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Chapter 2: Problem 46

Let \(e_{i}=[0, \ldots, 0,1,0, \ldots, 0],\) where 1 is the ith entry. Show (a) \(e_{i} A=A_{i},\) ith row of \(A\) (c) If \(e_{i} A=e_{i} B,\) for each \(i,\) then \(A=B\) (b) \( B e_{j}^{T}=B^{j}, j\) th column of \(B\) (d) If \(A e_{j}^{T}=B e_{j}^{T},\) for each \(j,\) then \(A=B\)

Short Answer

Expert verified
In summary, we have established the following relationships: (a) \(e_{i} A=A_{i}\): Multiplying a vector \(e_{i}\) with matrix \(A\) results in the ith row of \(A\). (b) \(B e_{j}^{T}=B^{j}\): Multiplying matrix \(B\) with a vector \(e_j^{T}\) results in the jth column of \(B\). (c) If \(e_{i} A=e_{i} B\) for each \(i\), then \(A=B\): When the ith rows of matrices \(A\) and \(B\) are equal for all \(i\), the matrices are equal. (d) If \(A e_{j}^{T}=B e_{j}^{T}\) for each \(j\), then \(A=B\): When the jth columns of matrices \(A\) and \(B\) are equal for all \(j\), the matrices are equal.

Step by step solution

01

Set up the multiplication

Since \(e_i\) is a row vector and has only 1 in its ith position (all others are 0), when multiplying \(e_i\) by matrix \(A\), only the ith row of \(A\) will contribute to the result. Given the nature of matrix multiplication, all other terms will be canceled out by the 0s in \(e_i\).
02

Perform the multiplication

Multiply \(e_i\) with \(A\), selecting the ith row of \(A\) and summing the products: \( (e_i)_{1} A_{i1} + (e_i)_{2} A_{i2} + \cdots + (e_i)_{i} A_{ii} + \cdots + (e_i)_{n} A_{in} = A_i \). Hence, \(e_{i} A=A_{i}\). (b) Show that \(Be_{j}^{T}=B^{j}\)
03

Set up the multiplication

Since \(e_j^{T}\) is a column vector and has only 1 in its jth position (all others are 0), when multiplying \(B\) by \(e_j^{T}\), only the jth column of \(B\) will contribute to the result. Given the nature of matrix multiplication, all other terms will be canceled out by the 0s in \(e_j^T\).
04

Perform the multiplication

Multiply \(B\) with \(e_j^T\), selecting the jth column of \(B\) and summing the products: \( B_{1j}(e_j^T)_{1} + B_{2j}(e_j^T)_{2} + \cdots + B_{ij}(e_j^T)_{i} + \cdots + B_{nj}(e_j^T)_{n} = B^j \). Hence, \(B e_{j}^{T}=B^{j}\). (c) Show that if \(e_{i}A=e_{i}B\) for each \(i\), then \(A=B\)
05

Utilizing Part (a)

Since we know that \(e_{i}A=A_{i}\) and \(e_{i}B=B_{i}\), if \(e_{i}A=e_{i}B\) for each \(i\), then \(A_{i}=B_{i}\) for each \(i\).
06

Conclude that A=B

Since all the rows of \(A\) are equal to the corresponding rows of \(B\), we can conclude that \(A=B\). (d) Show that if \(A e_{j}^{T} = B e_{j}^{T}\) for each \(j\), then \(A=B\)
07

Utilizing Part (b)

Since we know that \(A e_{j}^{T} = A^j\) and \(B e_{j}^{T} = B^j\), if \(A e_{j}^{T} = B e_{j}^{T}\) for each \(j\), then \(A^{j}=B^{j}\) for each \(j\).
08

Conclude that A=B

Since all the columns of \(A\) are equal to the corresponding columns of \(B\), we can conclude that \(A=B\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Basis Vectors
Standard basis vectors are essential in linear algebra because they help perform simple but significant operations like selecting rows or columns of matrices. A standard basis vector, typically denoted as \( e_i \) in an \( n \)-dimensional space, is a vector with a "1" in its \( i \)-th position and "0s" in all other positions. This means that if you have a 4-dimensional space, \( e_2 \) would be written as \([0, 1, 0, 0]\).

