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Chapter 2: Problem 49

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.

Short Answer

Expert verified
In summary, (a) if matrix \(A\) has a zero row, then the product matrix \(AB\) also has a zero row, since the dot product of the zero row in \(A\) with any column in \(B\) will result in a zero row; and (b) if matrix \(B\) has a zero column, then the product matrix \(AB\) also has a zero column, because the dot product of the zero column in \(B\) with any row in \(A\) will result in a zero column.

Step by step solution

01

a. If \(A\) has a zero row, then \(AB\) has a zero row.

Assume \(A\) has a zero row. Then there exists a row in matrix \(A\) with all elements equal to 0. Let the zero row be the \(i\)th row of \(A\). Now, we will find the \(i\)th row of the product matrix \(AB\). During matrix multiplication, to find the element in the \(i\)th row and \(j\)th column of the resulting matrix, we calculate the dot product of the \(i\)th row of the first matrix (\(A\)) with the \(j\)th column of the second matrix (\(B\)). Since all the elements of the \(i\)th row of matrix A are 0, the dot product of this row with any column of matrix \(B\) will be 0. Therefore, the \(i\)th row of \(AB\) will have all elements equal to 0, and hence, \(AB\) indeed has a zero row.
02

b. If \(B\) has a zero column, then \(AB\) has a zero column.

Assume \(B\) has a zero column. Then there exists a column in matrix \(B\) with all elements equal to 0. Let the zero column be the \(j\)th column of \(B\). Now, we will find the \(j\)th column of the product matrix \(AB\). During matrix multiplication, to find the element in the \(i\)th row and \(j\)th column of the resulting matrix, we calculate the dot product of the \(i\)th row of the first matrix (\(A\)) with the \(j\)th column of the second matrix (\(B\)). Since all the elements of the \(j\)th column of matrix \(B\) are 0, the dot product of this column with any row of matrix \(A\) will be 0. Therefore, the \(j\)th column of \(AB\) will have all elements equal to 0, and hence, \(AB\) indeed has a zero column.

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Most popular questions from this chapter

Prove that "is similar to" is an equivalence relation on $\mathrm{M}_{n \times n}(F)$.

Prove that if \(\\{x, y\\}\) is a basis for a vector space over \(C\), then so is $$ \left\\{\frac{1}{2}(x+y), \frac{1}{2 i}(x-y)\right\\} . $$

Let \(V\) be the solution space of an \(n\) th-order homogeneous linear differential equation with constant coefficients. Fix \(t_{0} \in R\), and define a mapping \(\Phi: V \rightarrow \mathrm{C}^{n}\) by $$ \Phi(x)=\left(\begin{array}{c} x\left(t_{0}\right) \\ x^{\prime}\left(t_{0}\right) \\ \vdots \\ x^{(n-1)}\left(t_{0}\right) \end{array}\right) \quad \text { for each } x \in \mathrm{V} . $$ (a) Prove that \(\Phi\) is linear and its null space is the zero subspace of \(\mathrm{V}\). Deduce that \(\Phi\) is an isomorphism. Hint: Use Exercise 14 . (b) Prove the following: For any \(n\) th-order homogeneous linear differential equation with constant coefficients, any \(t_{0} \in R\), and any complex numbers \(c_{0}, c_{1}, \ldots, c_{n-1}\) (not necessarily distinct), there exists exactly one solution, \(x\), to the given differential equation such that \(x\left(t_{0}\right)=c_{0}\) and \(x^{(k)}\left(t_{0}\right)=c_{k}\) for $k=1,2, \ldots, n-1$.

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