Linear Algebra Practice Questions

Let's analyze each statement:

I. The statement rank(AB) = rank(A)rank(B) is generally FALSE. For example, if A is a $2 \times 2$ matrix with rank 1 and B is a $2 \times 2$ matrix with rank 1, AB can have rank 0 or 1.
II. The statement det(AB) = det(A)det(B) is a fundamental property of determinants and is TRUE.
III. The statement rank(A+B) $\leq$ rank(A) + rank(B) is known as the subadditivity of rank and is a TRUE property of matrices.
IV. The statement det(A+B) $\leq$ det(A) + det(B) is generally FALSE. Consider

and

. Then $\det(A) = 0$, $\det(B) = 0$, but

, so $\det(A+B) = 1$. Here $1 \not\leq 0 + 0$.

Therefore, statements II and III are TRUE.

According to a fundamental theorem in linear algebra, a square matrix X has a multiplicative inverse (i.e., is invertible) if and only if its determinant is non-zero. The two conditions are defining properties of each other. Therefore, the statement "X is invertible" and the statement "Determinant of X is non-zero" are logically equivalent.

For any square matrix, invertibility is equivalent to determinant being non-zero.

The product of the eigenvalues of a matrix is equal to its determinant. The given matrix is a Vandermonde matrix. We can compute its determinant by applying row operations to simplify the matrix. Subtracting the first row from each of the other rows ($R_i \leftarrow R_i - R_1$ for $i=2,3,4$) creates zeros in the first column below the first element. Expanding the determinant along the first column and repeating this process leads to the result. The determinant is $(3-2)(4-2)(5-2)(4-3)(5-3)(5-4) = 1 \cdot 2 \cdot 3 \cdot 1 \cdot 2 \cdot 1 = 12$.

A matrix is diagonalizable if and only if it has a complete set of linearly independent eigenvectors. Since all eigenvectors of P are multiples of a single vector, P has only one linearly independent eigenvector. For a 2x2 matrix, this means it does not have a full set of two linearly independent eigenvectors, so it cannot be diagonalized (Statement III is true).

A matrix that is not diagonalizable must have at least one repeated eigenvalue (Statement II is true). Statement I is not necessarily true, as P could be invertible. For example, the matrix

has a repeated eigenvalue of 1 but is invertible.

First, we compute the matrix $A$ from the given vectors $u$ and $v$.
$A = uv^T = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix}$
The eigenvalues $\lambda$ are found using the characteristic equation $\lambda^2 - \text{trace}(A)\lambda + \det(A) = 0$.
The trace of A is the sum of the diagonal elements: $\text{trace}(A) = 1 + 2 = 3$.
The determinant of A is: $\det(A) = (1)(2) - (1)(2) = 0$.
Substituting these values into the characteristic equation gives:
$\lambda^2 - 3\lambda + 0 = 0$
$\lambda(\lambda - 3) = 0$
The eigenvalues are $\lambda = 0$ and $\lambda = 3$. The largest eigenvalue is 3.

For a quadratic equation $x^2 + Bx + C = 0$, the sum of the roots is $-B$ and the product of the roots is $C$. The given equation is $x^2 - (13)_b x + (36)_b = 0$, and the roots are $(5)_b$ and $(6)_b$.

Sum of roots: $(5)_b + (6)_b = (13)_b$.
In base 10, this is $5 + 6 = 1 \cdot b^1 + 3 \cdot b^0$, which simplifies to $11 = b + 3$. Solving for $b$ gives $b=8$.

Product of roots: $(5)_b \times (6)_b = (36)_b$.
In base 10, this is $5 \times 6 = 3 \cdot b^1 + 6 \cdot b^0$, which simplifies to $30 = 3b + 6$. Solving for $b$ gives $3b = 24$, so $b=8$.
Both conditions confirm the base is 8.

The condition $\sum_{i=1}^{n}c_{i}a_{i}=0$ with not all $c_i$ being zero is the definition of linear dependence for the column vectors $a_i$ of matrix A. A matrix with linearly dependent columns is singular, meaning its determinant is zero ($|A|=0$).

