \frac{1}{2}- \frac{1}{2}i$

$\frac{1}{-1 - i} =- \frac{1}{2}+ \frac{1}{2}i$

The set of eigenvalues of $A^{-1}$ is $\left \{ \frac{1}{2}+ \frac{1}{2}i, \frac{1}{2}- \frac{1}{2}i , -\frac{1}{2}+ \frac{1}{2}i, -\frac{1}{2}- \frac{1}{2}i , \right \}$ .

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Question

A matrix has $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ as its set of eigenvalues.

Give the set of eigenvalues of $A^{-1}$ .

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Answer

If is nonsingular, the set of eigenvalues of $A^{-1}$ is exactly the set of reciprocals of eigenvalues of . The eigenvalues of are $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ , so the eigenvalues of these numbers are the reciprocals of these. The reciprocal of is

$\frac{1}{1 + i} = \frac{1(1-i)}{(1 + i)(1-i)} = \frac{1 - i}{2} = \frac{1}{2}- \frac{1}{2}i$ .

Similarly,

$\frac{1}{1 - i} = \frac{1}{2}+ \frac{1}{2}i$

$\frac{1}{-1 + i} =- \frac{1}{2}- \frac{1}{2}i$

$\frac{1}{-1 - i} =- \frac{1}{2}+ \frac{1}{2}i$

The set of eigenvalues of $A^{-1}$ is $\left \{ \frac{1}{2}+ \frac{1}{2}i, \frac{1}{2}- \frac{1}{2}i , -\frac{1}{2}+ \frac{1}{2}i, -\frac{1}{2}- \frac{1}{2}i , \right \}$ .

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Question

$P_{N}$ is the set of all polynomials with degree or less.

Define a linear mapping $T: P_{3} \rightarrow P_{3}$ as follows:

$T(p) = \frac{1}{x}\int_{0}^{x}p(y) \textup{ d} y$

Is this mapping one-to-one and onto?

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Answer

$T: P_{3} \rightarrow P_{3}$ is a linear mapping of a vector space into itself, so it is possible for to be both one-to-one and onto.

A transformation is one-to-one if and only if for any in the domain,

implies that .

Suppose that for some $p, q \in P_{3}$ .

Since are third-degree polynomials:

$p(x) = a_{3}x^{3} + a_{2}x^{2}+ a_{1}x+ a_{0}$

$q(x) = b_{3}x^{3} + b_{2}x^{2}+ b_{1}x+ b_{0}$

for some scalar $\left \{ a_{0}, ...a_{3}, b_{0}, ...b_{3} \right \}$ .

, so

$\frac{1}{x}\int_{0}^{x}p(y) \textup{ d} y = \frac{1}{x}\int_{0}^{x}q(y) \textup{ d} y$

$\int_{0}^{x}p(y) \textup{ d} y = \int_{0}^{x}q(y) \textup{ d} y$

$\int_{0}^{x} (a_{3}y^{3} + a_{2}y^{2}+ a_{1}y+ a_{0}) \textup{ d} y = \int_{0}^{x} (b_{3}y^{3} + b_{2}y^{2}+ b_{1}y+ b_{0}) \textup{ d} y$

$\frac{a_{3}}{4}x^{4} + \frac{a_{2}}{3}x^{3}+ \frac{a_{1}}{2}x^{2}+ a_{0} x = \frac{b_{3}}{4}x^{4} + \frac{b_{2}}{3}x^{3}+ \frac{b_{1}}{2}x^{2}+ b_{0}x$

For these two polynomials to be equal, it must hold that all coefficients of the terms of equal degree are equal. We can immediately see from the first-degree terms that $a_{0} = b_{0}$ . Applying some simple algebra, it follows almost as quickly that $a_{1} = b_{1}$ , $a_{2} = b_{2}$ , and $a_{3} = b_{3}$ . Thus, , and is one-to-one.

A transformation is onto if, for each in the range, there exists in the domain such that .

