# Linear algebra identity matrix

1. Feb 9, 2014

### ME_student

1. The problem statement, all variables and given/known data

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

B. If the columns of A span R^n, the columns are linearly independent.

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

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

E. If A^T is not invertible, then A isn't invertible?

2. Relevant equations

3. The attempt at a solution

A. True

B. False

D. True I think.

As for the rest of them I would like to discuss them with you.

Last edited by a moderator: Feb 10, 2014
2. Feb 9, 2014

### vela

Staff Emeritus
You have a 50/50 chance of getting the right answer simply by guessing. You need to explain your reasoning.

3. Feb 9, 2014

### kduna

For E: Perhaps thinking about the contrapositive would be easier.
For C: If A is an nxn matrix with real coefficients, then A is the matrix corresponding to some transformation T from Rn to Rn. If Ax = b has a solution for all b $\in$ Rn, what can you say about the properties of T?

4. Feb 10, 2014

### ME_student

The contrapositive? I have never heard of that before in my class.

By the way, question A. is suppose to say identity matrix, not matrix.

A. True because two matrices will have the same solution set. Ax=0 is the trivial solution, when augmenting the identity matrix with zeros the solution set will be equal to that of Ax=b.

B.False because we could have three independent vectors in R^3, although only two of them span a plane in R^3 while the other vector depends on one of the two. This is the reason I said false.

C.True because an nxn matrix has rows=columns, thus there must be a solution for each b in R^n.

D. True because the non trivial solution has a linear dependence relation. For example, let say we have a 3x4 matrix row 3 would have all zeros meaning x_3=x_3 (free) which depends on one of the other two vectors. Therefore the only two pivot positions will be in row 1 and row 2.

E. True because A has to be an invertible matrix to make A^T to be invertible. (A^-1)^-1=A

5. Feb 10, 2014

### vela

Staff Emeritus
Is A supposed to be a square matrix? It seems like the question assumes it's nxn, but you didn't mention anything about the dimensions of A.

6. Feb 10, 2014

### ME_student

Yeah nxn is suppose to be a square matrix. In our textbook nxn means square matrix mxn means non-square matrix.

7. Feb 10, 2014

### pasmith

Then they're not independent, are they?

Any two linearly independent vectors in $\mathbb{R}^3$ will span a plane, but to span the whole of $\mathbb{R}^3$ you need another vector which doesn't lie in that plane. That vector is then linearly independent of both the first two.

It is not stated that the matrix must be invertible. Thus the following is a counterexample: there is no $\vec x \in \mathbb{R}^2$ such that
$$\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} \vec x = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$$

8. Feb 10, 2014

### vela

Staff Emeritus
What does Ax=b have to do with anything here? Also, x=0 is the trivial solution; Ax=0 could have more than x=0 as a solution depending on A.

I'm not following your logic. You seem to be contradicting yourself. You say you have three independent vectors but only two are independent.

I don't see how this follows.

So is A supposed to be square or not? It seems you're not supposed to assume A is square, which affects the answers to the previous parts.

You're just asserting what you're supposed to prove. How does A being invertible imply A^T is invertible?

9. Feb 10, 2014

### kduna

For the logical statement "If P, then Q." The contrapositive is "If not Q, then not P." The contrapositive is logically equivalent to the statement.

For question E: "If At is invertible, then A is invertible." The contrapositive of this is "If A is not invertible, then At is not invertible." As you can see, the contrapositive is much easier to work with.