For a system of linear equations $Ax=b$, if the matrix A is singular, the system has either no solution or infinitely many solutions.

We are given that $b = \sum_{i=1}^{n}a_{i}$. This means that $b$ is a linear combination of the columns of A. A specific solution exists: if we take $x$ to be a vector of all ones, then $Ax = \sum a_i = b$.

Since a solution exists and the matrix A is singular, there must be infinitely many solutions.

The sum of the squares of all elements of a real symmetric matrix A is equal to the sum of the squares of its eigenvalues, i.e., $\sum_{i,j} A_{ij}^2 = \sum_k \lambda_k^2$. We are given this sum is 50.

The rank of the matrix is 2, which means it has exactly two non-zero eigenvalues, say $\lambda_1$ and $\lambda_2$. All other $n-2$ eigenvalues are 0. Thus, $\lambda_1^2 + \lambda_2^2 = 50$.

(I) Since $n-2$ eigenvalues are 0 (assuming $n \ge 2$), and 0 is within the interval [-5, 5], at least one eigenvalue is in this range. So, statement (I) is true.

(II) Consider the case where $\lambda_1 = 5$ and $\lambda_2 = 5$. Then $\lambda_1^2 + \lambda_2^2 = 25 + 25 = 50$. The largest magnitude is 5, which is not strictly greater than 5. Therefore, statement (II) is not necessarily true.

First, we compute the sum of the matrices P and Q:
$P+Q = \begin{bmatrix} 1 & 1&-1 \\ 2&-3 & 4\\ 3 &-2 & 3 \end{bmatrix} + \begin{bmatrix} -1 & -2&-1 \\ 6 & 12& 6\\ 5&10 & 5 \end{bmatrix} = \begin{bmatrix} 0 & -1 & -2 \\ 8 & 9 & 10 \\ 8 & 8 & 8 \end{bmatrix}$
The rank of a matrix is the number of linearly independent rows. The determinant of P+Q is 0, which means its rank is less than 3. The first two rows are linearly independent. The third row is also not a linear combination of the first two. However, we can see that Row 2 - Row 3 = [0 1 2], which is -1 * Row 1. Thus, the rows are linearly dependent, and there are only 2 linearly independent rows. The rank is 2.

The characteristic polynomial is given by $\lambda^3 - 4\lambda^2 + a\lambda + 30 = 0$. Since $\lambda = 2$ is an eigenvalue, it must satisfy this equation.
Substituting $\lambda = 2$:
$(2)^3 - 4(2)^2 + a(2) + 30 = 0$
$8 - 16 + 2a + 30 = 0$
$22 + 2a = 0 \implies a = -11$.

The complete polynomial is $\lambda^3 - 4\lambda^2 - 11\lambda + 30 = 0$.
Since $\lambda=2$ is a root, $(\lambda-2)$ is a factor. Factoring the polynomial gives:
$(\lambda-2)(\lambda^2 - 2\lambda - 15) = 0$
$(\lambda-2)(\lambda-5)(\lambda+3) = 0$
The eigenvalues are $2, 5, -3$. Their absolute values are $2, 5, 3$. The largest among these is 5.

For a real matrix, complex eigenvalues must occur in conjugate pairs. Since $(2 + \sqrt{-1})$ or $(2+i)$ is an eigenvalue, its complex conjugate, $(2-i)$, must also be an eigenvalue. The three eigenvalues of the 3x3 matrix P are therefore $2+i$, $2-i$, and 3.

The determinant of a matrix is the product of its eigenvalues.
$\det(P) = (2+i)(2-i)(3)$
$\det(P) = (2^2 - i^2)(3) = (4 - (-1))(3) = (5)(3) = 15$

The determinant of a matrix is the product of its eigenvalues.
Given the eigenvalues of matrix A are 1, 2, and 4.
So, the determinant of A is $\det(A) = 1 \times 2 \times 4 = 8$.