When used in matrix multiplication, the power of a standard basis vector shines through its ability to "pick out" specific parts of a matrix. For instance, if \( e_i \) is multiplied by a matrix \( A \), it selects the \( i \)-th row of \( A \). This happens because the multiplication operates by summing the products of corresponding elements, and since all elements in \( e_i \) except the \( i \)-th are zero, only the \( i \)-th row remains in the result.

In the same vein, when \( e_i^T \) is used on the right side of a matrix, it selects the \( i \)-th column of the matrix it multiplies. This targeted selection tactic is handy for deconstructive analysis of matrices, providing a foundational tool for deeper mathematical discoveries.
Equality of Matrices
Equality of matrices occurs when two matrices are identical in both their dimensions and corresponding elements. Two matrices \( A \) and \( B \) are equal if the number of rows and columns in \( A \) matches those in \( B \), and each element \( A_{ij} \) corresponds exactly to \( B_{ij} \).

Determining matrix equality can involve evaluating every element one by one, but in practice, there are methods to simplify this comparison. For example, if \( e_i A = e_i B \) for each \( i \), as shown in part (c) of our exercise, then each row of matrix \( A \) equals the corresponding row of matrix \( B \), which implies that \( A = B \). Likewise, if \( A e_j^T = B e_j^T \) for each \( j \), each column in \( A \) matches the corresponding column in \( B \).

This method allows checking matrix equality efficiently, especially in systems where matrices are vast, and computations need to be streamlined. Understanding these conditions is vital as it ensures precise matrix manipulations necessary in various applications across science and engineering.
Rows and Columns in Matrices
Matrices are organized in rows and columns which are fundamental to understanding how matrix operations work. The arrangement dictates how matrices interact with other matrices or vectors. Each matrix is structured such that it has a certain number of rows (horizontal lines of elements) and columns (vertical lines of elements).

Consider a matrix \( A \) with dimensions \( m \times n \), meaning it has \( m \) rows and \( n \) columns. The element located in the \( i \)-th row and \( j \)-th column is denoted as \( A_{ij} \). Rows and columns serve critical roles in computation, including initiating operations like addition, subtraction, and particularly multiplication.

In matrix multiplication, rows from one matrix are multiplied by columns of another, following the rule that the number of columns in the first must equal the number of rows in the second. Thus, understanding the distribution of rows and columns helps in composing or deconstructing matrices, essential for solving systems of linear equations, analyzing networks, or transforming geometrical data.

To directly access a specific row or column, standard basis vectors, as previously mentioned, can be employed efficiently. Such operations simplify many computational tasks in linear algebra, enhancing both theoretical exploration and practical applications.

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Most popular questions from this chapter

Compute \(A B\) using block multiplication, where $$ A=\left[\begin{array}{ccc} 1 & 2 & 1 \\ 3 & 4 & 0 \\ 0 & 0 & 2 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{llll} 1 & 2 & 3 & 1 \\ 4 & 5 & 6 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right] $$ Here \(A=\left[\begin{array}{cc}E & F \\ 0_{1 \times 2} & G\end{array}\right]\) and \(B=\left[\begin{array}{cc}R & S \\ 0_{1 \times 3} & T\end{array}\right]\), where \(E, F, G, R, S, T\) are the given blocks, and \(0_{1 \times 2}\) and \(0_{1 \times 3}\) are zero matrices of the indicated sites. Hence, $$ A B=\left[\begin{array}{cc} E R & E S+F T \\ 0_{1 \times 3} & G T \end{array}\right]=\left[\left[\begin{array}{rrr} 9 & 12 & 15 \\ 19 & 26 & 33 \end{array}\right]{\left[\begin{array}{l} 3 \\ 7 \end{array}\right]+\left[\begin{array}{l} 1 \\ 0 \end{array}\right]} \\ {\left[\begin{array}{lll} 0 & 0 & 0 \end{array}\right]} &{2} \end{array}\right]=\left[\begin{array}{rrrr} 9 & 12 & 15 & 4 \\ 19 & 26 & 33 & 7 \\ 0 & 0 & 0 & 2 \end{array}\right] $$

