Matrix Structure and Transformations
Anton, Rorres, & Kaul §§1.7–1.9
Test 1 · Sep 17
October 9, 2026
Rotate · Reflect · Project · Stretch
\[\begin{gather*} \mat{A}\vct{x}=\vct{b} \;\xrightarrow{\mat{A}=\mat{P}\mat{L}\mat{U}}\; \underbrace{\mat{L}\vct{y}=\mat{P}^{\mathsf T}\vct{b}}_{\text{forward solve}} \;\longrightarrow\; \underbrace{\mat{U}\vct{x}=\vct{y}}_{\text{back solve}} \\ \mat{A}\vct{e}_j=\vct{A}_j \quad\implies\quad \mat{A}\vct{x}=\textstyle\sum_{j=1}^n x_j\vct{A}_j \end{gather*}\]
The pattern of entries can reveal how a matrix behaves before any calculation
For the diagonal matrix
\[ \mat{D}=\begin{bmatrix} d_1&0&\cdots&0\\ 0&d_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&d_n \end{bmatrix}, \]
each coordinate is scaled independently:
\[ \mat{D}\vct{x} = \begin{bmatrix} d_1x_1&d_2x_2&\cdots&d_nx_n \end{bmatrix}^{\mathsf T} \]
For a positive integer \(k\), \(\mat{D}^k\) has diagonal entries \(d_i^k\)
If every \(d_i\ne0\), \(\mat{D}^{-1}\) has diagonal entries \(1/d_i\)
\[ \mat{L}=\begin{bmatrix} \ell_{11}&0&0\\ \ell_{21}&\ell_{22}&0\\ \ell_{31}&\ell_{32}&\ell_{33} \end{bmatrix} \]
Zeros lie above the diagonal
\[ \mat{U}=\begin{bmatrix} u_{11}&u_{12}&u_{13}\\ 0&u_{22}&u_{23}\\ 0&0&u_{33} \end{bmatrix} \]
Zeros lie below the diagonal
A triangular matrix is invertible exactly when every diagonal entry is nonzero
Its inverse has the same triangular form
\[ \mat{U}= \begin{bmatrix} 1&2&1\\ 0&2&2\\ 0&0&1 \end{bmatrix} \]
\[ \left[ \begin{array}{rrr|rrr} 1&2&1&1&0&0\\ 0&2&2&0&1&0\\ 0&0&1&0&0&1 \end{array} \right] \]
Start at the bottom and work upward
\[ \left[ \begin{array}{rrr|rrr} 1&2&1&1&0&0\\ 0&2&2&0&1&0\\ 0&0&1&0&0&1 \end{array} \right] \xrightarrow{R_2\leftarrow\tfrac12R_2} \left[ \begin{array}{rrr|rrr} 1&2&1&1&0&0\\ 0&1&1&0&\tfrac12&0\\ 0&0&1&0&0&1 \end{array} \right] \]
All three diagonal pivots now equal \(1\)
\[ \left[ \begin{array}{rrr|rrr} 1&2&1&1&0&0\\ 0&1&1&0&\tfrac12&0\\ 0&0&1&0&0&1 \end{array} \right] \xrightarrow{\substack{R_1\leftarrow R_1-R_3\\R_2\leftarrow R_2-R_3}} \left[ \begin{array}{rrr|rrr} 1&2&0&1&0&-1\\ 0&1&0&0&\tfrac12&-1\\ 0&0&1&0&0&1 \end{array} \right] \]
Use the third pivot to clear both entries above it
\[ \left[ \begin{array}{rrr|rrr} 1&2&0&1&0&-1\\ 0&1&0&0&\tfrac12&-1\\ 0&0&1&0&0&1 \end{array} \right] \xrightarrow{R_1\leftarrow R_1-2R_2} \left[ \begin{array}{rrr|rrr} 1&0&0&1&-1&1\\ 0&1&0&0&\tfrac12&-1\\ 0&0&1&0&0&1 \end{array} \right] \]
Consider the linear boundary-value problem
\[ a(x)y''(x)+b(x)y'(x)+c(x)y(x)=d(x), \qquad y(0)=y(1)=0 \]
At the interior grid points \(x_i=ih\), \(i=1,\ldots,n\), where \(h=1/(n+1)\),
\[ y''(x_i)\approx\frac{y_{i-1}-2y_i+y_{i+1}}{h^2}, \qquad y'(x_i)\approx\frac{y_{i+1}-y_{i-1}}{2h} \]
Each centered-difference equation couples only the neighboring values \(y_{i-1}\), \(y_i\), and \(y_{i+1}\)
Write \(a_i=a(x_i)\), and similarly for \(b_i,c_i,d_i\). Then
