Linear Algebra Practice Questions

The eigenvalues $\lambda$ of a matrix A are the roots of its characteristic equation, $|A - \lambda I| = 0$. For the given matrix

, the equation is:
$(1-\lambda)(a-\lambda) - 4b = 0$
We are given two eigenvalues, $\lambda = -1$ and $\lambda = 7$. Substituting these values gives a system of two linear equations:

1. For $\lambda = -1$: $(1 - (-1))(a - (-1)) - 4b = 0 \implies 2(a+1) - 4b = 0 \implies 2b - a = 1$.
2. For $\lambda = 7$: $(1 - 7)(a - 7) - 4b = 0 \implies -6(a-7) - 4b = 0 \implies 2b + 3a = 21$.

Solving this system of equations yields $a = 5$ and $b = 3$.

The given system of linear equations is a homogeneous system. For such a system to have a non-trivial solution, the determinant of the coefficient matrix must be zero.
The coefficient matrix is:
$A = \begin{pmatrix} p & q & r \\ q & r & p \\ r & p & q \end{pmatrix}$
Setting its determinant to zero: $|A| = p(rq - p^2) - q(q^2 - rp) + r(qp - r^2) = 0$.
This simplifies to $3pqr - p^3 - q^3 - r^3 = 0$, which factors into $-(p+q+r)(p^{2+}q^{2+}r^{2-}pq-qr-rp) = 0$.
This condition is true if either $p+q+r=0$ or $p=q=r$.

The determinant of a matrix is zero if one column is a scalar multiple of another. In the given matrix, the third column is 15 times the first column ($C_3 = 15 \times C_1$). For example, $3 \times 15 = 45$ and $7 \times 15 = 105$. This means the columns are linearly dependent, and the determinant of the original matrix is 0.

The operations performed are:
(i) $R_2 \to R_2 + R_3$ (adding a multiple of one row to another).
(ii) $C_1 \to C_1 - C_3$ (adding a multiple of one column to another).
Neither of these elementary operations changes the value of the determinant. Therefore, the determinant of the resultant matrix remains 0.

Let's define a new function $g(x) = f(x) - f(x+1)$. We are given that $f(x)$ is continuous on $[0,2]$.
Therefore, $g(x)$ is continuous on $[0,1]$ since $x$ and $x+1$ both fall within $[0,2]$ for $x \in [0,1]$.
Evaluate $g(x)$ at the endpoints of the interval $[0,1]$:
$g(0) = f(0) - f(1) = -1 - 1 = -2$.
$g(1) = f(1) - f(2) = 1 - (-1) = 2$.
Since $g(0) = -2$ and $g(1) = 2$, and $g(x)$ is continuous on $[0,1]$, by the Intermediate Value Theorem, there must exist some $y \in (0,1)$ such that $g(y) = 0$.
This implies $f(y) - f(y+1) = 0$, or $f(y) = f(y+1)$.

Let the real eigenvalues of the $n \times n$ matrix be $\lambda_1, \lambda_2, \ldots, \lambda_n$. The trace of the matrix is the sum of its eigenvalues, and the determinant is the product of its eigenvalues.

The question states that the trace is positive (\text{Tr}(A) > 0$) and the determinant is negative ($\det(A) < 0$).

Focusing on the determinant, we have $\det(A) = \lambda_1 \cdot \lambda_2 \cdot \ldots \cdot \lambda_n < 0$.
Since all eigenvalues are real numbers, for their product to be negative, there must be an odd number of negative eigenvalues in the set $\{\lambda_1, \ldots, \lambda_n\}$. An odd count (1, 3, 5, ...) implies there must be at least one negative eigenvalue. The condition on the trace is compatible with this, but the negativity of the determinant is sufficient to prove the conclusion.

