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Computational Linear Algebra: Notes and Questions for MATH 3325

Nicholas Long

Section 2.1 Matrix Products and Equations

Subsection 2.1.1 Matrix-Vector Products

Definition 2.1.1.

We define a matrix-vector product as follows: If \(A\) is a \(m\) by \(n\) matrix,
\begin{equation*} A=\begin{bmatrix} a_{11} \amp a_{12} \amp ... \amp a_{1n} \\ a_{21}\amp a_{22}\amp ... \amp a_{2n} \\ \vdots \amp \vdots \amp \amp \vdots \\ a_{m1}\amp a_{m2} \amp ... \amp a_{mn} \end{bmatrix} \end{equation*}
and \(\vec{x}=\colvec{x_1\\x_2\\\vdots\\x_n} \in \mathbb{R}^n\text{,}\) then the matrix-vector product is given by
\begin{equation*} A\vec{x} = x_1 \colvec{a_{11}\\a_{21}\\\vdots\\a_{m1}}+x_2 \colvec{a_{12}\\a_{22}\\\vdots\\a_{m2}}+ \cdots x_n \colvec{a_{1n}\\a_{2n}\\\vdots\\a_{mn}} \end{equation*}
\(\mathbb{R}^m\text{.}\)

Investigation 2.1.1.

If \(A\) is a \(m\) by \(n\) matrix, then \(A\vec{x} \in \mathbb{R}^\diamondsuit\) for what value of \(\diamondsuit\text{?}\)
It should not surprise you that you can multiply a scalar multiple of a vector by a matrix by factoring out the scalar. In mathematical notation, \(A (k \vec{v}) = k (A\vec{v})\text{.}\) Additionally, you can apply the scalar multiplication to the matrix. In other words, \(A (k \vec{v}) = k (A\vec{v}) = (kA)\vec{v}\text{.}\) These kind of manipulations will be discussed more when we work with matrix operations later, but you may find these facts useful in your work right now. You should take time to write out the details of any of these arithmetic ideas that you think would be useful in your work.

Activity 2.1.2.

Let
\begin{equation*} A=\begin{bmatrix} 1\amp 3 \amp 4 \amp 5\\2 \amp 1 \amp 0 \amp 1\end{bmatrix} \end{equation*}
\begin{equation*} B=\begin{bmatrix} 1\amp 3 \amp 4 \\ 5 \amp 1 \amp 1 \\ -2 \amp 0 \amp 1\end{bmatrix} \end{equation*}
\begin{equation*} C=\begin{bmatrix} 1\amp 3 \\ 5 \amp 1 \\ -2 \amp 0 \end{bmatrix} \end{equation*}
\begin{equation*} \vec{x} =\colvec{ 4 \\ 2 \\ 2 \\ 1} \end{equation*}
\begin{equation*} \vec{y} =\colvec{ -5 \\ 1 \\ 2} \end{equation*}
\begin{equation*} \vec{z} =\colvec{ -3 \\ 2 } \end{equation*}
(a)
Write out the matrix vector product of \(A\) and \(\vec{x}\) as a linear combination of the columns of \(A\text{.}\)
(b)
Find \(A\vec{x}\)
(c)
Compute all of the other matrix vector products that will be allowed with the matrices \(A,B,C\) and \(\vec{x},\vec{y},\vec{z}\text{.}\)
The matrix \(A\) can be seen from a column vector form as \(A=[ \vec{a}_1 \; \vec{a}_2 \; \ldots \; \vec{a}_n]\) which means that we can think of the product of \(A\) and \(\vec{x}\) as \(A\vec{x}= [ \vec{a}_1 \; \vec{a}_2 \; \ldots \; \vec{a}_n] \colvec{x_1 \\x_2\\ \vdots \\x_n}=x_1\vec{a}_1+x_2 \vec{a}_2 +\ldots+x_n\vec{a}_n\text{.}\)

Investigation 2.1.3.

(a)
Write out the \(k\)-th component of the resulting vector of the product
\begin{equation*} \begin{bmatrix} a_{11} \amp a_{12} \amp ... \amp a_{1n} \\ a_{21}\amp a_{22}\amp ... \amp a_{2n} \\ \vdots \amp \vdots \amp \amp \vdots \\ a_{m1}\amp a_{m2} \amp ... \amp a_{mn} \end{bmatrix} \colvec{x_1\\x_2\\\vdots\\x_n} \end{equation*}
Solution.
\(a_{k1}x_1+a_{k2}x_2+\cdots + a_{kn}x_n\)
(b)
How can you express the result of the matrix-vector product in terms of \(\vec{x}\) and the rows of \(A\text{?}\)
Solution.
The \(k\)-th component of the matrix-vector product is the dot product of row \(k\) of \(A\) with \(\vec{x}\text{.}\)
(c)
How can you express the result of the matrix-vector product in terms of \(\vec{x}\) and the columns of \(A\text{?}\)
Solution.
One way to view this is as a linear combination of the columns of \(A\) with the coefficient on the \(k\)-th column of \(A\) being \(x_k\text{.}\)