\[ \alpha_i y_{i-1}+\beta_i y_i+\gamma_i y_{i+1}=d_i, \]
\[ \alpha_i=\frac{a_i}{h^2}-\frac{b_i}{2h}, \qquad \beta_i=-\frac{2a_i}{h^2}+c_i, \qquad \gamma_i=\frac{a_i}{h^2}+\frac{b_i}{2h} \]
Because \(y_0=y_{n+1}=0\), the unknowns \(\vct{y}=[y_1,\ldots,y_n]^{\mathsf T}\) satisfy
\[ \begin{bmatrix} \beta_1&\gamma_1&&&0\\ \alpha_2&\beta_2&\gamma_2&&\\ &\ddots&\ddots&\ddots&\\ &&\alpha_{n-1}&\beta_{n-1}&\gamma_{n-1}\\ 0&&&\alpha_n&\beta_n \end{bmatrix} \vct{y} = \begin{bmatrix}d_1\\d_2\\\vdots\\d_{n-1}\\d_n\end{bmatrix} \]
Take
\[ -y''(x)+y'(x)+y(x)=x, \qquad y(0)=y(1)=0 \]
With four interior points, \(h=1/5\). Multiplying the centered-difference equations by \(2\) gives
\[ \begin{bmatrix} 102&-45&0&0\\ -55&102&-45&0\\ 0&-55&102&-45\\ 0&0&-55&102 \end{bmatrix} \begin{bmatrix}y_1\\y_2\\y_3\\y_4\end{bmatrix} = \begin{bmatrix}2/5\\4/5\\6/5\\8/5\end{bmatrix} \]
\(\exstar\) Numerically solve the system for \(\vct{y}=[y_1,y_2,y_3,y_4]^{\mathsf T}\)
We already used \(\vct{a}^{\mathsf T}\) for a matrix row in Systems and Matrices
Now transpose any matrix:
For example,
\[ \mat{A}=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}, \qquad \mat{A}^{\mathsf T}=\begin{bmatrix}1&4\\2&5\\3&6\end{bmatrix} \]
The transpose exchanges the roles of rows and columns
A square matrix is symmetric when
\[ \mat{S}=\mat{S}^{\mathsf T} \]
For example,
\[ \mat{S}=\begin{bmatrix} 4&-1&2\\ -1&3&0\\ 2&0&5 \end{bmatrix} \]
has matching entries \(s_{ij}=s_{ji}\) across its diagonal
Symmetry records the same pairwise value in both directions
Elimination separates a matrix into simpler structured factors
Start with LU when no row exchanges are needed
Add a permutation matrix to obtain PLU
Reuse the triangular factors to solve systems
Recall the coefficient matrix from Systems and Matrices:
\[ \mat{A}= \begin{bmatrix} 1&2&-1\\ 2&-1&1\\ 3&1&2 \end{bmatrix} \]
The two column-elimination stages give
\[ \mat{M}_1= \begin{bmatrix}1&0&0\\-2&1&0\\-3&0&1\end{bmatrix}, \qquad \mat{M}_2= \begin{bmatrix}1&0&0\\0&1&0\\0&-1&1\end{bmatrix} \]
\[ \mat{U}=\mat{M}_2\mat{M}_1\mat{A} =\begin{bmatrix} 1&2&-1\\ 0&-5&3\\ 0&0&2 \end{bmatrix} \]
The elimination stages satisfy
\[ \mat{U}=\mat{M}_2\mat{M}_1\mat{A} \]
Undoing them in reverse order gives
\[ \mat{A}= \underbrace{\mat{M}_1^{-1}\mat{M}_2^{-1}}_{\mat{L}}\mat{U}, \qquad \mat{L}=\begin{bmatrix} 1&0&0\\2&1&0\\3&1&1 \end{bmatrix} \]
Without row exchanges, elimination gives \(\mat{A}=\mat{L}\mat{U}\)
Next we incorporate row exchanges and derive the general PLU pattern
A permutation matrix is obtained by reordering the rows of the identity matrix
Write \(\mat{P}_{jk}\) for one exchange of rows \(j\) and \(k\)
In PLU, \(\mat{P}\) records the combined row ordering
\[ \mat{P}=\begin{bmatrix} 0&0&1\\ 1&0&0\\ 0&1&0 \end{bmatrix}, \qquad \mat{P} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}x_3\\x_1\\x_2\end{bmatrix} \]
