Characteristic Polynomial - Linear Algebra

1. Let $P(x) = x^3 - 7x^2 + 14x - 8$ be the characteristic polynomial of a $3 \times 3$ matrix $A$. If matrix $B$ is obtained by applying the same similarity transformation to $A$ that would make the first column of the resulting matrix equal to $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, which of the following statements about the characteristic polynomial of $B$ is true?
   Answer: The characteristic polynomial of $B$ is the same as $P(x)$ Similarity transformations preserve the characteristic polynomial. If $B = Q^{-1}AQ$ for some invertible matrix $Q$, then $\det(B - \lambda I) = \det(Q^{-1}AQ - \lambda I) = \det(Q^{-1}(A - \lambda I)Q) = \det(Q^{-1})\det(A - \lambda I)\det(Q) = \det(A - \lambda I)$ since $\det(Q^{-1})\det(Q) = 1$. Therefore, $B$ has the same characteristic polynomial as $A$, which is $P(x) = x^3 - 7x^2 + 14x - 8$. Choice A incorrectly assumes we can factor out $(x-1)$ based on the first column information. Choice C gives a different polynomial entirely. Choice D is incorrect because similarity invariance allows us to determine this.
2. What is the characteristic polynomial of the matrix $A = \begin{pmatrix} 2 & 0 & 1 \\ 1 & 3 & 1 \\ 1 & 0 & 2 \end{pmatrix}$?
   Answer: $p(\lambda) = -\lambda^3 + 7\lambda^2 - 15\lambda + 9$ The characteristic polynomial is given by $p(\lambda) = \det(A - \lambda I)$. $A - \lambda I = \begin{pmatrix} 2-\lambda & 0 & 1 \\ 1 & 3-\lambda & 1 \\ 1 & 0 & 2-\lambda \end{pmatrix}$ We compute the determinant by cofactor expansion along the second column: $p(\lambda) = (3-\lambda) \det \begin{pmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{pmatrix}$ $p(\lambda) = (3-\lambda) [ (2-\lambda)^2 - 1 ]$ $p(\lambda) = (3-\lambda) [ (4 - 4\lambda + \lambda^2) - 1 ]$ $p(\lambda) = (3-\lambda) (\lambda^2 - 4\lambda + 3)$ $p(\lambda) = 3\lambda^2 - 12\lambda + 9 - \lambda^3 + 4\lambda^2 - 3\lambda$ $p(\lambda) = -\lambda^3 + 7\lambda^2 - 15\lambda + 9$ Distractor B represents the polynomial $\det(\lambda I - A)$, which is a common alternative definition but results in opposite signs. Distractor C results from an error in calculating the determinant, specifically in the expansion of $(2-\lambda)^2-1$. Distractor D incorrectly uses the diagonal entries as coefficients.
3. A square matrix $A$ is singular. Which statement must be true about its characteristic polynomial $p(\lambda) = \det(A - \lambda I)$?
   Answer: The constant term of $p(\lambda)$ is zero. A matrix $A$ is singular if and only if its determinant is zero, i.e., $\det(A) = 0$. The characteristic polynomial is $p(\lambda) = \det(A - \lambda I)$. The constant term of this polynomial is found by setting $\lambda = 0$: $p(0) = \det(A - 0 \cdot I) = \det(A)$. Since $A$ is singular, $\det(A) = 0$. Therefore, the constant term of its characteristic polynomial, $p(0)$, must be zero. This is equivalent to saying $\lambda=0$ is an eigenvalue. Distractor B is impossible; the leading coefficient of the characteristic polynomial of an $n \times n$ matrix is $(-1)^n$. Distractor C implies $p(1)=0$, which means $\lambda=1$ is an eigenvalue, which is not necessarily true for all singular matrices. Distractor D is false; $\lambda=0$ is a real root.
4. Let $A$ be a $2 \times 2$ matrix with $\mathrm{tr}(A) = 7$ and $\det(A) = 10$. What is the characteristic polynomial of $A$?
   Answer: $\lambda^2 - 7\lambda + 10$ For any $2 \times 2$ matrix $M$, the characteristic polynomial is given by the formula $p(\lambda) = \lambda^2 - \mathrm{tr}(M)\lambda + \det(M)$. Given $\mathrm{tr}(A) = 7$ and $\det(A) = 10$, we can substitute these values directly into the formula: $p(\lambda) = \lambda^2 - (7)\lambda + (10) = \lambda^2 - 7\lambda + 10$. Distractor A has an incorrect sign for the trace term. Distractor B incorrectly swaps the roles of the trace and the determinant. Distractor D swaps the roles and also has an incorrect sign for the determinant term.