We utilize the dimension theorem for vector subspaces, which states:
$\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W)$
Rearranging this formula to solve for the dimension of the intersection, we get:
$\dim(V_1 \cap V_2) = \dim(V_1) + \dim(V_2) - \dim(V_1 + V_2)$
We are given $\dim(V_1) = 4$ and $\dim(V_2) = 4$. Substituting these values gives:
$\dim(V_1 \cap V_2) = 4 + 4 - \dim(V_1 + V_2) = 8 - \dim(V_1 + V_2)$
To find the smallest possible dimension of the intersection, we must subtract the largest possible dimension of the sum, $\dim(V_1 + V_2)$.
Since $V_1 + V_2$ is a subspace of the 6-dimensional vector space $V$, its dimension cannot be greater than 6. So, $\dim(V_1 + V_2) \le 6$.
The smallest possible dimension of $V_1 \cap V_2$ occurs when $\dim(V_1 + V_2)$ is at its maximum, which is 6.
$\min(\dim(V_1 \cap V_2)) = 8 - 6 = 2$

[CORRECT_OPTION: 2]

Let the given matrix A be a product of a column vector and a row vector.
$A = \begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}\begin{bmatrix} 1 &9 &5 \end{bmatrix} = \begin{bmatrix} 2 \cdot 1 & 2 \cdot 9 & 2 \cdot 5\\ -4 \cdot 1 & -4 \cdot 9 & -4 \cdot 5\\ 7 \cdot 1 & 7 \cdot 9 & 7 \cdot 5 \end{bmatrix} = \begin{bmatrix} 2 & 18 & 10\\ -4 & -36 & -20\\ 7 & 63 & 35 \end{bmatrix}$
To find the determinant of A, we can observe the relationship between its rows.
Notice that the second row,

, is $-2$ times the first row,

. That is, $R_2 = -2R_1$.
A property of determinants states that if one row (or column) of a matrix is a scalar multiple of another row (or column), then the determinant of the matrix is zero.
Since $R_2 = -2R_1$, the rows are linearly dependent, which means the determinant of A is 0.

[CORRECT_OPTION: 0]

For a real symmetric matrix, eigenvectors corresponding to distinct eigenvalues are always orthogonal.
Orthogonality implies that their dot product is zero.
Since the matrix is a 4x4 symmetric positive definite matrix, it has 4 real eigenvalues.
If these eigenvalues are distinct, their corresponding eigenvectors will be orthogonal.
Therefore, the dot product of any pair of different eigenvectors will be 0.

[CORRECT_OPTION: 0]

We can solve this system of linear equations using Gaussian elimination on the augmented matrix.
The augmented matrix for the given system is:

Swap $R_1$ and $R_3$ to get a leading 1:

Perform row operations $R_2 \leftarrow R_2 - 4R_1$, $R_3 \leftarrow R_3 - 3R_1$, and $R_4 \leftarrow R_4 - R_1$:

To simplify the second row, we can swap $R_2$ and $R_3$ and multiply $R_2$ by -1 for a positive leading coefficient. Or, for simplicity as shown in the reference: Perform $R_3 \leftarrow 4R_3 - R_2$ and $R_4 \leftarrow 4R_4 - 3R_2$:

Finally, perform $R_4 \leftarrow R_4 + R_3$:

The rank of the coefficient matrix is 3, and the rank of the augmented matrix is also 3. Since the rank of the coefficient matrix equals the rank of the augmented matrix and this rank is equal to the number of variables (3), the system has a unique solution.

[CORRECT_OPTION: 1]

Since $f(x)$ is a non-zero polynomial of degree 3 with roots at $x = 1, x = 2,$ and $x = 3$, it can be written as $f(x) = C(x-1)(x-2)(x-3)$ for some non-zero constant $C$.
To determine the sign of $f(0)$, substitute $x=0$: $f(0) = C(0-1)(0-2)(0-3) = C(-1)(-2)(-3) = -6C$.
To determine the sign of $f(4)$, substitute $x=4$: $f(4) = C(4-1)(4-2)(4-3) = C(3)(2)(1) = 6C$.
Now, consider the product $f(0)f(4) = (-6C)(6C) = -36C^2$.
Since $C$ is a non-zero constant, $C^2$ must be positive. Therefore, $-36C^2$ must be negative.
Thus, $f(0)f(4) < 0$ must be true.