Every row and column of \(\mat{P}\) contains exactly one \(1\)
\[ \mat{P}^{-1}=\mat{P}^{\mathsf T}, \qquad \mat{P}^{\mathsf T}\mat{P}=\mat{I} \]
A single row-exchange matrix is symmetric
A general permutation matrix need not be
Write the rows of \(\mat{A}\) as \(\vct{a}_i^{\mathsf T}\) and its columns as \(\vct{A}_j\). For the displayed \(\mat{P}\),
\[ \mat{P}\mat{A} = \begin{bmatrix} 0&0&1\\ 1&0&0\\ 0&1&0 \end{bmatrix} \mat{A} = \begin{bmatrix} \vct{a}_3^{\mathsf T}\\ \vct{a}_1^{\mathsf T}\\ \vct{a}_2^{\mathsf T} \end{bmatrix} \]
\[ \mat{A}\mat{P} = \mat{A} \begin{bmatrix} 0&0&1\\ 1&0&0\\ 0&1&0 \end{bmatrix} = \begin{bmatrix}\vct{A}_2&\vct{A}_3&\vct{A}_1\end{bmatrix} \]
Because this \(\mat{P}\) is not symmetric, \(\mat{A}\mat{P}^{\mathsf T}\)—not \(\mat{A}\mat{P}\)—uses the same ordering as \(\mat{P}\mat{A}\)
Rightmost factor acts first
\[ \mat{U}=\mat{M}_2\mat{\Pi}_2\mat{M}_1\mat{\Pi}_1\mat{A} \]
Insert \(\mat{\Pi}_2^{\mathsf T}\mat{\Pi}_2=\mat{I}\)
\[ \mat{\Pi}_2\mat{M}_1 =\mat{\Pi}_2\mat{M}_1 \bigl(\mat{\Pi}_2^{\mathsf T}\mat{\Pi}_2\bigr) = \underbrace{\bigl(\mat{\Pi}_2\mat{M}_1\mat{\Pi}_2^{\mathsf T}\bigr)}_{\widetilde{\mat{M}}_1} \mat{\Pi}_2 \]
Move and relabel the elimination stage—do not commute the factors
The collected factors give
\[ \mat{U} =\mat{M}_2\widetilde{\mat{M}}_1 \underbrace{\bigl(\mat{\Pi}_2\mat{\Pi}_1\bigr)}_{\mat{P}^{\mathsf T}} \mat{A} =\mat{L}^{-1}\mat{P}^{\mathsf T}\mat{A}, \qquad \mat{L}=\widetilde{\mat{M}}_1^{-1}\mat{M}_2^{-1} \]
\[ \mat{U} =\mat{M}_q\mat{\Pi}_q \mat{M}_{q-1}\mat{\Pi}_{q-1} \cdots \mat{M}_1\mat{\Pi}_1\mat{A} \]
Every exchange after stage \(j\) acts only among rows \(j+1,\ldots,m\). Set
\[ \widetilde{\mat{M}}_j = \bigl(\mat{\Pi}_q\cdots\mat{\Pi}_{j+1}\bigr) \mat{M}_j \bigl(\mat{\Pi}_q\cdots\mat{\Pi}_{j+1}\bigr)^{\mathsf T} \]
Conjugation applies the same relabeling to rows and columns
Because \(\mat{\Pi}_q\cdots\mat{\Pi}_{j+1}\) fixes the first \(j\) rows and columns, conjugation only reorders the entries below the diagonal in column \(j\)
Therefore \(\widetilde{\mat{M}}_j\) and \(\widetilde{\mat{M}}_j^{-1}\) remain unit lower triangular
\[\begin{align*} \mat{P}^{\mathsf T} &=\mat{\Pi}_q\cdots\mat{\Pi}_1,\\ \mat{U} &=\widetilde{\mat{M}}_q\cdots\widetilde{\mat{M}}_1 \mat{P}^{\mathsf T}\mat{A},\\ \mat{L} &=\widetilde{\mat{M}}_1^{-1}\cdots\widetilde{\mat{M}}_q^{-1} \end{align*}\]
\[ \mat{A}= \begin{bmatrix} 0&0&1\\ 2&2&0\\ 1&3&1 \end{bmatrix} \]
\[ \mat{\Pi}_1\mat{A} = \begin{bmatrix} 2&2&0\\ 0&0&1\\ 1&3&1 \end{bmatrix} \]
Eliminate below the first pivot
\[ \mat{M}_1= \begin{bmatrix} 1&0&0\\ 0&1&0\\ -\tfrac12&0&1 \end{bmatrix}, \qquad \mat{M}_1\mat{\Pi}_1\mat{A} = \begin{bmatrix} 2&2&0\\ 0&0&1\\ 0&2&1 \end{bmatrix} \]
Next pivot candidate is zero — another row exchange
\(\mat{\Pi}_2=\mat{P}_{23}\) — exchange rows \(2\) and \(3\)
