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Answers for “如何将矩阵乘以向量?”

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The missing middle column is [0−169]\begin{bmatrix} 0 \\ -16 \\ 9 \end{bmatrix}. This is obtained by applying the transformation A to the middle column of B, which is [023]\begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}.

Conditions: A and B are the specific 3x3 matrices shown in the video.; The method of column-wise composition is used.

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A2×2A 2\times 2 transformation matrix maps the standard basis vectors to its own columns. Specifically, the first column of the matrix is the image of the vector [1,0]T[1, 0]^T, and the second column is the image of the vector [0,1]T[0, 1]^T.

Conditions: The matrix is 2×22\times 2.; Working in standard Cartesian coordinates.

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The mathematician generalizes the concept to any object that supports sensible addition and scalar multiplication operations.

Conditions: Abstract linear algebra context

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Rewriting a vector as a linear combination of standard basis vectors allows you to use the linearity of the transformation. Instead of computing the full matrix-vector product directly, you can apply the transformation to each basis vector separately (which corresponds to the columns of the matrix) and then combine the results using the original coefficients.

Conditions: The vectors are in R3R^3.; The coefficients are the coordinates of the vector relative to the standard basis.; A is linear: applying it preserves sums and scalar multiples.

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Subtraction is equivalent to adding the additive inverse. In matrix algebra, multiplying a matrix by the scalar -1 creates its additive inverse.

Conditions: Scalar multiplication by -1 is defined entrywise.; Matrix addition is defined entrywise.; A and B have the same dimensions.

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Multiplying a 3x3 matrix by a vector (x,y,z)(x, y, z) scales each column of the matrix (which represents a transformed basis vector) by the corresponding input coordinate and sums the results. This works because linear transformations preserve addition and scalar multiplication.

Conditions: Matrix is 3x3 representing a linear transformation; Input vector has coordinates (x,y,z)(x, y, z); Transformation is linear

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Matrix-vector multiplication in 3D works by scaling the columns of the matrix by the corresponding coordinates of the input vector and summing the results. The coordinates (x,y,z)(x, y, z) act as scalar multipliers for the transformed basis vectors (the columns), leveraging the linearity property that preserves addition and scalar multiplication.

Conditions: The matrix is a 3x3 transformation matrix.; The input vector has coordinates (x,y,z)(x, y, z).; The transformation is linear.

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The origin stays fixed because linearity requires T(av+bw)=aT(v)+bT(w)T(av+bw)=aT(v)+bT(w). If we set v=0v=0 and w=0w=0 (or simply consider the zero vector), T(0)=T(0⋅v)=0⋅T(v)=0T(0) = T(0\cdot v) = 0\cdot T(v) = 0.

Conditions: The transformation is linear.; Applying the definition of linearity to the zero vector.

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Geometrically, the determinant of a2×2a 2\times 2 matrix represents the factor by which the linear transformation scales areas. Specifically, it is the signed area of the parallelogram formed by the matrix's column vectors.

Conditions: The matrix is 2×22\times 2.; The transformation is linear.; Use ordinary Euclidean area in standard orthonormal coordinates.

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Matrix-vector multiplication is reframed as scaling the transformed basis vectors (which are the columns of the matrix) by the input vector's components and summing them. This constructs the final position within the skewed coordinate system defined by the matrix, rather than just following a tedious arithmetic recipe.

Conditions: Input and output coordinates use the fixed standard basis.; The matrix represents an active linear map moving vectors.

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The residual vector r⃗=Ax⃗∗−b⃗\vec{r} = A\vec{x}^* - \vec{b} is orthogonal to the column space C(A)C(A) because Ax⃗∗A\vec{x}^* is the orthogonal projection of b⃗\vec{b} onto C(A)C(A). Orthogonality to C(A)C(A) means r⃗\vec{r} is in the orthogonal complement C(A)⊥C(A)^\perp.

Conditions: Ax⃗∗A\vec{x}^* is the orthogonal projection of b⃗\vec{b} onto C(A)C(A); C(A)⊥=N(AT)C(A)^\perp = N(A^T) (Fundamental Theorem of Linear Algebra); Matrix multiplication distributes over subtraction