Let the given matrix be $A$.
$A = \begin{bmatrix} 1 & 0&0 & 0&1 \\ 0& 1& 1 & 1 & 0\\ 0& 1& 1& 1&0 \\ 0 & 1 & 1 & 1 & 0\\ 1&0 & 0&0 & 1 \end{bmatrix}$
To find the eigenvalues, we solve the characteristic equation $\det(A - \lambda I) = 0$.
The determinant can be simplified by observing that rows 2, 3, and 4 are identical. If a matrix has linearly dependent rows, its determinant is 0. This implies $\lambda=0$ is an eigenvalue. Since the question asks for the product of non-zero eigenvalues, we need to find the other eigenvalues.

Consider the rank of the matrix $A$. Since rows 2, 3, and 4 are identical, the rank of the matrix is at most 3. Therefore, there are at least $5 - \text{rank}(A)$ eigenvalues equal to 0. In this case, since rows 2, 3, and 4 are multiples of each other (they are identical), two eigenvalues are 0.

The product of all eigenvalues of a matrix is equal to its determinant. However, since the determinant is 0, we need another approach for non-zero eigenvalues.
Alternatively, the sum of the eigenvalues is the trace of the matrix (sum of diagonal elements), and the product of the non-zero eigenvalues equals the product of the non-zero eigenvalues of the submatrix obtained by removing linearly dependent rows/columns.

Let's find the eigenvalues by observation or row operations.
Consider vectors $v_1 = (1, 0, 0, 0, -1)^T$ and $v_2 = (0, 1, -1, 0, 0)^T$, $v_3 = (0, 1, 0, -1, 0)^T$.
$A v_1 = (0, 0, 0, 0, 0)^T$, so $\lambda = 0$ is an eigenvalue.

. So $\lambda = 2$ is an eigenvalue.

. So $\lambda = 3$ is an eigenvalue.

The matrix $A$ has three identical rows (rows 2, 3, 4). This means that $\lambda = 0$ is an eigenvalue with multiplicity at least 2.
We have found three distinct eigenvalues: $0, 2, 3$. We also know that $0$ has a multiplicity of at least 2.
Let the eigenvalues be $\lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_5$. We have found $0, 0, 2, 3$. The sum of eigenvalues is the trace of the matrix, which is $1+1+1+1+1 = 5$.
So, $\lambda_1 + \lambda_2 + \lambda_3 + \lambda_4 + \lambda_5 = 0+0+2+3 + \lambda_5 = 5$.
This implies $\lambda_5 = 0$.
So the eigenvalues are $0, 0, 0, 2, 3$.
The non-zero eigenvalues are $2$ and $3$.
Their product is $2 \times 3 = 6$.

[CORRECT_OPTION: 6]

The core operation within the nested loops is a swap of elements A[i][j] and A[j][i]. Let's analyze the three lines of code with original values a = A[i][j] and b = A[j][i].

1. Temp = a + C
2. A[i][j] = b
3. A[j][i] = Temp - C = (a + C) - C = a

This shows that the values of A[i][j] and A[j][i] are swapped. This operation is equivalent to swap(A[i][j], A[j][i]).

The nested loops for i = 1 to n and for j = 1 to n iterate through all possible pairs of indices (i, j). For any off-diagonal elements where i ≠ j, the code first performs swap(A[i][j], A[j][i]) when processing (i, j). Later, it will process (j, i) and perform swap(A[j][i], A[i][j]), which swaps the elements back to their original state. For diagonal elements where i = j, the swap operation swap(A[i][i], A[i][i]) has no effect.

Since every swap is reversed, the matrix is returned to its original form after the loops complete. The final output is the original matrix A.

Problem Explanation: Vandermonde Determinant Properties

The given matrix is a standard Vandermonde Matrix $A$:
$A = \begin{bmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{bmatrix}$
The determinant is given by: $\det(A) = (y - x)(z - x)(z - y)$.

To determine which option does NOT equal $\det(A)$, we evaluate the row and column operations applied to each. A determinant remains unchanged if we replace a row (or column) with the sum of itself and a multiple of another row (or column).

Analysis of Options

Option A (The Incorrect One)

The matrix is

.

