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