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If the columns of a transformation matrix are linearly dependent, it means the transformation collapses the space into a lower dimension. For example, a 2D plane might be squashed into a 1D line or a single point.

Conditions: The matrix represents a linear transformation.; Columns are linearly dependent.

If the columns of a transformation matrix are linearly dependent, it means the transformation collapses the space into a lower dimension. For example, a 2D plane might be squashed into a 1D line or a single point.

Conditions: The matrix represents a linear transformation.; Columns are linearly dependent.

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