FP3.6 - Further matrix algebra

A matrix represents a linear transformation by multiplying a column vector. If $T:\mathbb R^n\to\mathbb R^m$ has matrix $A$, then$$T(\mathbf x)=A\mathbf x.$$Linearity means $T(\mathbf u+\mathbf v)=T(\mathbf u)+T(\mathbf v)$ and $T(k\mathbf u)=kT(\mathbf u)$, so the origin maps to the origin.

The columns of $A$ are the images of the standard basis vectors. In three dimensions,$$A=\begin{pmatrix}|&|&|\\T(\mathbf e_1)&T(\mathbf e_2)&T(\mathbf e_3)\\|&|&|\end{pmatrix}.$$This gives the matrix immediately when the action on $\mathbf i,\mathbf j,\mathbf k$ is known, and explains why matrix multiplication forms the correct linear combination of those images.

Let$$A=\begin{pmatrix}1&0&2\\0&-1&1\\2&1&0\end{pmatrix}.$$Its columns show $T(\mathbf e_1)=(1,0,2)^T$, $T(\mathbf e_2)=(0,-1,1)^T$ and $T(\mathbf e_3)=(2,1,0)^T$. Hence$$T(1,2,-1)^T=A\begin{pmatrix}1\\2\\-1\end{pmatrix}=\begin{pmatrix}-1\\-3\\4\end{pmatrix}.$$The same result is the linear combination $T(\mathbf e_1)+2T(\mathbf e_2)-T(\mathbf e_3)$.

If images of enough independent vectors are supplied instead of basis vectors, put the input vectors as columns of a matrix $X$ and their images as columns of $Y$. The unknown transformation matrix satisfies $AX=Y$; when $X$ is invertible, $A=YX^{-1}$. Verify every supplied mapping after solving.

Matrix products encode successive transformations, but the rightmost matrix acts first. If $B$ is followed by $A$, then$$\mathbf x\xmapsto{B}B\mathbf x\xmapsto{A}A(B\mathbf x)=(AB)\mathbf x.$$This is why the product $AB$ represents $B$ followed by $A$.

A product $AB$ exists when the number of columns of $A$ equals the number of rows of $B$. The result has the rows of $A$ and columns of $B$. Each entry is a row-column dot product; dimension compatibility must be checked before interpreting a composition.

Let$$A=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad B=\begin{pmatrix}1&2\\0&1\end{pmatrix}.$$Then$$AB=\begin{pmatrix}1&2\\0&-1\end{pmatrix}$$represents the shear $B$ followed by reflection $A$. For $(1,1)^T$, $B$ gives $(3,1)^T$ and $A$ gives $(3,-1)^T$, matching $AB(1,1)^T$.

Reversing the order gives$$BA=\begin{pmatrix}1&-2\\0&-1\end{pmatrix},$$which sends $(1,1)^T$ to $(-1,-1)^T$. Thus $AB\ne BA$ in general. A quick test vector can reveal an order error even before the full product is interpreted.

The identity matrix $I$ leaves every vector unchanged, so $AI=IA=A$. If two successive transformations undo one another, their product is $I$ in the appropriate order; this becomes the basis of inverse transformations later in the Topic.

The transpose $A^{\mathsf T}$ is formed by exchanging rows and columns: its $(i,j)$ entry is the $(j,i)$ entry of $A$. Therefore an $m\times n$ matrix becomes an $n\times m$ matrix, and $(A^{\mathsf T})^{\mathsf T}=A$.

The product order reverses because$$(AB)^{\mathsf T}_{ij}=(AB)_{ji}=\sum_k A_{jk}B_{ki}=\sum_k(B^{\mathsf T})_{ik}(A^{\mathsf T})_{kj}.$$So the row-column pairing becomes $B^{\mathsf T}A^{\mathsf T}$, not $A^{\mathsf T}B^{\mathsf T}$.

For$$A=\begin{pmatrix}1&2\\0&-1\end{pmatrix},\qquad B=\begin{pmatrix}3&1\\2&4\end{pmatrix},$$we have $AB=\begin{pmatrix}7&9\\-2&-4\end{pmatrix}$ and$$(AB)^{\mathsf T}=\begin{pmatrix}7&-2\\9&-4\end{pmatrix}=B^{\mathsf T}A^{\mathsf T}.$$Direct multiplication confirms both the entries and reversed order.

Transposes convert column information into row information and are central to dot products and orthogonality. If a square matrix $P$ has orthonormal columns, then $P^{\mathsf T}P=I$, so $P^{-1}=P^{\mathsf T}$; this is used when symmetric matrices are diagonalised.

The determinant of a square matrix is a scalar that measures signed scale under its transformation. For a $3\times3$ matrix, expand along any row or column using signed $2\times2$ minors. A matrix is singular exactly when its determinant is zero; otherwise it is non-singular and has an inverse.

For$$M=\begin{pmatrix}2&1&0\\-1&3&2\\0&4&1\end{pmatrix},$$expansion along the first row gives$$\det M=2(3\cdot1-2\cdot4)-1((-1)\cdot1-2\cdot0)=-10+1=-9.$$Thus $M$ is non-singular and multiplies volumes by $9$, while reversing orientation.

For a matrix containing a parameter, calculate and factor the determinant, then solve $\det A=0$ to find exactly the singular values. Any later inverse formula must explicitly exclude those values. Substitute a candidate back into the determinant as a check.

