Matrix Representation of a Linear Transformation of the Vector Space $R^2$ to $R^2$
Problem 255
Let $B=\{\mathbf{v}_1, \mathbf{v}_2 \}$ be a basis for the vector space $\R^2$, and let $T:\R^2 \to \R^2$ be a linear transformation such that
\[T(\mathbf{v}_1)=\begin{bmatrix}
1 \\
-2
\end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix}
3 \\
1
\end{bmatrix}.\]
If $\mathbf{e}_1=\mathbf{v}_1+2\mathbf{v}_2 \text{ and } \mathbf{e}_2=2\mathbf{v}_1-\mathbf{u}_2$, where $\mathbf{e}_1, \mathbf{e}_2$ are the standard unit vectors in $\R^2$, then find the matrix of $T$ with respect to the basis $\{\mathbf{e}_1, \mathbf{e}_2\}$.
The matrix representation $A$ of the linear transformation $T$ with respect to the basis $\{\mathbf{e}_1, \mathbf{e}_2 \}$ is given by
\[A=\begin{bmatrix}
T(\mathbf{e}_1) & T(\mathbf{e}_2)
\end{bmatrix}. \tag{*}\]
To find the vectors $T(\mathbf{e}_1), T(\mathbf{e}_2)$ we use the linearity of $T$.
We have
\begin{align*}
T(\mathbf{e}_1)&=T(\mathbf{v}_1+2\mathbf{v}_2)\\
&=T(\mathbf{v}_1)+2T(\mathbf{v}_2) \qquad \text{ (by the linearity of $T$)}\\
&=\begin{bmatrix}
1 \\
-2
\end{bmatrix} +2\begin{bmatrix}
3 \\
1
\end{bmatrix}\\
&=\begin{bmatrix}
7 \\
0
\end{bmatrix}.
\end{align*}
Also we have
\begin{align*}
T(\mathbf{e}_2)&=T(2\mathbf{v}_1-\mathbf{u}_2)\\
&=2T(\mathbf{v}_1)-T(\mathbf{u}_2)\\
&=2\begin{bmatrix}
1 \\
-2
\end{bmatrix}-\begin{bmatrix}
3 \\
1
\end{bmatrix}\\
&=\begin{bmatrix}
-1 \\
-5
\end{bmatrix}.
\end{align*}
Therefore the matrix $A=\begin{bmatrix}
T(\mathbf{e}_1) & T(\mathbf{e}_2)
\end{bmatrix}$ of the linear transformation $T$ is given by
\[A=\begin{bmatrix}
7 & -1\\
0& -5
\end{bmatrix}.\]
Give a Formula For a Linear Transformation From $\R^2$ to $\R^3$
Let $\{\mathbf{v}_1, \mathbf{v}_2\}$ be a basis of the vector space $\R^2$, where
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
1
\end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.\]
The action of a linear transformation $T:\R^2\to \R^3$ on the […]
Determine linear transformation using matrix representation
Let $T$ be the linear transformation from the $3$-dimensional vector space $\R^3$ to $\R^3$ itself satisfying the following relations.
\begin{align*}
T\left(\, \begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix} \,\right)
=\begin{bmatrix}
1 \\
0 \\
1 […]
Give the Formula for a Linear Transformation from $\R^3$ to $\R^2$
Let $T: \R^3 \to \R^2$ be a linear transformation such that
\[T(\mathbf{e}_1)=\begin{bmatrix}
1 \\
4
\end{bmatrix}, T(\mathbf{e}_2)=\begin{bmatrix}
2 \\
5
\end{bmatrix}, T(\mathbf{e}_3)=\begin{bmatrix}
3 \\
6 […]
Give a Formula for a Linear Transformation if the Values on Basis Vectors are Known
Let $T: \R^2 \to \R^2$ be a linear transformation.
Let
\[
\mathbf{u}=\begin{bmatrix}
1 \\
2
\end{bmatrix}, \mathbf{v}=\begin{bmatrix}
3 \\
5
\end{bmatrix}\]
be 2-dimensional vectors.
Suppose that
\begin{align*}
T(\mathbf{u})&=T\left( \begin{bmatrix}
1 \\
[…]
A Matrix Representation of a Linear Transformation and Related Subspaces
Let $T:\R^4 \to \R^3$ be a linear transformation defined by
\[ T\left (\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \,\right) = \begin{bmatrix}
x_1+2x_2+3x_3-x_4 \\
3x_1+5x_2+8x_3-2x_4 \\
x_1+x_2+2x_3
\end{bmatrix}.\]
(a) Find a matrix $A$ such that […]
Prove a Given Subset is a Subspace and Find a Basis and Dimension
Let
\[A=\begin{bmatrix}
4 & 1\\
3& 2
\end{bmatrix}\]
and consider the following subset $V$ of the 2-dimensional vector space $\R^2$.
\[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\]
(a) Prove that the subset $V$ is a subspace of $\R^2$.
(b) Find a basis for […]
Vector Space of Polynomials and Coordinate Vectors
Let $P_2$ be the vector space of all polynomials of degree two or less.
Consider the subset in $P_2$
\[Q=\{ p_1(x), p_2(x), p_3(x), p_4(x)\},\]
where
\begin{align*}
&p_1(x)=x^2+2x+1, &p_2(x)=2x^2+3x+1, \\
&p_3(x)=2x^2, &p_4(x)=2x^2+x+1.
\end{align*}
(a) Use the basis […]
Let $\mathbf{a}$ and $\mathbf{b}$ be vectors in $\R^n$ such that their length are \[\|\mathbf{a}\|=\|\mathbf{b}\|=1\] and the inner product \[\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.\]...