Basic Notes for Matrices

• Matrix Multiplication: The meaning of $Ax$ (range of $A$) and $Ax=b$.

Most students, after finishing the course linear algebra, may not understand the matrix multiplication yet. Here I will show the readers, roughly speaking, the real and the most significant meaning of matrix multiplication.

Denotes a matrix $A$ in it's columns by $\left[ {\begin{array}{*{20}{c}}
{{A_1}}&{{A_2}}& \cdots &{{A_n}}
\end{array}} \right]$, where ${A_i}$ is the ith column of $A$. A vector $x = {\left[ {\begin{array}{*{20}{c}}
{{x_1}}&{{x_2}}& \cdots &{{x_n}}
\end{array}} \right]^T}$, where ${x_i} \in R$. Then \[Ax = \left[ {\begin{array}{*{20}{c}}
{{A_1}}&{{A_2}}& \cdots &{{A_n}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{x_1}}\\
{{x_2}}\\
 \vdots \\
{{x_n}}
\end{array}} \right] = {x_1}{A_1} + {x_2}{A_2} +  \cdots  + {x_n}{A_n} = \sum\limits_{i = 1}^n {{x_i}{A_i}} \]

which implies that the result of $Ax$ is a linear combination of the columns of $A$ with corresponding coefficients, the components of the vector $x$. That evidently indicate that the range of $A$, $\left\{ {v:v = Ax,x \in {R^n}} \right\} \buildrel \Delta \over = {\rm{range}}\left( A \right)$, is spanned by the columns of $A$. That is, if $v \in {\rm{range}}\left( A \right)$, then $v = {a_1}{A_1} + {a_2}{A_2} +  \cdots  + {a_n}{A_n}$ for some ${a_i}$. This may be the most significant for matrix multiplication. Since for matrix $B = \left[ {\begin{array}{*{20}{c}}
{{B_1}}&{{B_2}}& \cdots &{{B_n}}
\end{array}} \right]$ (where ${B_i}$ is the ith column of $B$), $AB = A\left[ {\begin{array}{*{20}{c}}
{{B_1}}&{{B_2}}& \cdots &{{B_n}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{A{B_1}}&{A{B_2}}& \cdots &{A{B_n}}
\end{array}} \right]$, that matrix multiply matrix may be reduced to that matrix multiply several related vectors.

　　1. Simplifying computation.

Que. Suppose $A = \left[ {\begin{array}{*{20}{c}}
1&2\\
{ - 1}&4\\
0&7
\end{array}} \right]$ and $x = \left[ {\begin{array}{*{20}{c}}
2\\
1
\end{array}} \right]$. Find $Ax$.

Ans. If we use the original algorithm, we cann't get the result until 6 times computation. However, if we use the algorithm above, we may complete the computation whthin 3 steps, \[Ax = \left[ {\begin{array}{*{20}{c}}
1&2\\
{ - 1}&4\\
0&7
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2\\
1
\end{array}} \right] = 2\left[ {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
0
\end{array}} \right] + 1\left[ {\begin{array}{*{20}{c}}
2\\
4\\
7
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2\\
{ - 2}\\
0
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
2\\
4\\
7
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
4\\
2\\
7
\end{array}} \right].\]

　　2. Helping to understand some concepts

The rank of a matrix $A$ is the size of the largest collection of linearly independent columns of $A$ (the column rank) or the size of the largest collection of linearly independent rows of $A$ (the row rank). For every matrix, the column rank is equal to the row rank.

The column space $C(A)$ of a matrix $A$ (sometimes called the range of a matrix) is the set of all possible linear combinations of its column vectors.

Moreover, it's obvious that the column rank of $A$ is the dimension of the column space of $A$.

It's well-known that if ${\rm{rank}}\left( A \right) = {\rm{rank}}\left( {A,b} \right)$ then the linear equations $Ax = b$ is consistent, which means that this equation has at least one solution. Now we will explain this proposition by the note above.

Since $Ax = b$ is consistent, there exists ${x_0} = {\left[ {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}& \cdots &{{a_n}}
\end{array}} \right]^T}$ such that \[Ax = \left[ {\begin{array}{*{20}{c}}
{{A_1}}&{{A_2}}& \cdots &{{A_n}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{a_0}}\\
{{a_1}}\\
 \vdots \\
{{a_n}}
\end{array}} \right] = {a_1}{A_1} + {a_2}{A_2} +  \cdots  + {a_n}{A_n} = b.\]

That indicates that the vector $b$ at the right hand side is in the column space of $A$, i.e. it can be represented by the linear combination of the columns of $A$. That is, $b$ is linearly dependent on the columns of $A$. Thus the size of the largest collection of linearly independent columns of $(A,b)$ is equal to that of $A$, which immediately gets that ${\rm{rank}}\left( {A,b} \right) = {\rm{rank}}\left( A \right)$ by the definition of rank.

