Let $T: \IR^2 \rightarrow \IR^3$ be given by

1. $\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}$

2. $\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}$

3. $\displaystyle \IR^2=\setBuilder{\left[\begin{array}{c}x \\ y\end{array}\right]}{x,y\in\IR}$

Let $T: V \rightarrow W$ be a linear transformation. The kernel of $T$ is an important subspace of $V$ defined by

Let $T: \IR^3 \rightarrow \IR^2$ be given by

1. $\displaystyle \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR}$

2. $\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\\ 0\end{array}\right]}{a\in\IR}$

3. $\displaystyle \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]}$

4. $\displaystyle \IR^3=\setBuilder{\left[\begin{array}{c}x \\ y\\z\end{array}\right]}{x,y,z\in\IR}$

Let $T: \IR^3 \rightarrow \IR^2$ be the linear transformation given by the standard matrix

Part 1.

Set $T\left(\left[\begin{array}{c}x\\y\\z\end{array}\right]\right) = \left[\begin{array}{c}0\\0\end{array}\right]$ to find a linear system of equations whose solution set is the kernel.

Let $T: \IR^3 \rightarrow \IR^2$ be the linear transformation given by the standard matrix

Part 2.

Use $\RREF(A)$ to solve this homogeneous system of equations and find a basis for the kernel of $T\text{.}$

Let $T: \IR^4 \rightarrow \IR^3$ be the linear transformation given by

Find a basis for the kernel of $T\text{.}$

Let $T: \IR^2 \rightarrow \IR^3$ be given by

1. $\displaystyle \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR}$

2. $\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\\ 0\end{array}\right]}{a,b\in\IR}$

3. $\displaystyle \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]}$

4. $\displaystyle \IR^3=\setBuilder{\left[\begin{array}{c}x \\ y\\z\end{array}\right]}{x,y,z\in\IR}$

Let $T: V \rightarrow W$ be a linear transformation. The image of $T$ is an important subspace of $W$ defined by

In the examples below, the left example's image is all of $\IR^2\text{,}$ but the right example's image is a planar subspace of $\IR^3\text{.}$

Let $T: \IR^3 \rightarrow \IR^2$ be given by

1. $\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}$

2. $\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}$

3. $\displaystyle \IR^2=\setBuilder{\left[\begin{array}{c}x \\ y\end{array}\right]}{x,y\in\IR}$

Let $T: \IR^4 \rightarrow \IR^3$ be the linear transformation given by the standard matrix

Since for a vector $\vec v =\left[\begin{array}{c}x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] \text{,}$ $T(\vec v)=T(x_1\vec e_1+x_2\vec e_2+x_3\vec e_3+x_4\vec e_4)\text{,}$ which of the following best describes the set of vectors

1. The set of vectors spans $\Im T$ but is not linearly independent.

2. The set of vectors is a linearly independent subset of $\Im T$ but does not span $\Im T\text{.}$

3. The set of vectors is linearly independent and spans $\Im T\text{;}$ that is, the set of vectors is a basis for $\Im T\text{.}$

Let $T: \IR^4 \rightarrow \IR^3$ be the linear transformation given by the standard matrix

Since the set $\setList{ \left[\begin{array}{c}3\\-1\\2\end{array}\right], \left[\begin{array}{c}4\\1\\1\end{array}\right], \left[\begin{array}{c}7\\0\\3\end{array}\right], \left[\begin{array}{c}1\\2\\-1\end{array}\right] }$ spans $\Im T\text{,}$ we can obtain a basis for $\Im T$ by finding $\RREF A = \left[\begin{array}{cccc} 1 & 0 & 1 & -1\\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]$ and only using the vectors corresponding to pivot columns:

Let $T:\IR^n\to\IR^m$ be a linear transformation with standard matrix $A\text{.}$

• The kernel of $T$ is the solution set of the homogeneous system given by the augmented matrix $\left[\begin{array}{c|c}A&\vec 0\end{array}\right]\text{.}$ Use the coefficients of its free variables to get a basis for the kernel.

• The image of $T$ is the span of the columns of $A\text{.}$ Remove the vectors creating non-pivot columns in $\RREF A$ to get a basis for the image.

Let $T: \IR^3 \rightarrow \IR^4$ be the linear transformation given by the standard matrix

Find a basis for the kernel and a basis for the image of $T\text{.}$

Let $T: \IR^n \rightarrow \IR^m$ be a linear transformation with standard matrix $A\text{.}$ Which of the following is equal to the dimension of the kernel of $T\text{?}$

1. The number of pivot columns

2. The number of non-pivot columns

3. The number of pivot rows

4. The number of non-pivot rows

Let $T: \IR^n \rightarrow \IR^m$ be a linear transformation with standard matrix $A\text{.}$ Which of the following is equal to the dimension of the image of $T\text{?}$

1. The number of pivot columns

2. The number of non-pivot columns

3. The number of pivot rows

4. The number of non-pivot rows

Combining these with the observation that the number of columns is the dimension of the domain of $T\text{,}$ we have the rank-nullity theorem:

The dimension of the domain of $T$ equals $\dim(\ker T)+\dim(\Im T)\text{.}$

The dimension of the image is called the rank of $T$ (or $A$) and the dimension of the kernel is called the nullity.

Let $T: \IR^3 \rightarrow \IR^4$ be the linear transformation given by the standard matrix