Review : Matrices & Vectors: Question 5-2.4 & Solution

Question 4

Work with real matrices and column vectors. Write $I_n$ for the $n\times n$ identity, $A^T$ for transpose, and $\|v\|=\sqrt{v^Tv}$ for Euclidean length. Show the reasoning behind every classification; do not use eigenvalue methods.

Use the ordered basis $u=(1,1)^T$ , $v=(1,-1)^T$ of the plane. Let $P$ have columns $u,v$ , so a coordinate column $c$ represents the physical vector $Pc$ . A transformation in standard coordinates is $T=\begin{pmatrix}2&1\\0&1\end{pmatrix}.$

Tasks

Find $P^{-1}$ and the coordinates of $w=(3,1)^T$ in the ordered basis.
Derive the matrix representing $T$ in this basis, explaining the order of all three operations.
Apply the coordinate matrix to your coordinate column and check the result against direct standard-coordinate multiplication.
For arbitrary coordinate columns, derive the relation between coordinate length and physical length. Compare the determinants of the two representations and interpret the physical area factor.

Original worksheet page 1: question and worked solution for 5-2-004

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Question 4 – Solution

Strategy. A vector and its coordinate column describe the same object in different languages. Convert in the correct order.

Step 1: Build and invert the basis matrix. Here $P=\begin{pmatrix}1&1\\1&-1\end{pmatrix},\qquad P^{-1}=\frac 12\begin{pmatrix}1&1\\1&-1\end{pmatrix}, \qquad [w]_{u,v}=P^{-1}w=\boxed{(2,1)^T}.$ Thus $w=2u+v$ , as the figure illustrates.

Step 2: Express the transformation in the new coordinates. Start with a coordinate column $c$ . Convert it to $Pc$ , apply $T$ , and convert the result back with $P^{-1}$ . Consequently $C=P^{-1}TP =\frac 12\begin{pmatrix}1&1\\1&-1\end{pmatrix} \begin{pmatrix}3&1\\1&-1\end{pmatrix} =\boxed{\begin{pmatrix}2&0\\1&1\end{pmatrix}}.$

Step 3: Verify both routes. The coordinate route gives $C(2,1)^T=(4,3)^T$ , representing $P(4,3)^T=(7,1)^T$ . Directly, $T(3,1)^T=(7,1)^T$ as well. Equivalently, $PC=TP$ verifies the agreement for every input, not only $w$ .

Step 4: Compare lengths and area factors. Since $P^TP=2I_2$ , $\boxed{\|Pc\|^2=c^TP^TPc=2\|c\|^2.}$ The basis vectors are perpendicular but have length $\sqrt 2$ , so physical length is $\sqrt 2$ times coordinate length. Also $\det C=\det(P^{-1})\det T\det P=\det T=2$ . The physical transformation doubles areas and preserves orientation, regardless of which basis is used to describe it. The negative determinant of $P$ describes the basis orientation; it does not change this area factor.

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 5-2-004

Review : Matrices & Vectors — Question 4

Question 4

Question 4 – Solution