• Matrix with rank 1.

A matrix, say $A$, with rank 1 is that the matrix has only one nozero linearly independent column which other columns depends on it. Suppose ${\rm{rank}}\left( A \right) = 1$ and ${A_1} \ne 0$ (this $0$ is the zero vector). Then we have ${A_i} = {a_i}{A_1}$ for $i = 2, \cdots ,n$. Thus \[A = \left[ {\begin{array}{*{20}{c}}
{{A_1}}&{{A_2}}& \cdots &{{A_n}}
\end{array}} \right] = {A_1}\left[ {\begin{array}{*{20}{c}}
1&{{a_2}}& \cdots &{{a_n}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{A_{11}}}\\
{{A_{21}}}\\
 \vdots \\
{{A_{n1}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{{a_2}}& \cdots &{{a_n}}
\end{array}} \right]\]

where ${{A_{i1}}}$ is the i1th component of the matrix $A$. We easily see that matrices with rank 1 can be represented by the multiplication of a column vector with a row vector.

Now let us discuss the eigenvectors and eigenvalues of these matrices. Multiplying the vector ${\left[ {\begin{array}{*{20}{c}}
{{A_{11}}}&{{A_{21}}}& \cdots &{{A_{n1}}}
\end{array}} \right]^T}$ on the right of the matrix $A$, we have \[\begin{array}{l}
A\left[ {\begin{array}{*{20}{c}}
{{A_{11}}}\\
{{A_{21}}}\\
 \vdots \\
{{A_{n1}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{A_{11}}}\\
{{A_{21}}}\\
 \vdots \\
{{A_{n1}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}& \cdots &{{a_n}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{A_{11}}}\\
{{A_{21}}}\\
 \vdots \\
{{A_{n1}}}
\end{array}} \right]\\
{\rm{            }} ~~~~~~~~~~~~~~~~~= \left[ {\begin{array}{*{20}{c}}
{{A_{11}}}\\
{{A_{21}}}\\
 \vdots \\
{{A_{n1}}}
\end{array}} \right]\left( {\left[ {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}& \cdots &{{a_n}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{A_{11}}}\\
{{A_{21}}}\\
 \vdots \\
{{A_{n1}}}
\end{array}} \right]} \right) = \mathop {\underline {\left( {\sum\limits_{i = 1}^n {{a_i}{A_{i1}}} } \right)} }\limits_{\mathop \lambda \limits^\parallel  } \left[ {\begin{array}{*{20}{c}}
{{A_{11}}}\\
{{A_{21}}}\\
 \vdots \\
{{A_{n1}}}
\end{array}} \right]
\end{array}\]

which implies that one eigenvector of $A$ is $\left[ {\begin{array}{*{20}{c}}
{{A_{11}}}\\
{{A_{21}}}\\
 \vdots \\
{{A_{n1}}}
\end{array}} \right]$ and one eigenvalue of $A$ is ${\lambda  = \sum\limits_{i = 1}^n {{a_i}{A_{i1}}} }$, which is the trace of $A$, where ${\rm{trace}}\left( A \right) = \sum\limits_{i = 1}^n {{A_{ii}}} $. Since ${\rm{rank}}\left( A \right) = 1$, the other eigenvalues of $A$ are all zeros.

Ex. Consider the matrix $A = \left[ {\begin{array}{*{20}{c}}
1&3&2\\
{ - 1}&{ - 3}&{ - 2}\\
2&6&4
\end{array}} \right]$. Then it's easy to see that $A = \left[ {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&3&2
\end{array}} \right]$. The uniquely nonzero eigenvalue of $A$ is $\left[ {\begin{array}{*{20}{c}}
1&3&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
2
\end{array}} \right] = 1 + \left( { - 3} \right) + 4 = 2 = {\rm{trace}}\left( A \right)$. And the corresponding eigenvector is $\left[ {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
2
\end{array}} \right]$.

转载于:https://www.cnblogs.com/aujun/p/3895992.html