What would you like to understand?

Each column of $a 2\times 2$ matrix can be read as the coordinates of a vector in the 2D plane. For the matrix $A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$, the first column $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$ represents the vector from the origin to the point (3,1), and the second column $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ represents the vector from the origin to the point (1,2).

Conditions: The matrix is $2\times 2$.; Working in standard Cartesian coordinates.

Each column of $a 2\times 2$ matrix can be read as the coordinates of a vector in the 2D plane. For the matrix $A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$, the first column $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$ represents the vector from the origin to the point (3,1), and the second column $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ represents the vector from the origin to the point (1,2).

Conditions: The matrix is $2\times 2$.; Working in standard Cartesian coordinates.

A 3D linear transformation is fully determined by tracking where the standard basis vectors ($\hat{i}$, $\hat{j}$, and $\hat{k}$) land. The coordinates of these three transformed vectors are recorded as column vectors to form a 3x3 matrix.

Conditions: Working in three-dimensional Cartesian space; Using the standard basis vectors aligned with x, y, and z axes; The transformation is linear (preserves grid lines parallel/evenly spaced and fixes origin)

A 3D linear transformation is fully determined by tracking where the standard basis vectors ($\hat{i}$, $\hat{j}$, and $\hat{k}$) land. The coordinates of these three transformed vectors are recorded as column vectors to form a 3x3 matrix.

Conditions: Working in three-dimensional Cartesian space; Using the standard basis vectors aligned with x, y, and z axes; The transformation is linear (preserves grid lines parallel/evenly spaced and fixes origin)

Starting from $Av = \lambda v$, we rewrite the right side as $(\lambda I)v$ and move all terms to one side to get $(A - \lambda I)v = 0$. Since we seek non-zero solutions for $v$, the matrix $(A - \lambda I)$ must squash space into a lower dimension (have a non-trivial null space).

Conditions: $v$ is a non-zero eigenvector; $A$ is a square matrix; $I$ is the identity matrix

Starting from $Av = \lambda v$, we rewrite the right side as $(\lambda I)v$ and move all terms to one side to get $(A - \lambda I)v = 0$. Since we seek non-zero solutions for $v$, the matrix $(A - \lambda I)$ must squash space into a lower dimension (have a non-trivial null space).

Conditions: $v$ is a non-zero eigenvector; $A$ is a square matrix; $I$ is the identity matrix