\[ \mat{U} =\mat{M}_2\mat{\Pi}_2\mat{M}_1\mat{\Pi}_1\mat{A} = \begin{bmatrix} 2&2&0\\ 0&2&1\\ 0&0&1 \end{bmatrix} \]
The entry below the second pivot is already zero, so \(\mat{M}_2=\mat{I}\)
The later exchange moves the first-stage multiplier from row \(3\) to row \(2\)
\[ \widetilde{\mat{M}}_1 =\mat{\Pi}_2\mat{M}_1\mat{\Pi}_2^{\mathsf T} = \begin{bmatrix} 1&0&0\\ -\tfrac12&1&0\\ 0&0&1 \end{bmatrix} \]
Since \(\mat{L}^{-1}=\mat{M}_2\widetilde{\mat{M}}_1\), invert in reverse order
Then use \(\mat{M}_2=\mat{I}\)
\[ \mat{L} =\bigl(\mat{M}_2\widetilde{\mat{M}}_1\bigr)^{-1} =\widetilde{\mat{M}}_1^{-1}\mat{M}_2^{-1} =\widetilde{\mat{M}}_1^{-1} = \begin{bmatrix} 1&0&0\\ \tfrac12&1&0\\ 0&0&1 \end{bmatrix} \]
The exchange matrices record the final row order
\[ \mat{P}^{\mathsf T} =\mat{\Pi}_2\mat{\Pi}_1 = \begin{bmatrix} 0&1&0\\ 0&0&1\\ 1&0&0 \end{bmatrix} \]
\(\mat{P}^{\mathsf T}\mat{A}=\mat{L}\mat{U}\) \(\qquad\Longleftrightarrow\qquad\) \(\mat{A}=\mat{P}\mat{L}\mat{U}\)
For \(\mat{A}\in\reals^{m\times n}\), set \(k=\min(m,n)\)
Compact PLU factors have shapes
\[ \mat{P}\in\reals^{m\times m}, \qquad \mat{L}\in\reals^{m\times k}, \qquad \mat{U}\in\reals^{k\times n}, \qquad \mat{A}=\mat{P}\mat{L}\mat{U} \]
For a tall matrix, the compact form omits the guaranteed zero rows at the bottom of full \(\mat{U}\) and the matching columns of full \(\mat{L}\)
For a square nonsingular matrix, if \(\mat{A}=\mat{P}\mat{L}\mat{U}\), then
\[ \mat{A}\vct{x}=\vct{b} \qquad\Longleftrightarrow\qquad \mat{L}\mat{U}\vct{x}=\mat{P}^{\mathsf T}\vct{b} \]
Introduce \(\vct{y}=\mat{U}\vct{x}\) and solve two triangular systems:
\[ \underbrace{\mat{L}\vct{y}=\mat{P}^{\mathsf T}\vct{b}}_{\text{forward substitution}}, \qquad \underbrace{\mat{U}\vct{x}=\vct{y}}_{\text{back substitution}} \]
First dense solve — factor and solve: \(O(n^3)\) operations
Each new \(\vct{b}\) with saved PLU — permute, forward solve, and back solve: \(O(n^2)\) operations, without forming an inverse
The Lecture 03 companion notebook carries out these steps with SciPy, including packed factors, pivot indices, multiple right-hand sides, residual checks, and measured solve times
We now study functions, or transformations, from \(\reals^n\) to \(\reals^m\)
Every \(\mat{A}\in\reals^{m\times n}\) defines one:
\[ T_{\mat{A}}:\reals^n\longrightarrow\reals^m, \qquad T_{\mat{A}}(\vct{x})=\mat{A}\vct{x} \]
One concrete matrix and one input
\[ \mat{A} =\begin{bmatrix}1&2\\3&-1\\-1&1\end{bmatrix}, \qquad \vct{x}=\begin{bmatrix}4\\-2\end{bmatrix}, \qquad \mat{A}\vct{x}=\begin{bmatrix}0\\14\\-6\end{bmatrix}=\vct{b} \]
\(\vct{x}\in\reals^2\)
\(\reals^2\) — real column vectors with two entries
Number of columns \(=\) input dimension \(2\)
\(\vct{b}=\mat{A}\vct{x}\in\reals^3\)
\(\reals^3\) — real column vectors with three entries
Number of rows \(=\) output dimension \(3\)
\(\mat{A}\in\reals^{3\times2}\) — two-entry inputs to three-entry outputs