Let $r(x) \in P_{3}$ . Then

$r(x) = c_{3}x^{3} + c_{2}x^{2}+ c_{1}x+ c_{0}$

for some scalar $\left \{ c_{0}, c_{1}, c_{2},c_{3} \right \}$ .

If , then, if is defined as before,

$\frac{1}{x}\int_{0}^{x}p(y) \textup{ d} y = r(x)$

$\frac{1}{x}\int_{0}^{x} (a_{3}y^{3} + a_{2}y^{2}+ a_{1}y+ a_{0}) \textup{ d} y = c_{3}x^{3} + c_{2}x^{2}+ c_{1}x+ c_{0}$

$\frac{1}{x} \left ( \frac{a_{3}}{4}x^{4} + \frac{a_{2}}{3}x^{3}+ \frac{a_{1}}{2}x^{2}+ a_{0} x \right ) = c_{3}x^{3} + c_{2}x^{2}+ c_{1}x+ c_{0}$

$\ \frac{a_{3}}{4}x^{3} + \frac{a_{2}}{3}x^{2}+ \frac{a_{1}}{2}x + a_{0} = c_{3}x^{3} + c_{2}x^{2}+ c_{1}x+ c_{0}$

Therefore,

$\ \frac{a_{3}}{4}= c_{3}$ , $\frac{a_{2}}{3}= c_{2}$ , $\frac{a_{1}}{2}= c_{1}$ , $a_{0} = c_{0}$

Or,

$a_{3} = 4c_{3}$ , $a_{2} =3 c_{2}$ , $a_{1} =2 c_{1}$ , $a_{0} = c_{0}$ .

Thus,

$p(x) = 4c_{3}x^{3}+ 3c_{2}x^{2}+ 2c_{1}x+ c_{0}$ is the polynomial in $P_{3}$ that maps into . Since such a polynomial exists in the domain for each range element, it follows that is onto.

The correct response is that is one-to-one and onto.

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Question

Let $m, n, N \in \mathbb{N}$ .

Define $M_{m \times n }$ as the set of all $m \times n$ matrices, and $P_{N}$ as the set of all polynomials in of dimension or less.

True or false: $M_{2 \times 2 }$ and $P_{4}$ are isomorphic linear spaces.

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Answer

Two linear spaces are isomorphic if and only if they are of the same dimension, or, equivalently, the size of a basis of each includes the same number of elements.

$M_{2 \times 2}$ is the set of $2 \times 2$ matrices; each such matrix includes 4 elements, so it has four elements in one of its bases; for example, one such basis is

$\left \{ \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix} \right \}$

$P_{4}$ is the set of all polynomials in of degree 4 or less - that is, all polynomials of the form

$p(x) = a_{0}+ a_{1}x+ a_{2}x^{2} + a_{3}x^{3}+ a_{4}x^{4}$

Each polynomial is defined by five coefficients, so the dimension of $P_{4}$ is 5. Each basis of $P_{4}$ has five elements; for example, one such basis is

$\left \{ 1, x, x^{2},x^{3}, x^{4}\right \}$ .

Since $\dim (M_{2 \times 2}) \ne \dim (P_{4})$ , the spaces are not isomorphic.

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Question

The characteristic equation of a three-by-three matrix is

$\lambda ^{3}+ 10\lambda - 57= 0$

Which of the following is its dominant eigenvalue?

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Answer

The eigenvalues of a matrix are the zeroes of its characteristic equation.

We can try to extract one or more zeroes using the Rational Zeroes Theorem, which states that any rational zero must be the positive or negative quotient of a factor of the constant and a factor of the leading coefficient. Since the constant is 57, and the leading coefficient of the polynomial is 1, any rational zeroes must be one of the set $\left \{ 1,3,19 , 57 \right \}$ divided by 1, with positive or native numbers taken into account. Thus, any rational zeroes must be in the set

$\left \{\pm 1, \pm 3, \pm19 ,\pm 57 \right \}$

By trial and error, it can be determined that 3 is a solution of the equation:

$\lambda ^{3}+ 10\lambda - 57= 0$

$3^{3}+ 10 (3) - 57= 0$

True.