Let \(A=\left[\begin{array}{rr}1 & 3 \\ 2 & -1\end{array}\right]\) and \(B=\left[\begin{array}{rrr}2 & 0 & -4 \\ 3 & -2 & 6\end{array}\right]\). Find: (a) \(A B\), (b) \(B A\) (a) Because \(A\) is a \(2 \times 2\) matrix and \(B\) a \(2 \times 3\) matrix, the product \(A B\) is defined and is a \(2 \times 3\) matrix. To obtain the entries in the first row of \(A B\), multiply the first row \([1,3]\) of \(A\) by the columns \(\left[\begin{array}{l}2 \\\ 3\end{array}\right],\left[\begin{array}{r}0 \\\ -2\end{array}\right],\left[\begin{array}{r}-4 \\ 6\end{array}\right]\) of \(B\), respectively, as follows: $$ A B=\left[\begin{array}{rr} 1 & 3 \\ 2 & -1 \end{array}\right]\left[\begin{array}{rrr} 2 & 0 & -4 \\ 3 & -2 & 6 \end{array}\right]=\left[\begin{array}{lll} 2+9 & 0-6 & -4+18 \end{array}\right]=\left[\begin{array}{lll} 11 & -6 & 14 \end{array}\right] $$ To obtain the entries in the second row of \(A B\), multiply the second row \([2,-1]\) of \(A\) by the columns of \(B\) : $$ A B=\left[\begin{array}{rr} 1 & 3 \\ 2 & -1 \end{array}\right]\left[\begin{array}{rrr} 2 & 0 & -4 \\ 3 & -2 & 6 \end{array}\right]=\left[\begin{array}{ccc} 11 & -6 & 14 \\ 4-3 & 0+2 & -8-6 \end{array}\right] $$ Thus, $$ A B=\left[\begin{array}{rrr} 11 & -6 & 14 \\ 1 & 2 & -14 \end{array}\right] $$ (b) The size of \(B\) is \(2 \times 3\) and that of \(A\) is \(2 \times 2\). The inner numbers 3 and 2 are not equal; hence, the product \(B A\) is not defined.

Prove Theorem \(2.1\) (i) and (v): (i) \((A+B)+C=A+(B+C)\), (v) \(k(A+B)=k A+k B\). Suppose \(A=\left[a_{i j}\right], B=\left[b_{i j}\right], C=\left[c_{i j}\right]\). The proof reduces to showing that corresponding \(i j\)-entries in each side of each matrix equation are equal. [We prove only (i) and (v), because the other parts of Theorem \(2.1\) are proved similarly.] (i) The ij-entry of \(A+B\) is \(a_{i j}+b_{i j}\); hence, the ij-entry of \((A+B)+C\) is \(\left(a_{i j}+b_{i j}\right)+c_{i j}\). On the other hand, the \(i j\)-entry of \(B+C\) is \(b_{i j}+c_{i j} ;\) hence, the ij-entry of \(A+(B+C)\) is \(a_{i j}+\left(b_{i j}+c_{i j}\right) .\) However, for scalars in \(K\) $$ \left(a_{i j}+b_{i j}\right)+c_{i j}=a_{i j}+\left(b_{i j}+c_{i j}\right) $$ Thus, \((A+B)+C\) and \(A+(B+C)\) have identical \(i j\)-entries. Therefore, \((A+B)+C=A+(B+C)\). (v) The ij-entry of \(A+B\) is \(a_{i j}+b_{i j}\); hence, \(k\left(a_{i j}+b_{i j}\right)\) is the \(i j\)-entry of \(k(A+B)\). On the other hand, the ijentries of \(k A\) and \(k B\) are \(k a_{i j}\) and \(k b_{i j}\), respectively. Thus, \(k a_{i j}+k b_{i j}\) is the ij-entry of \(k A+k B\). However, for scalars in \(K\), $$ k\left(a_{i j}+b_{i j}\right)=k a_{i j}+k b_{i j} $$ Thus, \(k(A+B)\) and \(k A+k B\) have identical \(i\) j-entries. Therefore, \(k(A+B)=k A+k B\).

Prove Theorem 2.3: (i) \((A+B)^{T}=A^{T}+B^{T}\), (ii) \(\left(A^{T}\right)^{T}=A\), (iii) \((k A)^{T}=k A^{T}\).

Show (a) If \(A\) has a zero row, then \(A B\) has a zero row. (b) If \(B\) has a zero column, then \(A B\) has a zero column.
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