If a $3\times 3$ matrix is not invertible, how do you prove the rest of the invertible matrix theorem?

Question

This is the exact question/instructions I am to follow. I answered part (a) and found the matrix I used is not invertible. Now I am stuck.

propose a specific $n \times n$ matrix $A$, where $n \ge 3$. Select four statements from the invertible matrix theorem and show that all four statements are true or false, depending on the matrix you selected. Make sure to clearly explain and justify your work. Also, make sure to label your statements using the notation used in the notes (e.g., part (a), part (f), etc.). In responding to your classmates’ posts, verify that their work is correct and select three additional statements they did not originally address and show that these statements are true (or false). Again, make sure to show your work.

My matrix: $\pmatrix{1& 1& 1\\ 1& 1& 0\\ 0& 0& 1}$

I'm trying to answer parts a,b,c, and k of:

Let A be a square n x n matrix. Then, the following statements are equivalent. That is, for a given A, the statements are either all true or all false.

• a. A is an invertible matrix.
• b. A is row equivalent to the n x n identity matrix.
• c. A has n pivot positions
• d. The equation Ax = 0 has only the trivial solution.
• e. The columns of A for a linearly independent set.
• f. The linear transformation x ↦ Ax is one-to-one.
• g. The equation Ax = b has at least one solution for each b in ℝn.
• h. The columns of A span ℝn.
• i. The linear transformation x ↦ Ax maps ℝn onto ℝn.
• j. There is an n x n matrix C such that CA = I.
• k. There is an n x n matrix D such that AD = I.
• l. AT is an invertible matrix.

Answer 1

$A$ is an invertible matrix.

Find $B$ such that $AB = BA = I$ or show that it is impossible.

$A$ is row equivalent to the n x n identity matrix.

Reduce $A$ to reduced row echelon form. See if is identity.

$A$ has n pivot positions

Pretty much the same as above.

The equation $Ax = 0$ has only the trivial solution.

Solve the system.

The columns of $A$ for a linearly independent set.

Write down your columns as vectors in $\Bbb R^n$. See if $\{x_1,x_2,\ldots,x_n\}$ is linearly dependent or independent by your usual means.

The linear transformation $x ↦ Ax$ is one-to-one.

Solve $Av = 0$ as above. Pick one solution $v_0$ and define $y = x + v_0$ for any vector $x$. Note that $x= y$ if and only if $v_0 = 0$.

The equation $Ax = b$ has at least one solution for each $b$ in $ℝ^n$.

If $A$ is invertible, $A^{-1}b$ is the solution. If not, find $b$ such that $Ax = b$ has no solutions (hint: reduce extended matrix $(A|b)$ to row echelon form. If there are zero rows in transformed $A$, choose $b$ such that its transformed component in that row is non-zero).

The columns of $A$ span $ℝ^n$.

Write down columns of $A$ as vectors in $\Bbb R^n$. See if they generate $\Bbb R^n$ or not. Since there are $n$ columns, it is equivalent to check if columns are linearly dependent or not.

The linear transformation $x ↦ Ax$ maps $ℝ^n$ onto $ℝ^n$.

You must check surjectivity of $A$, i.e. for all $b\in ℝ^n$, you must find solution to $Ax = b$ or find $b$ such that there is no solution.

There is an n x n matrix $C$ such that $CA = I$.

Write general matrix $C$, multiply by $A$ and show that coefficients of $C$ can be chosen to get identity or disprove it.

There is an n x n matrix $D$ such that $AD = I$.

As above.

$AT$ is an invertible matrix.

As the first one.

Answer 2

So now you need to pick three other statements in the invertable matrix theorem and show that they don't hold for that matrix. For example, we can compute that the determinant of your matrix to find that it is $(1+0+0)-(1+0+0)=0$ by multiplying across the diagonals. For the other two, you could look at the row rank of the matrix, or show that it's non-surjective.

As mentioned in the comments, many many things are equivalent to a matrix being invertable, so double check to make sure you're using ones that you've talked about in class! A list of 23 potential statements can be found here.

Answer 3

d. The equation Ax = 0 has only the trivial solution.

suppose x = (1,-1,0) Ax = 0

e. The columns of A for a linearly independent set.

the first two column vectors are the same. v_1 - v_2 = 0

etc.