• Reasoning: There is no single set of linear row/column operations that transforms matrix $A$ into this form while preserving the determinant.
• Verification: If we let $x=0, y=1, z=2$, then $\det(A) = 2$. Substituting these into Option A gives

, which has a determinant of -2.

• Conclusion: Since $-2 \neq 2$, Option A is not equal to $\det(A)$.

Option B

The matrix is

.

• Operation: This is obtained by $C_2 \to C_2 + C_1$ and $C_3 \to C_3 + C_1$.
• Result: These operations preserve the determinant. $\det(B) = \det(A)$.

Option C

The matrix is

.

• Operation: This is obtained by $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$.
• Result: Row subtractions preserve the determinant. $\det(C) = \det(A)$.

Option D

The matrix is

.

• Operation: This is obtained by $R_1 \to R_1 + R_2$ and $R_2 \to R_2 + R_3$.
• Result: Row additions preserve the determinant. $\det(D) = \det(A)$.

Final Answer: Option A is the only matrix whose determinant differs from the original.

To find the eigenvalues of $A^{19}$, we first find the eigenvalues of matrix A.
The matrix A is given by:
$A = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$
The characteristic equation is $\det(A - \lambda I) = 0$:
$\begin{vmatrix} 1-\lambda & 1 \\ 1 & -1-\lambda \end{vmatrix} = 0$
$(1-\lambda)(-1-\lambda) - (1)(1) = 0$
$-(1-\lambda)(1+\lambda) - 1 = 0$
$- (1 - \lambda^2) - 1 = 0$
$\lambda^2 - 1 - 1 = 0$
$\lambda^2 - 2 = 0$
The eigenvalues of A are $\lambda_1 = \sqrt{2}$ and $\lambda_2 = -\sqrt{2}$.

If $\lambda$ is an eigenvalue of A, then $\lambda^k$ is an eigenvalue of $A^k$.
Therefore, the eigenvalues of $A^{19}$ are $(\sqrt{2})^{19}$ and $(-\sqrt{2})^{19}$.
$(\sqrt{2})^{19} = 2^{19/2} = 2^{9} \cdot 2^{1/2} = 512\sqrt{2}$
$(-\sqrt{2})^{19} = -(2^{19/2}) = -2^{9} \cdot 2^{1/2} = -512\sqrt{2}$
Thus, the eigenvalues of $A^{19}$ are $512\sqrt{2}$ and $-512\sqrt{2}$.

The final answer is $\boxed{\text{D}}$.

To find the eigenvalues of a triangular matrix (either upper or lower triangular), we can simply read them from its main diagonal elements.

The given matrix is:

This is an upper triangular matrix because all elements below the main diagonal are zero.
The eigenvalues of a triangular matrix are its diagonal entries.
The diagonal entries of matrix $A$ are 1, 4, and 3.
Therefore, the eigenvalues of the matrix are 1, 4, and 3.

To find the values of X and Y, we can use two fundamental properties that relate a matrix to its eigenvalues:

1. Sum of Eigenvalues = Trace of the Matrix
   The trace of a matrix is the sum of the elements on its main diagonal. For matrix

, the trace is $2+Y$.
The given eigenvalues are $\lambda_1 = 4$ and $\lambda_2 = 8$. Their sum is $4+8=12$.
Equating the sum of eigenvalues to the trace:
$12 = 2 + Y$
Solving for Y, we get:
$Y = 12 - 2 = 10$

2. Product of Eigenvalues = Determinant of the Matrix
   The determinant of a 2x2 matrix

is $ad-bc$. For matrix A, the determinant is $(2 \times Y) - (3 \times X) = 2Y - 3X$.
The product of the given eigenvalues is $4 \times 8 = 32$.
Equating the product of eigenvalues to the determinant:
$32 = 2Y - 3X$

Now, we can substitute the value of Y we found in the first step into this equation:
$32 = 2(10) - 3X$
$32 = 20 - 3X$
$3X = 20 - 32$
$3X = -12$
Solving for X, we get:
$X = -4$

Thus, the values are $X = -4$ and $Y = 10$.