It follows that $\lambda - 3$ is a factor. Divide $\lambda ^{3}+ 10\lambda - 57$ by $\lambda - 3$ ; the quotient can be found to be $\lambda ^{2} +3 \lambda+19$ , which is prime.

The characteristic equation is factorable as

$(\lambda ^{2} +3 \lambda+19)(\lambda - 3) = 0$

We already know that $\lambda_{1}= 3$ is an eigenvalue; the other two eigenvalues are the zeroes of $\lambda ^{2} +3 \lambda+19$ , which can be found by way of the quadratic formula:

$\lambda _{2}= \frac{-3+ \sqrt{3^{2}-4(1)(19)}}{2} = \frac{-3 + \sqrt{-67}}{2} =- \frac{3} {2}+ \frac{ \sqrt{67}}{2} i$

$\lambda _{3}= \frac{-3- \sqrt{3^{2}-4(1)(19)}}{2} = \frac{-3 - \sqrt{-67}}{2} =- \frac{3} {2}-\frac{ \sqrt{67}}{2} i$

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. Calculate the absolute values of all three eigenvalues:

$\left |\lambda_{1} \right |= \left |3 \right | = 3$

$\left |\lambda _{2} \right| =\left | - \frac{3} {2}+ \frac{ \sqrt{67}}{2} i \right |$

$= \sqrt{\left ( - \frac{3} {2} \right )^{2}+\left ( \frac{ \sqrt{67}}{2} \right ) ^{2}}$

$= \sqrt{\frac{9}{4}+\frac{67}{4}}$

$= \sqrt{\frac{76}{4}}$

$= \sqrt{19}$

$\approx 4.4$

$\left |\lambda _{3} \right| =\left | - \frac{3} {2}- \frac{ \sqrt{67}}{2} i \right |$

$= \sqrt{\left ( - \frac{3} {2} \right )^{2}+\left (- \frac{ \sqrt{67}}{2} \right ) ^{2}}$

$= \sqrt{\frac{9}{4}+\frac{67}{4}}$

$= \sqrt{\frac{76}{4}}$

$= \sqrt{19}$

$\approx 4.4$

Therefore,

$\left |\lambda _{2} \right| = \left |\lambda _{3} \right| > \left |\lambda _{1} \right|$ .

Since no single eigenvalue has absolute value strictly greater than that of the other two, the matrix has no dominant eigenvalue.

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Question

Let $m, n, N \in \mathbb{N}$ .

Define $M_{m \times n }$ as the set of all $m \times n$ matrices, and $P_{N}$ as the set of all polynomials in of dimension or less.

True or false: $M_{2 \times 2 }$ and $P_{4}$ are isomorphic linear spaces.

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Answer

Two linear spaces are isomorphic if and only if they are of the same dimension, or, equivalently, the size of a basis of each includes the same number of elements.

$M_{2 \times 2}$ is the set of $2 \times 2$ matrices; each such matrix includes 4 elements, so it has four elements in one of its bases; for example, one such basis is

$\left \{ \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix} \right \}$

$P_{4}$ is the set of all polynomials in of degree 4 or less - that is, all polynomials of the form

$p(x) = a_{0}+ a_{1}x+ a_{2}x^{2} + a_{3}x^{3}+ a_{4}x^{4}$

Each polynomial is defined by five coefficients, so the dimension of $P_{4}$ is 5. Each basis of $P_{4}$ has five elements; for example, one such basis is

$\left \{ 1, x, x^{2},x^{3}, x^{4}\right \}$ .

Since $\dim (M_{2 \times 2}) \ne \dim (P_{4})$ , the spaces are not isomorphic.

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