Linear Algebra Chapter 2

If the nxn matrices E and F have the property that EF=​I, then E and F commute.

According the Invertible Matrix​ Theorem, E and F must be invertible and inverses. So FE=I and I=EF. ​Thus, E and F commute.

If the columns of a 7 times 7 matrix D are linearly​ independent, what can you say about the solutions of Dx=b​?

Equation Dx=b has a solution for each b in set of real numbers R^7. According to the Invertible Matrix​ Theorem, a matrix is invertible if the columns of the matrix form a linearly independent​ set; this would mean that the equation Dx=b has at least one solution for each b in set of real numbers R^n.

If A and B are nxn and​ invertible, then A^-1 B^-1 is the inverse of AB.

False; B^-1 A^-1 is the inverse of AB.

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

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

Explain why the columns of an n x n matrix A are linearly independent when A is invertible

If A is​ invertible, then the equation Ax=0 has the unique solution x=0. Since Ax=0 has only the trivial​ solution, the columns of A must be linearly independent.

Explain why the columns of A^2 span set of real numbers R^n whenever the columns of an nxn 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 R Superscript n

Suppose A is nxn and the equation Ax=b has a solution for each b in set of real numbers R^n. Explain why A must be invertible.

If the equation Ax=b has a solution for each b in set of real numbers R Superscript n​, then A has a pivot position in each row. Since A is​ square, the pivots must be on the diagonal of A. It follows that A is row equivalent to I n. ​Therefore, A is invertible.

If A is​ invertible, then the columns of Upper A^-1 are linearly independent

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 6x6 and the equation Cx=v is consistent for every v in set of real numbers R^6​, is it possible that for some v​, the equation Cx=v has more than one​ solution?

It is not possible. Since Cx=v is consistent for every v in set of real numbers R^6​, according to the Invertible Matrix Theorem that makes the 6x6 matrix invertible. Since it is​ invertible, Cx=v has a unique solution.

Is it possible for a 5x5 matrix to be invertible when its columns do not span set of real numbers R^ 5​?

It is not​ possible; according to the Invertible Matrix Theorem an nxn matrix cannot be invertible when its columns do not span set of real numbers R^n.

Let A and B be nxn 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

Suppose A is n x n and the equation Ax=0 has only the trivial solution. Explain why A has n pivot columns and A is row equivalent to I n.

Suppose A is n x n and the equation Ax=0 has only the trivial solution. Then there are no free variables in this​ equation, thus A has n pivot columns. Since A is square and the n pivot positions must be in different​ rows, the pivots in an echelon form of A must be on the main diagonal. Hence A is row equivalent to the n x n identity​ matrix, I n.

Suppose the last column of AB is entirely zero but B itself has no column of zeros. What can you say about the columns of​ A?

The columns of A are linearly dependent because if the last column in B is denoted b p​, then the last column of AB can be rewritten as Ab p=0. Since b p is not all​ zeros, then any solution to A b p=0 can not be the trivial solution.

If an nxn matrix K cannot be row reduced to I n​, what can you say about the columns of​ K

The columns of K are linearly dependent and the columns do not span set of real numbers R^n. According to the Invertible Matrix​ Theorem, if a matrix cannot be row reduced to I n that matrix is non invertible.

Suppose the first two​ columns, b 1 and b 2​, of B are equal. What can you say about the columns of AB​ (if AB is​ defined)?

The first two columns of AB are Ab 1 and Ab 2. They are equal since b 1 and b 2 are equal.

Can a square matrix with two identical columns be​ invertible?

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.

Each column of AB is a linear combination of the columns of B using weights from the corresponding column of A

The statement is false. The definition of AB states that each column of AB is a linear combination of the columns of A using weights from the corresponding column of B.

The transpose of a product of matrices equals the product of their transposes in the same order.

The statement is false. The transpose of a product of matrices equals the product of their transposes in the reverse order

AB+AC=​A(B+C)

The statement is true. The distributive law for matrices states that ​A(B+​C)=AB+AC.

Suppose H is an nxn matrix. If the equation Hx=c is inconsistent for some c in set of real numbers R^n​, what can you say about the equation Hx=0​

The statement that Hx=c is inconsistent for some c is equivalent to the statement that Hx=c has no solution for some c. From​ this, all of the statements in the Invertible Matrix Theorem are​ false, including the statement that Hx=0 has only the trivial solution.​ Thus, Hx=0 has a nontrivial solution.

In order for a matrix B to be the inverse of​ A, both equations AB=I and BA=I must be true

True, by definition of invertible.

If Upper 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

If the equation Ax=0 has a nontrivial​ solution, then A has fewer than n pivot positions

True; by the Invertible Matrix Theorem if the equation Ax=0 has a nontrivial​ solution, then matrix A is not invertible.​ Therefore, A has fewer than n pivot positions.

If the equation Ax=0 has only the trivial​ solution, then A is row equivalent to the n x n identity matrix.

True; by the Invertible Matrix Theorem if the equation Ax=0 has only the trivial​ solution, then the matrix is invertible.​ Thus, A must also be row equivalent to the n x n identity matrix.

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

True; the Invertible Matrix Theorem states that if the columns of A span set of real numbers R^n​, then matrix A is invertible.​ Therefore, the columns are linearly independent.

Give a formula for (ABx)^T​, where x is a vector and A and B are matrices of appropriate size.

X^T B^T A^T