We use two properties of determinants:

1. $\det(A^{-1}) = \frac{1}{\det(A)}$
2. $\det(M^T) = \det(M)$

Therefore, $\det((A^{-1})^T) = \det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{8} = 0.125$.

The expression is $(x^{3}+x^{4}+x^{5}+...)^{3}$. By factoring out $x^3$, we get $(x^3(1+x+x^{2+}...))^3$, which simplifies to $x^9(1+x+x^{2+}...)^3$. The infinite series is the geometric series for $(1-x)^{-1}$. The expression becomes $x^9(1-x)^{-3}$. To find the coefficient of $x^{12}$, we need the coefficient of $x^3$ from the expansion of $(1-x)^{-3}$. Using the binomial theorem for a negative exponent, the coefficient of $x^r$ in $(1-x)^{-n}$ is $\binom{n+r-1}{r}$. For $n=3$ and $r=3$, the coefficient is $\binom{3+3-1}{3} = \binom{5}{3} = 10$.

Let's analyze each statement:
I. "If $m < n$, then all such systems have a solution" is false. A system can be inconsistent. For example, $x+y+z=1$ and $x+y+z=2$ has no solution, with $m=2, n=3$.

II. "If $m > n$, then none of these systems has a solution" is false. A system can have redundant equations and be consistent. For example, $x+y=2$ and $2x+2y=4$ and $x-y=0$ has a solution $x=1, y=1$, with $m=3, n=2$.

III. "If $m = n$, then there exists a system which has a solution" is true. We can easily construct such a system, for example, $x=1, y=2$ is a system with $m=n=2$ that has a solution.

The eigenvalues (λ) are the roots of the characteristic equation det(A - λI) = 0.
For the given matrix A, the equation is:
$\begin{vmatrix} 4-\lambda & 5 \\ 2 & 1-\lambda \end{vmatrix} = 0$
Expanding the determinant gives:
$(4-\lambda)(1-\lambda) - (5)(2) = 0$
$\lambda^2 - 5\lambda + 4 - 10 = 0$
$\lambda^2 - 5\lambda - 6 = 0$
Factoring the quadratic equation, we get $(\lambda-6)(\lambda+1) = 0$.
The eigenvalues are $\lambda = 6$ and $\lambda = -1$. The larger of these is 6.

The LU decomposition is given by the equation $A = LU$. We are given

and the structure of L and U, where U has 1s on its diagonal.
$\begin{bmatrix} 2 & 2\\ 4&9 \end{bmatrix} = \begin{bmatrix} l_{11} & 0\\ l_{21}&l_{22} \end{bmatrix} \begin{bmatrix} 1 & u_{12}\\ 0&1 \end{bmatrix}$
By matrix multiplication:

1. $l_{11} \cdot 1 = 2 \implies l_{11} = 2$.
2. $l_{21} \cdot 1 = 4 \implies l_{21} = 4$.
3. $l_{11} \cdot u_{12} = 2 \implies 2 \cdot u_{12} = 2 \implies u_{12} = 1$.
4. $l_{21} \cdot u_{12} + l_{22} \cdot 1 = 9 \implies 4 \cdot 1 + l_{22} = 9 \implies l_{22} = 5$.

To find the eigenvectors $X$ corresponding to an eigenvalue $\lambda$, we solve the equation $(A - \lambda I)X = 0$. Given the eigenvalue $\lambda = 1$, we have:
$(A - I)X = \begin{bmatrix} 1-1 & -1 & 2 \\ 0 & 1-1 & 0 \\ 1 & 2 & 1-1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
This gives the system of equations: $-x_2 + 2x_3 = 0$ and $x_1 + 2x_2 = 0$. From the first equation, $x_2 = 2x_3$. Substituting this into the second equation gives $x_1 = -2x_2 = -4x_3$. The eigenvector is of the form $\alpha(-4, 2, 1)$ for any non-zero scalar $\alpha \in \mathbb{R}$.