Subsection 2.1.2 The Matrix Equation

Based on the above definition of the matrix vector product, if
\begin{equation*} A=\begin{bmatrix} a_{11} \amp a_{12} \amp ... \amp a_{1n} \\ a_{21}\amp a_{22}\amp ... \amp a_{2n} \\ \vdots \amp \vdots \amp \amp \vdots \\ a_{m1}\amp a_{m2} \amp ... \amp a_{mn} \end{bmatrix} \end{equation*}
and \(\vec{b}= \colvec{b_1\\b_2\\\vdots\\b_m}\text{,}\) then by Investigation 1.6.1, \(A\vec{x} = \vec{b}\) has the same solution set as the system
\begin{equation*} \begin{array}{rcrcrcrcr} a_{11} x_1 \amp +\amp a_{12} x_2 \amp +\amp ... \amp +\amp a_{1n}x_n \amp =\amp b_1 \\ % -1 3 a_{21} x_1 \amp +\amp a_{22} x_2 \amp +\amp ... \amp +\amp a_{2n}x_n \amp =\amp b_2 \\ \vdots \amp \amp \vdots \amp \amp \amp \amp \vdots \amp \amp \vdots \\ a_{m1} x_1 \amp +\amp a_{m2} x_2 \amp +\amp ... \amp +\amp a_{mn}x_n \amp =\amp b_m \end{array} \end{equation*}

Investigation 2.1.4.

Write each of the following as a matrix equation, a vector equation, and system of equations. You need to write out the exact corresponding vector equation, matrix equation, and system of equations, not some equivalent form.
(a)
\(\begin{bmatrix} 1 \amp 2\amp 3\\4\amp 5\amp 6\\7\amp 8\amp 9 \end{bmatrix} \colvec{x_1\\x_2\\x_3} =\colvec{a\\b\\c}\)
(b)
\(a_1 \colvec{1\\2\\5}+\colvec{3\\0\\-1}+a_2 \colvec{-1\\0\\2} =\colvec{4\\5\\0}\)
(c)
\begin{align*} 2x_1+3x_2\phantom{+3x_3}\amp =7\\ -x_1+x_2+4x_3\amp =0\\ 5x_1-6x_2-x_3\amp =2 \end{align*}

Subsection 2.1.3 Linear Transformations

Our definition of a matrix vector product suggests that the matrix-vector product 2.1.1 of a \(m\) by \(n\) matrix will transform vectors from \(\mathbb{R}^n\) to vectors in \(\mathbb{R}^m\text{.}\) In this way, we can define a function as follows.

Definition 2.1.2.

Let \(A\) be a \(m\) by \(n\) matrix. Then we define \(T_A : \mathbb{R}^n \rightarrow \mathbb{R}^m\text{,}\) where \(T_A(\vec{x})=A\vec{x}\text{.}\)
Using what we know of solving linear systems, vector equations, and thier relationship to matrix vector products, we note the following:
  • If there is no soltuion to the matrix equation \(A\vec{x}=\vec{b}\text{,}\) then we say that \(\vec{b}\) is NOT in the range of the linear transformation \(T_A\text{.}\)
  • If there is always a solution to the matrix equation \(A\vec{x}=\vec{b}\text{,}\) then we say that the map \(T_A\) completely covers \(\mathbb{R}^m\text{.}\) In other words, the range of \(T_A\) is all of \(\mathbb{R}^m\text{.}\)
  • If whenever a solution exists, the solution is unique, then we say that the map \(T_A\) is one-to-one.
Linear Transformations are very important ways to understand how a vector space is changed under the operation given. Linear transformations are important because they preserve linear combinations. In other words,
\begin{equation*} T_A(a_1\vec{v}_1 +a_2 \vec{v}_2)= a_1 T_A(\vec{v}_1) +a_2 T_A(\vec{v}_2) \end{equation*}

Activity 2.1.6.

Let \(A=\begin{bmatrix} 2 \amp 0 \amp 0 \\0 \amp -1 \amp 1 \end{bmatrix}\text{.}\)
(a)
The linear transformation \(T_A\) takes \(\mathbb{R}^{m}\) to \(\mathbb{R}^{n}\) for what values of \(m\) and \(n\text{?}\)
(b)
Compute \(T_A \left( \colvec{1\\0\\0 } \right)\text{.}\)
(c)
Compute \(T_A \left( \colvec{1\\1\\0} \right)\text{.}\)
(d)
Compute \(T_A \left( \colvec{a\\b\\c} \right)\text{.}\)
(e)
Compute \(T_A\) for any other vectors you might want to and write a few sentences about how vectors are transformed by \(T_A\text{.}\)

Activity 2.1.7.

Let \(B=\begin{bmatrix} 1 \amp 1 \\ 0 \amp -1 \\ 2 \amp 1 \end{bmatrix}\text{.}\)
(a)
The linear transformation \(T_B\) takes \(\mathbb{R}^{m}\) to \(\mathbb{R}^{n}\) for what values of \(m\) and \(n\text{?}\)
(b)
Compute \(T_B \left( \colvec{1\\0 } \right)\text{.}\)
(c)
Compute \(T_B \left( \colvec{1\\1} \right)\text{.}\)
(d)
Compute \(T_B \left( \colvec{a\\b} \right)\text{.}\)
(e)
Compute \(T_B\) for any other vectors you might want to and write a few sentences about how vectors are transformed by \(T_B\text{.}\)
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