5. Consider the matrix $A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ k & -1 & 2 \end{pmatrix}$. For what value of $k$ is $\lambda=1$ an eigenvalue of $A$?
   Answer: $k = 0$ If $\lambda=1$ is an eigenvalue, then by definition $\det(A - \lambda I) = 0$. In this case, we need to solve $\det(A - 1 \cdot I) = 0$. $A - I = \begin{pmatrix} 0-1 & 1 & 0 \\ 0 & 0-1 & 1 \\ k & -1 & 2-1 \end{pmatrix} = \begin{pmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ k & -1 & 1 \end{pmatrix}$ Now, we compute the determinant: $\det(A - I) = -1 \begin{vmatrix} -1 & 1 \\ -1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 0 & 1 \\ k & 1 \end{vmatrix} + 0 \begin{vmatrix} 0 & -1 \\ k & -1 \end{vmatrix}$ $= -1((-1)(1) - (1)(-1)) - 1((0)(1) - (1)(k)) + 0$ $= -1(-1 + 1) - 1(-k)$ $= -1(0) + k = k$ For $\lambda=1$ to be an eigenvalue, we must have $\det(A-I)=0$, which means $k=0$. Distractor A, B, and D result from various calculation errors or misconceptions, such as incorrectly calculating the determinant or confusing the eigenvalue with the determinant.
6. A $3 \times 3$ matrix $B$ has eigenvalues of $2, 2,$ and $-3$. What is the trace of matrix $B$?
   Answer: $1$ The trace of a matrix is the sum of its diagonal entries. A fundamental property of eigenvalues is that the sum of the eigenvalues of a matrix is equal to its trace. Given the eigenvalues $\lambda_1 = 2$, $\lambda_2 = 2$, and $\lambda_3 = -3$, the trace of $B$ is: $\mathrm{tr}(B) = \lambda_1 + \lambda_2 + \lambda_3 = 2 + 2 + (-3) = 1$. Distractor A is the product of the eigenvalues, which equals the determinant of the matrix. Distractor C is the repeated eigenvalue. Distractor D is the sum of the distinct eigenvalues, incorrectly ignoring the multiplicity of the eigenvalue 2.
7. A $4 \times 4$ matrix $M$ has characteristic polynomial $q(\lambda) = \lambda^4 - 10\lambda^3 + 35\lambda^2 - 50\lambda + 24$. After factoring, it's found that $q(\lambda) = (\lambda-1)(\lambda-2)(\lambda-3)(\lambda-4)$. If we construct matrix $N = M^2 - 5M + 6I$, and we know that $N$ is singular (non-invertible), which of the following statements must be true?
   Answer: $M$ must have eigenvalue $2$ or eigenvalue $3$ For $N = M^2 - 5M + 6I$ to be singular, it must have eigenvalue $0$. If $\lambda$ is an eigenvalue of $M$, then the corresponding eigenvalue of $N$ is $\lambda^2 - 5\lambda + 6$. For $N$ to be singular, we need $\lambda^2 - 5\lambda + 6 = 0$ for at least one eigenvalue $\lambda$ of $M$. Factoring: $\lambda^2 - 5\lambda + 6 = (\lambda - 2)(\lambda - 3) = 0$. So $\lambda = 2$ or $\lambda = 3$. Since the characteristic polynomial tells us that $M$ has eigenvalues $1, 2, 3, 4$, and we need $M$ to have eigenvalue $2$ or $3$ for $N$ to be singular, this condition is satisfied. Both eigenvalues $2$ and $3$ are indeed eigenvalues of $M$. Therefore, $N$ will have at least one zero eigenvalue (in fact, it will have two zero eigenvalues corresponding to the eigenvalues $2$ and $3$ of $M$), making $N$ singular. Choice A correctly identifies this requirement. Choice B gives the wrong eigenvalues. Choice C contradicts the given information. Choice D is unnecessary since the given eigenvalues are consistent with $N$ being singular.
8. A $3 \times 3$ matrix $B$ has the characteristic polynomial $p(\lambda) = (\lambda-3)^2(\lambda+1)$. Which of the following statements about $B$ must be true?
   Answer: The trace of $B$ is 5. The roots of the characteristic polynomial are the eigenvalues of the matrix. The roots of $p(\lambda) = (\lambda-3)^2(\lambda+1)=0$ are $\lambda=3$ (with algebraic multiplicity 2) and $\lambda=-1$. The eigenvalues are $\{3, 3, -1\}$. The trace of a matrix is the sum of its eigenvalues: $\mathrm{tr}(B) = 3 + 3 + (-1) = 5$. This statement is true. Distractor A is false; the determinant is the product of the eigenvalues: $\det(B) = 3 \times 3 \times (-1) = -9$. Distractor C is false; a matrix is not invertible (singular) if and only if 0 is an eigenvalue. Since 0 is not an eigenvalue, $B$ is invertible. Distractor D is not guaranteed to be true; the eigenvalue 3 has algebraic multiplicity 2, but its geometric multiplicity (the number of linearly independent eigenvectors) could be 1 or 2. If it is 1, the matrix is not diagonalizable and only has two linearly independent eigenvectors.
