In exercise 11 and 12, the matrices are all \(n \times n\). Each part of the exercise is an implication of the form “If “statement 1” then “statement 2”.”Mark the implication as True if the truth of “statement 2”always follows whenever “statement 1” happens to be true. An implication is False if there is an instance in which “statement 2” is false but “statement 1” is true. Justify each answer.

a. If the equation \[A{\bf{x}} = {\bf{0}}\] has only the trivial solution, then \(A\) is row equivalent to the \(n \times n\) identity matrix.

b. If the columns of \(A\) span \({\mathbb{R}^n}\), then the columns are linearly independent.

c. If \(A\) is an \(n \times n\) matrix, then the equation \(A{\bf{x}} = {\bf{b}}\) has at least one solution for each \({\bf{b}}\) in \({\mathbb{R}^n}\).

d. If the equation \[A{\bf{x}} = {\bf{0}}\] has a non trivial solution, then \[A\] has fewer than \(n\) pivot positions.

e. If \({A^T}\) is not invertible, then \(A\) is not invertible.

Short Answer

Expert verified

a. True

b. True

c. False

d. True

e. True

Step by step solution

01

Check for statement (a)

If the equation \(A{\bf{x}} = 0\) has only a trivial solution, A is row equivalent to the \(n \times n\) identity matrix.

Since both the statements are true, the implication is true.

02

Check for statement (b)

If the columns of A span \({\mathbb{R}^n}\), then they are linearly independent.

Since both the given statements are true, the implication is true.

03

Check for statement (c)

Statement (c): The matrix A has to be invertible.

The given statement is true if A is invertible.

So, the implication is false.

04

Check for statement (d)

The equation \(A{\bf{x}} = 0\) will have a non-trivial solution if the number of pivots is less than n.

As both the statements are true, the implication is true.

05

Check for statement (e)

Both the given statements are true, i.e., if \({A^T}\) is not invertible, then A is not invertible.

So, the implication is true.

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Most popular questions from this chapter

Explain why the columns of an \(n \times n\) matrix Aspan \({\mathbb{R}^{\bf{n}}}\) when

Ais invertible. (Hint:Review Theorem 4 in Section 1.4.)

Use partitioned matrices to prove by induction that for \(n = 2,3,...\), the \(n \times n\) matrices \(A\) shown below is invertible and \(B\) is its inverse.

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\1&1&0&{}&0\\1&1&1&{}&0\\ \vdots &{}&{}& \ddots &{}\\1&1&1& \ldots &1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&0\\ \vdots &{}& \ddots & \ddots &{}\\0&{}& \ldots &{ - 1}&1\end{array}} \right]\]

For the induction step, assume A and Bare \(\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrices, and partition Aand B in a form similar to that displayed in Exercises 23.

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

7. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&Z\\{\bf{0}}&{\bf{0}}\\B&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

Show that if the columns of Bare linearly dependent, then so are the columns of AB.

Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?

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