Systems and Matrices — solve \(\mat{A}\vct{x}=\vct{b}\) for \(\vct{x}\)
For the same matrix, \(\mat{A}=[\vct{A}_1\ \vct{A}_2]\)
\[ \mat{A}\vct{x} =x_1\vct{A}_1+x_2\vct{A}_2 =\vct{b} \]
Linear combination — a weighted sum of vectors
\[ 4\begin{bmatrix}1\\3\\-1\end{bmatrix} -2\begin{bmatrix}2\\-1\\1\end{bmatrix} =\begin{bmatrix}0\\14\\-6\end{bmatrix} =\vct{b} \]
Solving \(\mat{A}\vct{x}=\vct{b}\) — finding column weights in a linear combination that produces \(\vct{b}\)
Same \(\mat{A}\), input vectors \(\vct{u}=\vct{e}_1\) and \(\vct{v}=\vct{e}_2\), weights \(2\) and \(3\)
\[ 2\vct{u}+3\vct{v} =\begin{bmatrix}2\\3\end{bmatrix} \]
\[ \mat{A}(2\vct{u}+3\vct{v}) =\begin{bmatrix}8\\3\\1\end{bmatrix} \]
\[ \mat{A}\vct{u} =\begin{bmatrix}1\\3\\-1\end{bmatrix}, \qquad \mat{A}\vct{v} =\begin{bmatrix}2\\-1\\1\end{bmatrix} \]
\[ 2\mat{A}\vct{u}+3\mat{A}\vct{v} =\begin{bmatrix}8\\3\\1\end{bmatrix} \]
A rule \(T\) is a linear transformation when, for every \(\vct{u},\vct{v}\) and all scalars \(a,b\), \(T(a\vct{u}+b\vct{v})=aT(\vct{u})+bT(\vct{v})\)
\(j\)th standard basis vector \(\vct{e}_j\) — a \(1\) in position \(j\), zeros elsewhere
Every vector — a linear combination of the standard basis vectors
\[ \vct{x}=x_1\vct{e}_1+\cdots+x_n\vct{e}_n \]
Transform the weighted sum = use the same weights on the transformed basis vectors
\[ T(\vct{x}) =x_1T(\vct{e}_1)+\cdots+x_nT(\vct{e}_n) \]
Basis images — columns of the standard matrix
\[ \mat{A} =\begin{bmatrix} T(\vct{e}_1)&\cdots&T(\vct{e}_n) \end{bmatrix} \]
Basis images — enough to determine every output
Concrete rule behind the running matrix
\[ T(x,y)=(x+2y,\ 3x-y,\ -x+y) \]
Basis images
\[ T(\vct{e}_1)=T(1,0)=\begin{bmatrix}1\\3\\-1\end{bmatrix}, \qquad T(\vct{e}_2)=T(0,1)=\begin{bmatrix}2\\-1\\1\end{bmatrix} \]
Basis images as columns
\[ \mat{A}=\begin{bmatrix}1&2\\3&-1\\-1&1\end{bmatrix}, \qquad T(\vct{x})=\mat{A}\vct{x} \]
Dimensions \(\reals^2\to\reals^3\) — a \(3\times2\) standard matrix
The images of the coordinate directions reveal the geometry
Shown above — diagonal stretch \(\mat{D}\), reflection \(\mat{H}\), projection \(\mat{Q}\), and \(90^\circ\) rotation \(\mat{R}_{90}\)
\[ \mat{D}=\begin{bmatrix}2&0\\0&\tfrac12\end{bmatrix}, \qquad \mat{D} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}2x\\y/2\end{bmatrix} \]
\[ \begin{bmatrix}-1&0\\0&1\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}-x\\y\end{bmatrix} \]
Diagonal structure — independent actions on the coordinate directions
Reflection across the line \(y=x\) exchanges the coordinates:
\[ \mat{H}=\begin{bmatrix}0&1\\1&0\end{bmatrix}, \qquad \mat{H}\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}y\\x\end{bmatrix} \]
Its columns are the reflected basis vectors:
\[ \mat{H}\vct{e}_1=\vct{e}_2, \qquad \mat{H}\vct{e}_2=\vct{e}_1 \]
Reflecting twice restores every vector:
\[ \mat{H}^2=\mat{I}, \qquad \mat{H}^{-1}=\mat{H} \]
A reflection changes geometry without losing information
For a unit vector \(\vct{u}\), the Householder reflection across \(\vct{u}^{\mathsf T}\vct{x}=0\) is
\[ \mat{H}=\mat{I}-2\vct{u}\vct{u}^{\mathsf T}, \qquad \mat{H}\vct{x}=\vct{x}-2\vct{u} \bigl(\vct{u}^{\mathsf T}\vct{x}\bigr) \]
Householder reflections can zero several vector entries in one step, a key idea behind QR factorization
\(\exstar\) For the Householder matrix \(\mat{H}\) defined above, show that \(\mat{H}^2=\mat{I}\) Use \(\vct{u}^{\mathsf T}\vct{u}=1\)
Every reflection satisfies \(\mat{H}^2=\mat{I}\), but this condition is only necessary:
\[ (-\mat{I}_2)^2=\mat{I}_2, \qquad -\mat{I}_2\text{ is rotation through }180^\circ \]
To prove that \(\mat{A}\) is a reflection, find a unit vector \(\vct{u}\) such that
\[ \mat{A}=\mat{I}-2\vct{u}\vct{u}^{\mathsf T} \]
For reflection across \(y=x\), a unit normal is
\[ \vct{u}=\frac{1}{\sqrt{2}}\begin{bmatrix}1\\-1\end{bmatrix}, \qquad \mat{I}-2\vct{u}\vct{u}^{\mathsf T} =\begin{bmatrix}0&1\\1&0\end{bmatrix} \]
\(\mat{A}^2=\mat{I}\) is necessary; a Householder representation is sufficient
Projection onto the \(x\)-axis keeps the first coordinate and removes the second:
\[ \mat{Q}=\begin{bmatrix}1&0\\0&0\end{bmatrix}, \qquad \mat{Q}\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}x\\0\end{bmatrix} \]
This projection is singular: it collapses a direction and cannot be undone
A plane in \(\reals^3\) with nonzero normal \(\vct{u}\): \(\vct{u}^{\mathsf T}\vct{x}=0\)
Remove the normal component to reach the plane:
\[\begin{align*} \vct{y}&=\vct{x}-\alpha\vct{u}, &0=\vct{u}^{\mathsf T}\vct{y} &=\vct{u}^{\mathsf T}\vct{x}-\alpha\vct{u}^{\mathsf T}\vct{u}, &\alpha&=\frac{\vct{u}^{\mathsf T}\vct{x}}{\vct{u}^{\mathsf T}\vct{u}} \end{align*}\]
Thus \(\vct{y}=\mat{Q}\vct{x}\) with projection matrix
\[ \mat{Q}=\mat{I}-\frac{\vct{u}\vct{u}^{\mathsf T}}{\vct{u}^{\mathsf T}\vct{u}} \]
A counterclockwise rotation through angle \(\theta\) sends
\[ \vct{e}_1\longmapsto \begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}, \qquad \vct{e}_2\longmapsto \begin{bmatrix}-\sin\theta\\\cos\theta\end{bmatrix} \]
These images become the columns:
\[ \mat{R}_{\theta} =\begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix} \]
The columns show where the two coordinate directions go
Since \(\cos 90^\circ=0\) and \(\sin 90^\circ=1\),
\[ \mat{R}_{90} =\begin{bmatrix}0&-1\\1&0\end{bmatrix}, \qquad \mat{R}_{90}\begin{bmatrix}2\\1\end{bmatrix} =\begin{bmatrix}-1\\2\end{bmatrix} \]
Rotating backward undoes the action:
\[ \mat{R}_{90}^{-1} =\mat{R}_{-90} =\mat{R}_{90}^{\mathsf T} \]
This is the purple arrow in the comparison picture
A Givens rotation changes only coordinates \(i\) and \(j\) and can make one selected component zero:
\[ \mat{G}(i,j,\theta) = \begin{bmatrix} 1&0&\cdots&0&\cdots&\cdots&0\\ 0&\ddots&&\vdots&&&\vdots\\ \vdots&&c&\cdots&-s&&\vdots\\ 0&\cdots&\vdots&\ddots&\vdots&\cdots&0\\ \vdots&&s&\cdots&c&&\vdots\\ \vdots&&&\vdots&&\ddots&0\\ 0&\cdots&\cdots&0&\cdots&0&1 \end{bmatrix} \in\reals^{n\times n} \qquad \begin{matrix} c=\cos\theta\\ s=\sin\theta \end{matrix} \]
A real square matrix is orthogonal when
\[ \mat{A}^{-1}=\mat{A}^{\mathsf T} \]
\(\exstar\) For \(\mat{A}\in\reals^{2\times2}\), show that this identity holds exactly when its columns are an orthonormal pair—each has length \(1\), and their dot product is \(0\)
Show that it also holds exactly when its rows are an orthonormal pair Hint: the entries of \(\mat{A}^{\mathsf T}\mat{A}\) are dot products of columns; the entries of \(\mat{A}\mat{A}^{\mathsf T}\) are dot products of rows Verify the identity for the rotation \(\mat{R}_{\theta}\) above and the reflection across \(y=x\), \(\mat{H}=\begin{bmatrix}0&1\\1&0\end{bmatrix}\)
Matrix multiplication records a sequence of actions
Suppose
\[ T_{\mat{A}}:\reals^n\to\reals^k, \qquad T_{\mat{B}}:\reals^k\to\reals^m \]
First apply \(T_{\mat{A}}\), then \(T_{\mat{B}}\):
\[ (T_{\mat{B}}\circ T_{\mat{A}})(\vct{x}) =\mat{B}(\mat{A}\vct{x}) =(\mat{B}\mat{A})\vct{x} \]
The rightmost matrix acts first: \(T_{\mat{B}}\circ T_{\mat{A}}=T_{\mat{B}\mat{A}}\)
Let \(\mat{R}\) rotate \(90^\circ\) counterclockwise and let \(\mat{Q}\) project onto the \(x\)-axis:
\[ \mat{R}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}, \qquad \mat{Q}=\begin{bmatrix}1&0\\0&0\end{bmatrix} \]
\[ \mat{Q}\mat{R}\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}-y\\0\end{bmatrix} \]
\[ \mat{R}\mat{Q}\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}0\\x\end{bmatrix} \]
Usually \(\mat{Q}\mat{R}\ne\mat{R}\mat{Q}\)
Planar rotations about the same origin are a useful commuting exception
If \(T_{\mat{A}}\) and \(T_{\mat{B}}\) are invertible operators on \(\reals^n\), then
\[ (T_{\mat{B}}\circ T_{\mat{A}})^{-1} =T_{\mat{A}^{-1}}\circ T_{\mat{B}^{-1}} \]
The corresponding matrix rule from Inverses is
\[ (\mat{B}\mat{A})^{-1} =\mat{A}^{-1}\mat{B}^{-1} \]
Undo the last action first
The PLU factorization
\[ \mat{A}=\mat{P}\mat{L}\mat{U} \]
describes the action of \(\mat{A}\) as a sequence:
\[ \vct{x} \xrightarrow{\ \mat{U}\ } \mat{U}\vct{x} \xrightarrow{\ \mat{L}\ } \mat{L}\mat{U}\vct{x} \xrightarrow{\ \mat{P}\ } \mat{P}\mat{L}\mat{U}\vct{x} =\mat{A}\vct{x} \]
Factorization is representation by simpler actions
Multiplication composes them
One matrix, computational and geometric views
Factors decompose the computation; geometric actions explain what it does
Matrix transformations can
Determinants will attach one number to a square matrix action and answer:
© 2026 Fred J. Hickernell · Illinois Tech · assisted by ChatGPT and Codex · Matrix Transformations · MATH 332 — Fall 2026 Website · \(\exstar\) = exercise