2.3

An m*n upper triangular matrix is one whose entries below the main diagonal are​ zeros, as is shown in the matrix to the right. When is a square upper triangular matrix​ invertible? Justify your answer.

A square upper triangular matrix is invertible when all entries on its main diagonal are nonzero. If all of the entries on its main diagonal are​ nonzero, then the nxn matrix has n pivot positions.

If A is an nxn ​matrix, then the equation Ax=b has at least one solution for each b in set of real numbers (ℝn).

False; by the Invertible Matrix Theorem Ax=b has at least one solution for each b in set of real numbers (ℝn) only if a matrix is invertible.

If the linear transformation ​(x​) maps to↦ Ax maps set of real numbers ℝn into set of real numbers ℝn​, then A has n pivot positions.

False; the linear transformation ​(x​) maps to↦ Ax will always map set of real numbers ℝn into set of real numbers ℝn for any n×n matrix. According to the Invertible Matrix Theorem A has n pivot positions only if x(x) ↦ Ax maps set of real numbers ℝn onto set of real numbers ℝn.

Explain why the columns of A^2 span set of real numbers ℝn whenever the columns of an n×n matrix A are linearly independent.

If the columns of A are linearly independent and A is​ square, then A is​ invertible, by the IMT.​ Thus, A^2​, which is the product of invertible​ matrices, is also invertible.​ So, by the​ IMT, the columns of A^2 span set of real numbers ℝn.

If A is​ invertible, then the columns of A^(−1) are linearly independent. Explain why.

It is a known theorem that if A is invertible then A−1 must also be invertible. According to the Invertible Matrix​ Theorem, if a matrix is invertible its columns form a linearly independent set.​ Therefore, the columns of A−1 are linearly independent.

If C is 6×6 matrix and the equation Cx=v is consistent for every v in set of real numbers ℝ6​, is it possible that for some v​, the equation Cx=v has more than one​ solution? Why or why​ not?

It is not possible. Since Cx=v is consistent for every v in set of real numbers ℝ6​, according to the Invertible Matrix Theorem that makes the 6×6 matrix invertible. Since it is​ invertible, Cx=v has a unique solution.

Let A and B be nx n matrices. Show that if AB is invertible so is B.

Let W be the inverse of AB. Then WAB=I and ​(WA)B=I. ​Therefore, matrix B is invertible by part​ (j) of the IMT.

Can a square matrix with two identical columns be​ invertible? Why or why​ not?

The matrix is not invertible. If a matrix has two identical columns then its columns are linearly dependent. According to the Invertible Matrix Theorem this makes the matrix not invertible.

if there is a b in set of real numbers Rn such that the equation Ax=b is​ inconsistent, then the transformation x maps to x↦ Ax is not​ one-to-one.

True; according to the Invertible Matrix Theorem if there is a b in set of real numbers ℝn such that the equation Ax=b is​ inconsistent, then equation Ax=b does not have at least one solution for each b in set of real numbers ℝn and this makes A not invertible.

If the equation Ax=b has at least one solution for each b in set of real numbers ℝn​, then the solution is unique for each b.

True; by the Invertible Matrix Theorem if Ax=b has at least one solution for each b in set of real numbers ℝn​, then matrix A is invertible. If A is​ invertible, then according to the invertible matrix theorem the solution is unique for each b.

If the columns of A are linearly​ independent, then the columns of A span set of real numbers ℝn.

True; by the Invertible Matrix Theorem if the columns of A are linearly​ independent, then the columns of A must span set of real numbers ℝn.

If there is a nxn matrix D such that ADequals=​I, then there is also an nxn matrix C such that CA=I.

True; by the Invertible Matrix​ Theorem, if there is an nxn matrix D such that AD=I, then it must be true that there is also an nxn matrix C such that CA=I.

Is it possible for a 5x5 matrix to be invertible when its columns do not span set of real numbers ℝ5​? Why or why​ not?

it is not​ possible; according to the Invertible Matrix Theorem an n×n matrix cannot be invertible when its columns do not span set of real numbers ℝn.

if A^(T) is not​ invertible, then A is not invertible.

true; by the Invertible Matrix Theorem if A^(T) is not invertible all statements in the theorem are​ false, including A is invertible.​ Therefore, A is not invertible.