9. The characteristic polynomial of a $3 \times 3$ matrix $A$ is given by $p(\lambda) = -\lambda^3 - 2\lambda^2 + 5\lambda + 6$. What is the determinant of matrix $A$?
   Answer: $6$ The characteristic polynomial is defined as $p(\lambda) = \det(A - \lambda I)$. To find the determinant of $A$, we can evaluate the characteristic polynomial at $\lambda = 0$. $p(0) = \det(A - 0 \cdot I) = \det(A)$. Substituting $\lambda = 0$ into the given polynomial: $p(0) = -(0)^3 - 2(0)^2 + 5(0) + 6 = 6$. Therefore, $\det(A) = 6$. Distractor A is the coefficient of $\lambda^2$, which relates to the trace of the matrix, not the determinant. Distractor B is the leading coefficient. Distractor D is the sum of the coefficients, which is $p(1)$, not $p(0)$.
10. The characteristic polynomial of a matrix $A$ is $p(\lambda) = -\lambda^3 + 5\lambda^2 - 8\lambda + 4$. What are the eigenvalues of the matrix $A+2I$?
    Answer: $\{3, 4, 4\}$ First, we find the eigenvalues of $A$ by finding the roots of its characteristic polynomial $p(\lambda) = -\lambda^3 + 5\lambda^2 - 8\lambda + 4 = 0$. By the Rational Root Theorem, we test integer factors of 4. We find that $p(1) = -1+5-8+4=0$ and $p(2) = -8+20-16+4=0$. Since the sum of the roots must be $-5/(-1)=5$, and we have found roots 1 and 2, the third root $\lambda_3$ must satisfy $1+2+\lambda_3=5$, so $\lambda_3=2$. The eigenvalues of $A$ are $\{1, 2, 2\}$. If $\lambda$ is an eigenvalue of $A$, then $\lambda+k$ is an eigenvalue of $A+kI$. For this problem, $k=2$. So, the eigenvalues of $A+2I$ are obtained by adding 2 to each eigenvalue of $A$: $\{1+2, 2+2, 2+2\} = \{3, 4, 4\}$. Distractor A lists the original eigenvalues of $A$, forgetting to perform the shift. Distractor C represents the eigenvalues of $A-2I$, a shift in the wrong direction. Distractor D represents the eigenvalues of $2A$, a scaling instead of a shift.
11. For what value of $k$ does the matrix $A = \begin{pmatrix} 3 & 7 \\ k & 5 \end{pmatrix}$ have an eigenvalue of $\lambda = 10$?
    Answer: $k = 5$ If $\lambda = 10$ is an eigenvalue of $A$, then the matrix $A - \lambda I$ must be singular, meaning its determinant is zero. Here, $A - 10I$. $A - 10I = \begin{pmatrix} 3-10 & 7 \\ k & 5-10 \end{pmatrix} = \begin{pmatrix} -7 & 7 \\ k & -5 \end{pmatrix}$ Now, we set the determinant of this matrix to zero: $\det(A - 10I) = (-7)(-5) - (7)(k) = 0$ $35 - 7k = 0$ $35 = 7k$ $k = 5$ Distractor A results from a sign error in the calculation ($ -35 - 7k = 0$). Distractor B occurs if one incorrectly sets the determinant of the original matrix $A$ equal to the eigenvalue ($15 - 7k = 10$). Distractor D results from a simple calculation error.
12. Two $3 \times 3$ matrices $A$ and $B$ are similar. If the characteristic polynomial of $A$ is $p_A(\lambda) = -\lambda^3 + 4\lambda^2 - 5\lambda + 2$, what is the characteristic polynomial of $B$, denoted $p_B(\lambda)$?
    Answer: $p_B(\lambda) = -\lambda^3 + 4\lambda^2 - 5\lambda + 2$ A fundamental property of similar matrices is that they share the same characteristic polynomial. If $A$ and $B$ are similar, there exists an invertible matrix $P$ such that $B = P^{-1}AP$. The characteristic polynomial of $B$ is: $p_B(\lambda) = \det(B - \lambda I) = \det(P^{-1}AP - \lambda P^{-1}P) = \det(P^{-1}(A - \lambda I)P)$. Using the property $\det(XYZ) = \det(X)\det(Y)\det(Z)$, we get: $p_B(\lambda) = \det(P^{-1})\det(A - \lambda I)\det(P) = \det(P)^{-1} p_A(\lambda) \det(P) = p_A(\lambda)$. Therefore, the characteristic polynomials are identical. Distractor B is the negative of the correct polynomial, which is incorrect. Distractor C is subtly wrong; because the characteristic polynomials are identical, they cannot be different. Distractor D is an incomplete statement; while the constant terms (the determinants) are the same, all other corresponding coefficients (like those related to the trace) are also the same.
