MAT223H1 Linear Algebra I: study guide and tutoring

Start with $\mathbf v=a\mathbf b_1+b\mathbf b_2$. A matrix’s columns describe what it does to the basis vectors, expressed in the output basis.

1. Check the basis and find the coordinates

   The columns of $P=\begin{pmatrix}1&2\\1&1\end{pmatrix}$ are the proposed basis vectors. Since $\det P=1-2=-1\ne0$, they form a basis of $\mathbb R^2$.

   The equations $a+2b=5$ and $a+b=3$ give $b=2$, $a=1$. Thus $[\mathbf v]_B=\begin{pmatrix}1\\2\end{pmatrix}$: one copy of $\mathbf b_1$ plus two of $\mathbf b_2$ gives the original vector $(5,3)$.

2. Build the matrix in that basis

   Compute $T(\mathbf b_1)=(1,1)=\mathbf b_1$ and $T(\mathbf b_2)=(4,2)=2\mathbf b_2$. Their $B$-coordinate columns are $(1,0)$ and $(0,2)$, so $[T]_B=D=\begin{pmatrix}1&0\\0&2\end{pmatrix}$.

   Both inputs are nonzero, so these same equations show that $\mathbf b_1$ and $\mathbf b_2$ are eigenvectors with eigenvalues 1 and 2. The map scales each of those directions.

3. Transform, convert and check

   In $B$-coordinates, $[T(\mathbf v)]_B=D[\mathbf v]_B=\begin{pmatrix}1\\4\end{pmatrix}$. Converting back gives $T(\mathbf v)=\mathbf b_1+4\mathbf b_2=(9,5)$.

   Check directly: $T(5,3)=(15-6,5)=(9,5)$. The standard-coordinate matrix is $A=\begin{pmatrix}3&-2\\1&0\end{pmatrix}$; it acts on $(5,3)$, while $D$ acts on the $B$-coordinate column $(1,2)$.

   The vector has not changed when you replace its standard coordinates $(5,3)$ by its $B$-coordinates $(1,2)$. Applying $T$ changes the vector. Mixing those two coordinate systems is the error to watch for.