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Chapter 1: Problem 14

Let \(A\) be an \(m \times n\) matrix. Explain why the matrix multiplications \(A^{T} A\) and \(A A^{T}\) are possible.

Short Answer

Expert verified
The matrix multiplications \(A^T A\) and \(A A^T\) are possible for an \(m \times n\) matrix A because the number of columns in the first matrix is equal to the number of rows in the second matrix for both cases. For the multiplication \(A^T A\), the dimensions are \((n \times m)(m \times n)\), resulting in an \(n \times n\) product. For the multiplication \(A A^T\), the dimensions are \((m \times n)(n \times m)\), resulting in an \(m \times m\) product.

Step by step solution

01

Find dimensions of \(A^T\) matrix

For an \(m 脳 n\) matrix A, the transpose of A, denoted by \(A^T\) has dimensions \(n 脳 m\). This is done by interchanging the rows and columns of the original matrix.
02

Check the condition for matrix multiplication of \(A^T A\)

Now, let's check the possibility of multiplying \(A^T A\): We have \(A^T\) with dimensions \((n 脳 m)\), and matrix A with dimensions \((m 脳 n)\). The number of columns in \(A^T\) is equal to the number of rows in A, which is m. Since this condition is satisfied, matrix multiplication of \(A^T A\) is possible.
03

Find dimensions of the product matrix \(A^T A\)

Since the inner dimensions of \(A^T A\) match, we can now find the dimensions of the product matrix. Here, the dimensions of the resulting matrix will be equal to the outer dimensions after multiplication - this is \((n 脳 m) 脳 (m 脳 n)\). The product will have dimensions \((n 脳 n)\).
04

Check the condition for matrix multiplication of \(A A^T\)

Now, let's check the possibility of multiplying \(A A^T\): We have matrix A with dimensions \((m 脳 n)\) and \(A^T\) with dimensions \((n 脳 m)\). The number of columns in A is equal to the number of rows in \(A^T\), which is n. Since this condition is satisfied, matrix multiplication of \(A A^T\) is possible.
05

Find dimensions of the product matrix \(A A^T\)

Since the inner dimensions of \(A A^T\) match, we can now find the dimensions of the product matrix. Here, the dimensions of the resulting matrix will be equal to the outer dimensions after multiplication - this is \((m 脳 n) 脳 (n 脳 m)\). The product will have dimensions \((m 脳 m)\). In conclusion, both matrix multiplications \(A^T A\) and \(A A^T\) are possible as the conditions for matrix multiplication are satisfied in each case. The resulting matrices will have dimensions \((n 脳 n)\) and \((m 脳 m)\) respectively.

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Most popular questions from this chapter

Consider the matrix \\[ A=\left(\begin{array}{lllll} 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 \end{array}\right) \\] (a) Draw a graph that has \(A\) as its adjacency matrix. Be sure to label the vertices of the graph. (b) By inspecting the graph, determine the number of walks of length 2 from \(V_{2}\) to \(V_{3}\) and from \(V_{2}\) to \(V_{5}\) (c) Compute the second row of \(A^{3},\) and use it to determine the number of walks of length 3 from \(V_{2}\) to \(V_{3}\) and from \(V_{2}\) to \(V_{5}\).

Given the linear systems $$\begin{aligned} &\text { (a) } \quad x_{1}+2 x_{2}+x_{3}=2\\\ &\begin{array}{l} -x_{1}-x_{2}+2 x_{3}=3 \\ 2 x_{1}+3 x_{2}=0 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (b) } \quad x_{1}+2 x_{2}+x_{3}=-1\\\ &\begin{array}{l} -x_{1}-x_{2}+2 x_{3}=2 \\ 2 x_{1}+3 x_{2}=-2 \end{array} \end{aligned}$$ solve both systems by computing the row echelon form of an augmented matrix \((A | B)\) and performing back substitution twice.

In general, matrix multiplication is not commutative (i.e., \(A B \neq B A\) ). However, in certain special cases the commutative property does hold. Show that (a) if \(D_{1}\) and \(D_{2}\) are \(n \times n\) diagonal matrices, then \(D_{1} D_{2}=D_{2} D_{1}\) (b) if \(A\) is an \(n \times n\) matrix and $$B=a_{0} I+a_{1} A+a_{2} A^{2}+\cdots+a_{k} A^{k}$$ where \(a_{0}, a_{1}, \ldots, a_{k}\) are scalars, then \(A B=B A\)

An \(n \times n\) matrix \(A\) is said to be an involution if \(A^{2}=I .\) Show that if \(G\) is any matrix of the form \\[ G=\left(\begin{array}{rr} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array}\right) \\] then \(G\) is an involution.

For each of the pairs of matrices that follow, determine whether it is possible to multiply the first matrix times the second. If it is possible, perform the multiplication. (a) \(\left[\begin{array}{rrr}3 & 5 & 1 \\ -2 & 0 & 2\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 1 & 3 \\ 4 & 1\end{array}\right]\) (b) \(\left[\begin{array}{cc}4 & -2 \\ 6 & -4 \\ 8 & -6\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]\) (c) \(\left[\begin{array}{lll}1 & 4 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 2\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 1 & 1 \\ 4 & 5\end{array}\right]\) (d) \(\left[\begin{array}{ll}4 & 6 \\ 2 & 1\end{array}\right]\left[\begin{array}{lll}3 & 1 & 5 \\ 4 & 1 & 6\end{array}\right]\) (e) \(\left[\begin{array}{lll}4 & 6 & 1 \\ 2 & 1 & 1\end{array}\right]\left[\begin{array}{lll}3 & 1 & 5 \\ 4 & 1 & 6\end{array}\right]\) (f) \(\left[\begin{array}{r}2 \\ -1 \\\ 3\end{array}\right]\left[\begin{array}{llll}3 & 2 & 4 & 5\end{array}\right]\)
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