Swapping two rows changes the determinant's sign; multiplying one row by $k$ multiplies the determinant by $k$; adding a multiple of one row to another leaves it unchanged. These rules can simplify an evaluation, but every operation's effect must be tracked.

A square matrix $A$ has an inverse only when $\det A\ne0$. The inverse satisfies$$AA^{-1}=A^{-1}A=I$$and can be calculated from cofactors:$$A^{-1}=\frac1{\det A}\operatorname{adj}A.$$The adjugate is the transpose of the cofactor matrix.

For$$A=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix},\qquad \det A=1,$$the inverse is$$A^{-1}=\begin{pmatrix}1&-1&1\\0&1&-1\\0&0&1\end{pmatrix}.$$Multiplying $AA^{-1}$ gives $I$, including the cancellation $1-1=0$ and $1-1=0$ in the upper off-diagonal entries.

For invertible $A$ and $B$,$$(AB)^{-1}=B^{-1}A^{-1}.$$Indeed, $(AB)(B^{-1}A^{-1})=AIA^{-1}=I$. The order reverses because the most recently applied transformation must be undone first.

The equation $A\mathbf x=\mathbf b$ has the unique solution $\mathbf x=A^{-1}\mathbf b$ when $A$ is non-singular. In exact work, retain fractions and parameter restrictions until the final result, then verify by substitution.

If a transformation sends $\mathbf x$ to $\mathbf y=A\mathbf x$ and $A$ is non-singular, its inverse sends the image back to the original vector:$$\mathbf x=A^{-1}\mathbf y.$$No inverse transformation exists when $A$ is singular because distinct inputs have been collapsed to the same output.

If $B$ acts first and $A$ second, the combined matrix is $AB$. To reverse the combination, undo $A$ first and then $B$:$$\mathbf y=AB\mathbf x\quad\Longrightarrow\quad \mathbf x=B^{-1}A^{-1}\mathbf y.$$Thus $(AB)^{-1}=B^{-1}A^{-1}$ is both an algebraic and procedural rule.

A linear transformation maps every point and direction vector of a line. To find the preimage of an image line$$\mathbf y=\mathbf p+\lambda\mathbf d,$$apply $A^{-1}$ to both fixed vectors:$$\mathbf x=A^{-1}\mathbf p+\lambda A^{-1}\mathbf d.$$The same parameter remains because matrix multiplication is linear.

Let$$A=\operatorname{diag}\left(2,-1,\frac12\right)$$and suppose the image line is$$\mathbf y=(4,1,3)^T+\lambda(2,-2,1)^T.$$Since $A^{-1}=\operatorname{diag}(1/2,-1,2)$, the original line is$$\mathbf x=(2,-1,6)^T+\lambda(1,2,2)^T.$$Applying $A$ to its point and direction recovers the image line.

Check an inverse transformation in both directions on a general vector or on the defining point/direction data. For a combination, multiply the proposed inverse by the combined matrix in the correct order and require the identity matrix.

A non-zero vector $\mathbf v$ is an eigenvector of $A$ when the transformation changes only its scale (and possibly direction):$$A\mathbf v=\lambda\mathbf v.$$The scalar $\lambda$ is the corresponding eigenvalue. The zero vector is never an eigenvector.

For$$A=\begin{pmatrix}4&1\\2&3\end{pmatrix},$$the characteristic equation is$$(4-\lambda)(3-\lambda)-2=\lambda^2-7\lambda+10=0,$$so $\lambda=5$ or $2$.

For $\lambda=5$, $(A-5I)\mathbf v=0$ gives $y=x$, so one eigenvector is $(1,1)^T$ and a normalised choice is $(1,1)^T/\sqrt2$. For $\lambda=2$, the equation gives $y=-2x$, so a normalised choice is $(1,-2)^T/\sqrt5$. Direct multiplication verifies both pairs.

If a vector is supplied, multiply it by $A$ and compare corresponding non-zero components. It is an eigenvector only if one common scalar $\lambda$ works in every component, including components that are zero. If an eigenvalue is supplied, substitute it into the nullspace equations rather than recomputing every root.

The same method applies to $3\times3$ matrices: the characteristic determinant is cubic and each eigenvector comes from a homogeneous three-variable system. Row-reduce without forcing the free parameter to zero, then choose a convenient non-zero scale before normalising.

A real symmetric matrix $A=A^{\mathsf T}$ has an orthonormal basis of eigenvectors. Put those unit eigenvectors into the columns of an orthogonal matrix $P$. Then$$P^{\mathsf T}P=I,\qquad P^{-1}=P^{\mathsf T},\qquad D=P^{\mathsf T}AP$$is diagonal.

Writing the column equations together gives $AP=PD$. Multiplying on the left by $P^{\mathsf T}=P^{-1}$ yields $P^{\mathsf T}AP=D$. Orthogonality is what makes the change of coordinates preserve lengths and turns the inverse into a transpose.

For$$A=\begin{pmatrix}2&1\\1&2\end{pmatrix},$$the eigenvalue $3$ has unit eigenvector $(1,1)^T/\sqrt2$, and eigenvalue $1$ has unit eigenvector $(1,-1)^T/\sqrt2$. Therefore$$P=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix},\qquad D=\begin{pmatrix}3&0\\0&1\end{pmatrix},$$and direct multiplication gives $P^{\mathsf T}AP=D$.

Permuting the columns of $P$ is allowed only when the diagonal entries of $D$ are permuted in exactly the same way. Changing the sign of any eigenvector column leaves orthogonality and the resulting diagonal entry unchanged.