Linear Algebra

The matrix $$\begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix}$$ represents a reflection about the $x$-axis because it satisfies the following equation: $$\begin{bmatrix} x' \ y' \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}$$, where $$(x', y')$$ are the coordinates of a point after the reflection and $$(x, y)$$ are the coordinates of the point before the reflection.

The matrix $$\begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$$ represents a shear transformation that maps the line $y = x$ to the line $y = 2x$ because it satisfies the following equation: $$\begin{bmatrix} x' \ y' \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}$$, where $$(x', y')$$ are the coordinates of a point after the shear transformation and $$(x, y)$$ are the coordinates of the point before the shear transformation.

The matrix $$\begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}$$ represents a projection onto the $x$-axis because it satisfies the following equation: $$\begin{bmatrix} x' \ y' \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}$$, where $$(x', y')$$ are the coordinates of a point after the projection and $$(x, y)$$ are the coordinates of the point before the projection.

The matrix $$\begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix}$$ represents a change of basis from the standard basis to the basis $${\begin{bmatrix} 1 \ 2 \end{bmatrix}, \begin{bmatrix} 3 \ 4 \end{bmatrix}}$$ because it satisfies the following equation: $$\begin{bmatrix} x' \ y' \end{bmatrix} = \begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}$$, where $$(x', y')$$ are the coordinates of a point in the new basis and $$(x, y)$$ are the coordinates of the point in the old basis.

The matrix $$\begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix}$$ represents the transpose of the matrix $$\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$$ because it is obtained by interchanging the rows and columns of the original matrix.

The determinant of the matrix $$\begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix}$$ is $$-2$$ because it is calculated by subtracting the product of the elements on the main diagonal from the product of the elements on the other diagonal.

The trace of the matrix $$\begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix}$$ is $$7$$ because it is calculated by adding the elements on the main diagonal.

The rank of the matrix $$\begin{bmatrix} 1 & 2 \ 3 & 4 \ 5 & 6 \end{bmatrix}$$ is $$2$$ because it is the maximum number of linearly independent rows or columns in the matrix.

The null space of the matrix $$\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$$ is $${\begin{bmatrix} -2 \ 1 \end{bmatrix}}$$ because it is the set of all vectors that are orthogonal to the row space of the matrix.

The column space of the matrix $$\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$$ is $${\begin{bmatrix} 1 \ 3 \end{bmatrix}, \begin{bmatrix} 2 \ 4 \end{bmatrix}}$$ because it is the set of all linear combinations of the columns of the matrix.

The row space of the matrix $$\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$$ is $${\begin{bmatrix} 1 \ 2 \end{bmatrix}, \begin{bmatrix} 3 \ 4 \end{bmatrix}}$$ because it is the set of all linear combinations of the rows of the matrix.

The eigenvalues of the matrix $$\begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix}$$ are $$2$$ and $$7$$ because they are the roots of the characteristic polynomial of the matrix.

The eigenvectors of the matrix $$\begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix}$$ are $$\begin{bmatrix} 1 \ 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 \ 2 \end{bmatrix}$$ because they are the vectors that are not changed by the matrix except for a scalar multiple.

Gauss-Jordan Elimination is an ancient algorithm that efficiently solves systems of linear equations. It is widely used in modern AI for tasks like linear regression, where it helps find the best-fit line or hyperplane for a given dataset.

In a Hilbert space, the inner product of two vectors is a complex number.