Key methods

Finding the Null Space of a Matrix

1. Set up the equation $A\mathbf{x}=0$.
2. Reduce $A$ to its Reduced Row Echelon Form (RREF) using row operations.
3. Identify free variables (columns in RREF without a pivot).
4. Express the solution $\mathbf{x}$ in terms of free variables to find the basis of the null space.

Determining if a Vector is in the Null Space of a Matrix

1. Given matrix $A$ and vector $\mathbf{v}$, set up $A\mathbf{v}=0$.
2. Solve the equation; if it yields the zero vector, then $\mathbf{v}$ is in the null space.

Finding the Column Space of a Matrix

1. Write down matrix $A$.
2. Reduce $A$ to RREF using row operations.
3. Identify pivot columns in RREF.
4. Select corresponding columns from $A$; they form a basis for the column space.

Determining if a Vector is in the Column Space of a Matrix

1. Form the augmented matrix $[A|\mathbf{b}]$ with matrix $A$ and vector $\mathbf{b}$.
2. Reduce to RREF.
3. If the system is consistent (no row of form $[0,0,...,0|1]$), then $\mathbf{b}$ is in the column space.

Finding the Row Space of a Matrix

1. Write down matrix $A$.
2. Reduce $A$ to RREF using row operations.
3. Non-zero rows in RREF form a basis for the row space.

Finding the Rank of a Matrix

1. Reduce the matrix to RREF.
2. Count the number of pivot columns; this number is the rank of $A$.

Finding the Dimension of a Vector Space

1. Identify a basis for the vector space.
2. Count the number of vectors in this basis; this count is the dimension of the vector space.

Finding the Inner Product

1. For vectors $\mathbf{u}$ and $\mathbf{v}$, calculate the sum of the products of their corresponding components, ${\sum }_{i=1}^n{u}_i{v}_i$.

Finding the Vector Length (Magnitude)

1. For vector $\mathbf{v}$, calculate the square root of the sum of the squares of its components, $\sqrt{{\sum }_{i=1}^n{v}_i^2}$.

Finding the Vector Distance

1. The distance between vectors $\mathbf{u}$ and $\mathbf{v}$ is the magnitude of $\mathbf{u}-\mathbf{v}$, calculated as $\sqrt{{\sum }_{i=1}^n(u_i-v_i{)}^2}$.

Finding the Eigenvalues and Eigenvectors

1. Set up the equation $(A-\lambda I)\mathbf{x}=0$, where $\lambda$ is an eigenvalue and $\mathbf{x}$ is an eigenvector.
2. Solve the characteristic equation $\det (A-\lambda I)=0$ for $\lambda$.
3. For each $\lambda$, solve $(A-\lambda I)\mathbf{x}=0$ to find the corresponding eigenvector $\mathbf{x}$.

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Let’s extend the explanation with an example for each concept. Consider the matrix $A$ and vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ as follows:

$A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right),\mathbf{u}=\left(\begin{array}{c}1\\ 0\end{array}\right),\mathbf{v}=\left(\begin{array}{c}2\\ 1\end{array}\right),\mathbf{w}=\left(\begin{array}{c}5\\ 11\end{array}\right)$

Example for Finding the Null Space of a Matrix

1. We set up $A\mathbf{x}=0$.
2. After reducing $A$, the RREF is: $\text{RREF}(A)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$
3. There are no free variables, implying the null space contains only the zero vector, $0$.

Example for Determining if a Vector is in the Null Space of a Matrix

• To check if $\mathbf{u}$ is in the null space, we solve $A\mathbf{u}=0$. Given $A\mathbf{u}=\left(\begin{array}{c}1\\ 3\end{array}\right)\ne 0$, $\mathbf{u}$ is not in the null space.

Example for Finding the Column Space of a Matrix

• The columns of $A$ are already linearly independent. Thus, the column space of $A$ is spanned by its columns: $\left(\begin{array}{c}1\\ 3\end{array}\right)$ and $\left(\begin{array}{c}2\\ 4\end{array}\right)$.

Example for Determining if a Vector is in the Column Space of a Matrix

• To check if $\mathbf{w}$ is in the column space, we set up and solve $A\mathbf{x}=\mathbf{w}$. Since $\mathbf{w}$ can be expressed as a linear combination of the columns of $A$, $\mathbf{w}=1\left(\begin{array}{c}1\\ 3\end{array}\right)+2\left(\begin{array}{c}2\\ 4\end{array}\right)$, it is in the column space. and $\mathbf{v}$ is $\sqrt{(1-2{)}^2+(0-1{)}^2}=\sqrt{2}$.

Example for Finding the Eigenvalues and Eigenvectors

1. Solve $(A-\lambda I)\mathbf{x}=0$ for $A$.
2. The characteristic equation is $\det (A-\lambda I)=0$, leading to ${\lambda }^2-5\lambda +4=0$, with solutions ${\lambda }_1=1$ and ${\lambda }_2=4$.
3. For ${\lambda }_1=1$, solve $(A-I)\mathbf{x}=0$ to find an eigenvector, and similarly for ${\lambda }_2=4$.

This provides a step-by-step example for each concept, illustrating how to apply the theoretical aspects to practical situations.