13. Given the matrix $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$, what is the characteristic polynomial of the matrix $A^2$?
    Answer: $\lambda^2 - 29\lambda + 100$ First, find the eigenvalues of $A$. The characteristic polynomial of $A$ is $p_A(\lambda) = \lambda^2 - \mathrm{tr}(A)\lambda + \det(A)$. $\mathrm{tr}(A) = 4+3=7$. $\det(A) = (4)(3)-(1)(2)=10$. So, $p_A(\lambda) = \lambda^2 - 7\lambda + 10 = (\lambda-2)(\lambda-5) = 0$. The eigenvalues of $A$ are $\lambda_1=2$ and $\lambda_2=5$. The eigenvalues of $A^2$ are the squares of the eigenvalues of $A$, which are $\mu_1 = 2^2=4$ and $\mu_2 = 5^2=25$. The characteristic polynomial of $A^2$ is $p_{A^2}(\lambda) = \lambda^2 - \mathrm{tr}(A^2)\lambda + \det(A^2)$. The new trace is the sum of the new eigenvalues: $\mathrm{tr}(A^2) = 4+25=29$. The new determinant is the product of the new eigenvalues: $\det(A^2) = 4 \times 25 = 100$. Thus, the characteristic polynomial of $A^2$ is $\lambda^2 - 29\lambda + 100$. Distractor A is the characteristic polynomial of $A$, not $A^2$. Distractor B is derived from incorrectly squaring each element of $A$ to get $A^2$. Distractor C is derived from squaring the trace and determinant of $A$ to find the new coefficients.
14. Let $M$ be the matrix $M = \begin{pmatrix} 4 & 1 & -2 \\ 0 & -1 & 3 \\ 0 & 0 & 4 \end{pmatrix}$. Which of the following is the characteristic polynomial of $M$?
    Answer: $p(\lambda) = -(\lambda-4)^2(\lambda+1)$ For a triangular matrix (either upper or lower), the eigenvalues are its diagonal entries. The diagonal entries of $M$ are $4, -1,$ and $4$. The eigenvalues are $\lambda_1 = 4$, $\lambda_2 = -1$, and $\lambda_3 = 4$. The characteristic polynomial, $p(\lambda) = \det(M - \lambda I)$, can be written in factored form using the eigenvalues: $p(\lambda) = (-1)^3 (\lambda - \lambda_1)(\lambda - \lambda_2)(\lambda - \lambda_3)$. $p(\lambda) = -(\lambda - 4)(\lambda - (-1))(\lambda - 4) = -(\lambda-4)^2(\lambda+1)$. Distractor A corresponds to the polynomial $\det(\lambda I - M)$, which is monic but not the standard definition $\det(M - \lambda I)$. Distractor C has an incorrect sign on the $\lambda$ term. Distractor D incorrectly suggests that all non-zero entries of the matrix contribute as roots to the polynomial.
15. A $3 \times 3$ matrix $M$ has characteristic polynomial $p(\lambda) = -\lambda^3 + 6\lambda^2 - 11\lambda + 6$. If one eigenvalue of $M$ is known to be $\lambda_1 = 2$, and the matrix $M + kI$ (where $k$ is a constant) has $\lambda = 0$ as an eigenvalue, what are the possible values of $k$?
    Answer: $k = -1, -2, -3$ Since $\lambda_1 = 2$ is an eigenvalue, we can factor the characteristic polynomial: $p(\lambda) = -(\lambda - 2)(\lambda^2 - 4\lambda + 3) = -(\lambda - 2)(\lambda - 1)(\lambda - 3)$. So the eigenvalues of $M$ are $1, 2, 3$. If $M + kI$ has $\lambda = 0$ as an eigenvalue, then the eigenvalues of $M + kI$ are the eigenvalues of $M$ shifted by $k$: $1+k, 2+k, 3+k$. For $\lambda = 0$ to be an eigenvalue, we need one of $1+k = 0$, $2+k = 0$, or $3+k = 0$. This gives $k = -1, -2,$ or $k = -3$. Choice A is correct. Choice B gives positive values which would shift eigenvalues away from zero. Choice C is missing $k = -3$. Choice D has positive values and is missing $k = -2$.

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